How to get the id while uploading a file












-1














this is my gallery.php;



<br>
<center><form action="galleryuploadxd.php" method="post" enctype="multipart/form-data">
Category
<select>
<option value="">Select...</option>
<option id_image="1" value="1">Admin Images</option>
<option id_image="2" value="2">User Images</option>
</select>
<br><br>

Select image to upload:
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="submit" value="Upload Image" name="submit">
</form></center>




So i want the option tag id_images to be written in my database. I did something like this.



galleryuploadxd.php;



<?php
session_start();
if(isset($_SESSION['sess_user_id']) && $_SESSION['sess_user_id'] != "") {
} else {
header('location:index.php');
}
?>
<body style="background-color: lightgray"></body>
<center><img src="../images/x.png"></center>

<?php
include "db.php";

$id = $_GET['id_image'];
$target_file2 = "random-dir/";
$target_dir = "randomdir/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$target_file3 = $target_file2 .$target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));

if(isset($_POST["submit"])) {
$check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
if($check !== false) {
echo "<br>";
echo "<br>";
echo "<h1><center>File is an image - " . $check["mime"] . "." ."</center></h1>";
$uploadOk = 1;
} else {
echo "<br>";
echo "<h1><center>File is not an image.</center></h1>";
$uploadOk = 0;
}
}

if (file_exists($target_file)) {
echo "<br>";
echo "<h1><center>Sorry, file already exists.</center></h1>";
echo "<h1><a href = gallery-edit.php>Go back </a></h1>";
$uploadOk = 0;
}
// Check file size
if ($_FILES["fileToUpload"]["size"] > 500000) {
echo "<br>";
echo "<h1><center>Sorry, your file is too large.</center></h1>";
echo "<h1><a href = gallery-edit.php>Go back </a></h1>";
$uploadOk = 0;
}
if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg"
&& $imageFileType != "gif" ) {
echo "<br>";
echo "<h1><center>Sorry, only JPG, JPEG, PNG & GIF files are allowed.</center></h1>";
$uploadOk = 0;
}
if ($uploadOk == 0) {
echo "<br>";
echo "<h1><center>Sorry, your file was not uploaded.</center></h1>";
echo "<br>";
echo "<br>";
echo "<br>";
echo "<h1><center><a href = gallery-edit.php>Go back </a></center></h1>";


} else {
if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) {
echo "<h1><center>The file ". basename( $_FILES["fileToUpload"]["name"]). " has been uploaded.</center></h1>";
echo " <meta http-equiv="refresh" content="5;url=gallery-edit.php" />";
echo "<center><h1>You Will be redicted to user gallery in 5 seconds...</h1></center>";
echo "<center><h1>If your browser doesn't support redict please<a href=gallery-edit.php> click here </h1></a></center>";
$sql = "INSERT INTO categories (id,image_link) VALUES (:id,:image_link)";
$stmt = $db->prepare($sql);

$stmt->bindValue(':image_link', $target_file3);
$stmt->bindValue(':id', $id);
$result = $stmt->execute();
} else {
echo "<br>";
echo "<center>Sorry, there was an error uploading your file.</center>";
echo " <center><a href = gallery-edit.php>Go back </a></center>";
}
}
?>


So in the galleryuploadxd.php i say in the beginning $id = $_GET['id_image']; and after in the end



$sql = "INSERT INTO categories (id,image_link) VALUES (:id,:image_link)";
$stmt = $db->prepare($sql);

$stmt->bindValue(':image_link', $target_file3);
$stmt->bindValue(':id', $id);
$result = $stmt->execute();


i get the error that id cannot be null why doesn't it get the id can anyone help me?



the error;




Fatal error: Uncaught PDOException: SQLSTATE[23000]: Integrity constraint violation: 1048 Column 'id' cannot be null in C:xampphtdocsxxxxxxx-admingalleryuploadxd.php:75 Stack trace: #0 C:xampphtdocsxxxxxxx-admingalleryuploadxd.php(75): PDOStatement->execute() #1 {main} thrown in C:xampphtdocsxxxxxxx-admingalleryuploadxd.php on line 75




It tells me I can't get the id. how can I get the id on the option thats all? I need i'm trying to categorize the uploads thats why i'm doing something like this.



Thanks for the helps!










share|improve this question
























  • The message seems very self explanatory to me. I would say that $id has no value
    – RiggsFolly
    Nov 20 at 19:25






  • 1




    You need a name="something" in the <select> tag and not in the <option> tag
    – RiggsFolly
    Nov 20 at 19:26










  • If you POST data you will find it in the $_POST array and not the $_GET array
    – RiggsFolly
    Nov 20 at 19:27










  • @RiggsFolly but i need every select tag with different id how can i do that?
    – andrew
    Nov 20 at 19:29










  • <select name="user_type"> then $_POST['user_type']` hold the ONE option selected.
    – RiggsFolly
    Nov 20 at 19:31
















-1














this is my gallery.php;



<br>
<center><form action="galleryuploadxd.php" method="post" enctype="multipart/form-data">
Category
<select>
<option value="">Select...</option>
<option id_image="1" value="1">Admin Images</option>
<option id_image="2" value="2">User Images</option>
</select>
<br><br>

Select image to upload:
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="submit" value="Upload Image" name="submit">
</form></center>




So i want the option tag id_images to be written in my database. I did something like this.



galleryuploadxd.php;



<?php
session_start();
if(isset($_SESSION['sess_user_id']) && $_SESSION['sess_user_id'] != "") {
} else {
header('location:index.php');
}
?>
<body style="background-color: lightgray"></body>
<center><img src="../images/x.png"></center>

<?php
include "db.php";

$id = $_GET['id_image'];
$target_file2 = "random-dir/";
$target_dir = "randomdir/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$target_file3 = $target_file2 .$target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));

if(isset($_POST["submit"])) {
$check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
if($check !== false) {
echo "<br>";
echo "<br>";
echo "<h1><center>File is an image - " . $check["mime"] . "." ."</center></h1>";
$uploadOk = 1;
} else {
echo "<br>";
echo "<h1><center>File is not an image.</center></h1>";
$uploadOk = 0;
}
}

if (file_exists($target_file)) {
echo "<br>";
echo "<h1><center>Sorry, file already exists.</center></h1>";
echo "<h1><a href = gallery-edit.php>Go back </a></h1>";
$uploadOk = 0;
}
// Check file size
if ($_FILES["fileToUpload"]["size"] > 500000) {
echo "<br>";
echo "<h1><center>Sorry, your file is too large.</center></h1>";
echo "<h1><a href = gallery-edit.php>Go back </a></h1>";
$uploadOk = 0;
}
if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg"
&& $imageFileType != "gif" ) {
echo "<br>";
echo "<h1><center>Sorry, only JPG, JPEG, PNG & GIF files are allowed.</center></h1>";
$uploadOk = 0;
}
if ($uploadOk == 0) {
echo "<br>";
echo "<h1><center>Sorry, your file was not uploaded.</center></h1>";
echo "<br>";
echo "<br>";
echo "<br>";
echo "<h1><center><a href = gallery-edit.php>Go back </a></center></h1>";


} else {
if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) {
echo "<h1><center>The file ". basename( $_FILES["fileToUpload"]["name"]). " has been uploaded.</center></h1>";
echo " <meta http-equiv="refresh" content="5;url=gallery-edit.php" />";
echo "<center><h1>You Will be redicted to user gallery in 5 seconds...</h1></center>";
echo "<center><h1>If your browser doesn't support redict please<a href=gallery-edit.php> click here </h1></a></center>";
$sql = "INSERT INTO categories (id,image_link) VALUES (:id,:image_link)";
$stmt = $db->prepare($sql);

$stmt->bindValue(':image_link', $target_file3);
$stmt->bindValue(':id', $id);
$result = $stmt->execute();
} else {
echo "<br>";
echo "<center>Sorry, there was an error uploading your file.</center>";
echo " <center><a href = gallery-edit.php>Go back </a></center>";
}
}
?>


So in the galleryuploadxd.php i say in the beginning $id = $_GET['id_image']; and after in the end



$sql = "INSERT INTO categories (id,image_link) VALUES (:id,:image_link)";
$stmt = $db->prepare($sql);

$stmt->bindValue(':image_link', $target_file3);
$stmt->bindValue(':id', $id);
$result = $stmt->execute();


i get the error that id cannot be null why doesn't it get the id can anyone help me?



the error;




Fatal error: Uncaught PDOException: SQLSTATE[23000]: Integrity constraint violation: 1048 Column 'id' cannot be null in C:xampphtdocsxxxxxxx-admingalleryuploadxd.php:75 Stack trace: #0 C:xampphtdocsxxxxxxx-admingalleryuploadxd.php(75): PDOStatement->execute() #1 {main} thrown in C:xampphtdocsxxxxxxx-admingalleryuploadxd.php on line 75




It tells me I can't get the id. how can I get the id on the option thats all? I need i'm trying to categorize the uploads thats why i'm doing something like this.



Thanks for the helps!










share|improve this question
























  • The message seems very self explanatory to me. I would say that $id has no value
    – RiggsFolly
    Nov 20 at 19:25






  • 1




    You need a name="something" in the <select> tag and not in the <option> tag
    – RiggsFolly
    Nov 20 at 19:26










  • If you POST data you will find it in the $_POST array and not the $_GET array
    – RiggsFolly
    Nov 20 at 19:27










  • @RiggsFolly but i need every select tag with different id how can i do that?
    – andrew
    Nov 20 at 19:29










  • <select name="user_type"> then $_POST['user_type']` hold the ONE option selected.
    – RiggsFolly
    Nov 20 at 19:31














-1












-1








-1


1





this is my gallery.php;



<br>
<center><form action="galleryuploadxd.php" method="post" enctype="multipart/form-data">
Category
<select>
<option value="">Select...</option>
<option id_image="1" value="1">Admin Images</option>
<option id_image="2" value="2">User Images</option>
</select>
<br><br>

Select image to upload:
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="submit" value="Upload Image" name="submit">
</form></center>




So i want the option tag id_images to be written in my database. I did something like this.



galleryuploadxd.php;



<?php
session_start();
if(isset($_SESSION['sess_user_id']) && $_SESSION['sess_user_id'] != "") {
} else {
header('location:index.php');
}
?>
<body style="background-color: lightgray"></body>
<center><img src="../images/x.png"></center>

<?php
include "db.php";

$id = $_GET['id_image'];
$target_file2 = "random-dir/";
$target_dir = "randomdir/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$target_file3 = $target_file2 .$target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));

if(isset($_POST["submit"])) {
$check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
if($check !== false) {
echo "<br>";
echo "<br>";
echo "<h1><center>File is an image - " . $check["mime"] . "." ."</center></h1>";
$uploadOk = 1;
} else {
echo "<br>";
echo "<h1><center>File is not an image.</center></h1>";
$uploadOk = 0;
}
}

if (file_exists($target_file)) {
echo "<br>";
echo "<h1><center>Sorry, file already exists.</center></h1>";
echo "<h1><a href = gallery-edit.php>Go back </a></h1>";
$uploadOk = 0;
}
// Check file size
if ($_FILES["fileToUpload"]["size"] > 500000) {
echo "<br>";
echo "<h1><center>Sorry, your file is too large.</center></h1>";
echo "<h1><a href = gallery-edit.php>Go back </a></h1>";
$uploadOk = 0;
}
if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg"
&& $imageFileType != "gif" ) {
echo "<br>";
echo "<h1><center>Sorry, only JPG, JPEG, PNG & GIF files are allowed.</center></h1>";
$uploadOk = 0;
}
if ($uploadOk == 0) {
echo "<br>";
echo "<h1><center>Sorry, your file was not uploaded.</center></h1>";
echo "<br>";
echo "<br>";
echo "<br>";
echo "<h1><center><a href = gallery-edit.php>Go back </a></center></h1>";


} else {
if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) {
echo "<h1><center>The file ". basename( $_FILES["fileToUpload"]["name"]). " has been uploaded.</center></h1>";
echo " <meta http-equiv="refresh" content="5;url=gallery-edit.php" />";
echo "<center><h1>You Will be redicted to user gallery in 5 seconds...</h1></center>";
echo "<center><h1>If your browser doesn't support redict please<a href=gallery-edit.php> click here </h1></a></center>";
$sql = "INSERT INTO categories (id,image_link) VALUES (:id,:image_link)";
$stmt = $db->prepare($sql);

$stmt->bindValue(':image_link', $target_file3);
$stmt->bindValue(':id', $id);
$result = $stmt->execute();
} else {
echo "<br>";
echo "<center>Sorry, there was an error uploading your file.</center>";
echo " <center><a href = gallery-edit.php>Go back </a></center>";
}
}
?>


So in the galleryuploadxd.php i say in the beginning $id = $_GET['id_image']; and after in the end



$sql = "INSERT INTO categories (id,image_link) VALUES (:id,:image_link)";
$stmt = $db->prepare($sql);

$stmt->bindValue(':image_link', $target_file3);
$stmt->bindValue(':id', $id);
$result = $stmt->execute();


i get the error that id cannot be null why doesn't it get the id can anyone help me?



the error;




Fatal error: Uncaught PDOException: SQLSTATE[23000]: Integrity constraint violation: 1048 Column 'id' cannot be null in C:xampphtdocsxxxxxxx-admingalleryuploadxd.php:75 Stack trace: #0 C:xampphtdocsxxxxxxx-admingalleryuploadxd.php(75): PDOStatement->execute() #1 {main} thrown in C:xampphtdocsxxxxxxx-admingalleryuploadxd.php on line 75




It tells me I can't get the id. how can I get the id on the option thats all? I need i'm trying to categorize the uploads thats why i'm doing something like this.



Thanks for the helps!










share|improve this question















this is my gallery.php;



<br>
<center><form action="galleryuploadxd.php" method="post" enctype="multipart/form-data">
Category
<select>
<option value="">Select...</option>
<option id_image="1" value="1">Admin Images</option>
<option id_image="2" value="2">User Images</option>
</select>
<br><br>

Select image to upload:
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="submit" value="Upload Image" name="submit">
</form></center>




So i want the option tag id_images to be written in my database. I did something like this.



galleryuploadxd.php;



<?php
session_start();
if(isset($_SESSION['sess_user_id']) && $_SESSION['sess_user_id'] != "") {
} else {
header('location:index.php');
}
?>
<body style="background-color: lightgray"></body>
<center><img src="../images/x.png"></center>

<?php
include "db.php";

$id = $_GET['id_image'];
$target_file2 = "random-dir/";
$target_dir = "randomdir/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$target_file3 = $target_file2 .$target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));

if(isset($_POST["submit"])) {
$check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
if($check !== false) {
echo "<br>";
echo "<br>";
echo "<h1><center>File is an image - " . $check["mime"] . "." ."</center></h1>";
$uploadOk = 1;
} else {
echo "<br>";
echo "<h1><center>File is not an image.</center></h1>";
$uploadOk = 0;
}
}

if (file_exists($target_file)) {
echo "<br>";
echo "<h1><center>Sorry, file already exists.</center></h1>";
echo "<h1><a href = gallery-edit.php>Go back </a></h1>";
$uploadOk = 0;
}
// Check file size
if ($_FILES["fileToUpload"]["size"] > 500000) {
echo "<br>";
echo "<h1><center>Sorry, your file is too large.</center></h1>";
echo "<h1><a href = gallery-edit.php>Go back </a></h1>";
$uploadOk = 0;
}
if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg"
&& $imageFileType != "gif" ) {
echo "<br>";
echo "<h1><center>Sorry, only JPG, JPEG, PNG & GIF files are allowed.</center></h1>";
$uploadOk = 0;
}
if ($uploadOk == 0) {
echo "<br>";
echo "<h1><center>Sorry, your file was not uploaded.</center></h1>";
echo "<br>";
echo "<br>";
echo "<br>";
echo "<h1><center><a href = gallery-edit.php>Go back </a></center></h1>";


} else {
if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) {
echo "<h1><center>The file ". basename( $_FILES["fileToUpload"]["name"]). " has been uploaded.</center></h1>";
echo " <meta http-equiv="refresh" content="5;url=gallery-edit.php" />";
echo "<center><h1>You Will be redicted to user gallery in 5 seconds...</h1></center>";
echo "<center><h1>If your browser doesn't support redict please<a href=gallery-edit.php> click here </h1></a></center>";
$sql = "INSERT INTO categories (id,image_link) VALUES (:id,:image_link)";
$stmt = $db->prepare($sql);

$stmt->bindValue(':image_link', $target_file3);
$stmt->bindValue(':id', $id);
$result = $stmt->execute();
} else {
echo "<br>";
echo "<center>Sorry, there was an error uploading your file.</center>";
echo " <center><a href = gallery-edit.php>Go back </a></center>";
}
}
?>


So in the galleryuploadxd.php i say in the beginning $id = $_GET['id_image']; and after in the end



$sql = "INSERT INTO categories (id,image_link) VALUES (:id,:image_link)";
$stmt = $db->prepare($sql);

$stmt->bindValue(':image_link', $target_file3);
$stmt->bindValue(':id', $id);
$result = $stmt->execute();


i get the error that id cannot be null why doesn't it get the id can anyone help me?



the error;




Fatal error: Uncaught PDOException: SQLSTATE[23000]: Integrity constraint violation: 1048 Column 'id' cannot be null in C:xampphtdocsxxxxxxx-admingalleryuploadxd.php:75 Stack trace: #0 C:xampphtdocsxxxxxxx-admingalleryuploadxd.php(75): PDOStatement->execute() #1 {main} thrown in C:xampphtdocsxxxxxxx-admingalleryuploadxd.php on line 75




It tells me I can't get the id. how can I get the id on the option thats all? I need i'm trying to categorize the uploads thats why i'm doing something like this.



Thanks for the helps!







php forms pdo






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 20 at 19:34









Funk Forty Niner

80.5k1247100




80.5k1247100










asked Nov 20 at 19:22









andrew

32




32












  • The message seems very self explanatory to me. I would say that $id has no value
    – RiggsFolly
    Nov 20 at 19:25






  • 1




    You need a name="something" in the <select> tag and not in the <option> tag
    – RiggsFolly
    Nov 20 at 19:26










  • If you POST data you will find it in the $_POST array and not the $_GET array
    – RiggsFolly
    Nov 20 at 19:27










  • @RiggsFolly but i need every select tag with different id how can i do that?
    – andrew
    Nov 20 at 19:29










  • <select name="user_type"> then $_POST['user_type']` hold the ONE option selected.
    – RiggsFolly
    Nov 20 at 19:31


















  • The message seems very self explanatory to me. I would say that $id has no value
    – RiggsFolly
    Nov 20 at 19:25






  • 1




    You need a name="something" in the <select> tag and not in the <option> tag
    – RiggsFolly
    Nov 20 at 19:26










  • If you POST data you will find it in the $_POST array and not the $_GET array
    – RiggsFolly
    Nov 20 at 19:27










  • @RiggsFolly but i need every select tag with different id how can i do that?
    – andrew
    Nov 20 at 19:29










  • <select name="user_type"> then $_POST['user_type']` hold the ONE option selected.
    – RiggsFolly
    Nov 20 at 19:31
















The message seems very self explanatory to me. I would say that $id has no value
– RiggsFolly
Nov 20 at 19:25




The message seems very self explanatory to me. I would say that $id has no value
– RiggsFolly
Nov 20 at 19:25




1




1




You need a name="something" in the <select> tag and not in the <option> tag
– RiggsFolly
Nov 20 at 19:26




You need a name="something" in the <select> tag and not in the <option> tag
– RiggsFolly
Nov 20 at 19:26












If you POST data you will find it in the $_POST array and not the $_GET array
– RiggsFolly
Nov 20 at 19:27




If you POST data you will find it in the $_POST array and not the $_GET array
– RiggsFolly
Nov 20 at 19:27












@RiggsFolly but i need every select tag with different id how can i do that?
– andrew
Nov 20 at 19:29




@RiggsFolly but i need every select tag with different id how can i do that?
– andrew
Nov 20 at 19:29












<select name="user_type"> then $_POST['user_type']` hold the ONE option selected.
– RiggsFolly
Nov 20 at 19:31




<select name="user_type"> then $_POST['user_type']` hold the ONE option selected.
– RiggsFolly
Nov 20 at 19:31












2 Answers
2






active

oldest

votes


















0














why you don't set the name of your select input



<select name="id_image">
<option value="">Select...</option>
<option value="1">Admin Images</option>
<option value="2">User Images</option>
</select>


use this as $id = $_POST['id_image'];






share|improve this answer





















  • No error this time but i choose admin images and upload an image i don't see it in my table i have id and image_link i need the value 1 when i choose admin images to be id in my table that's all
    – andrew
    Nov 20 at 19:43










  • did u changed the value of $id
    – Mahfuzar Rahman
    Nov 20 at 19:45










  • i need the id to upon choosing from the options if they choose admin images the value will be 1 and user images needs to be 2
    – andrew
    Nov 20 at 19:49










  • change $id = $_GET['id_image']; to $id = $_POST['id_image'];
    – Mahfuzar Rahman
    Nov 20 at 19:51










  • is your categories table have only (id,image_link)
    – Mahfuzar Rahman
    Nov 20 at 19:52



















-1














In your form the method is POST and you try to retreive the records with



$id = $_GET['id_image'];


you should do:



$id = $_POST['id_image'];





share|improve this answer





















  • i tried that too same fatal error.
    – andrew
    Nov 20 at 19:28










  • Would be useful to add what I said in my comment about where the name="" attribute belongs to make this a complete answer
    – RiggsFolly
    Nov 20 at 19:28












  • But beware the code vampire, code this bad will normally require a complete rewrite and the OP wont credit you until you have fixed all their errors
    – RiggsFolly
    Nov 20 at 19:29













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2 Answers
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2 Answers
2






active

oldest

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0














why you don't set the name of your select input



<select name="id_image">
<option value="">Select...</option>
<option value="1">Admin Images</option>
<option value="2">User Images</option>
</select>


use this as $id = $_POST['id_image'];






share|improve this answer





















  • No error this time but i choose admin images and upload an image i don't see it in my table i have id and image_link i need the value 1 when i choose admin images to be id in my table that's all
    – andrew
    Nov 20 at 19:43










  • did u changed the value of $id
    – Mahfuzar Rahman
    Nov 20 at 19:45










  • i need the id to upon choosing from the options if they choose admin images the value will be 1 and user images needs to be 2
    – andrew
    Nov 20 at 19:49










  • change $id = $_GET['id_image']; to $id = $_POST['id_image'];
    – Mahfuzar Rahman
    Nov 20 at 19:51










  • is your categories table have only (id,image_link)
    – Mahfuzar Rahman
    Nov 20 at 19:52
















0














why you don't set the name of your select input



<select name="id_image">
<option value="">Select...</option>
<option value="1">Admin Images</option>
<option value="2">User Images</option>
</select>


use this as $id = $_POST['id_image'];






share|improve this answer





















  • No error this time but i choose admin images and upload an image i don't see it in my table i have id and image_link i need the value 1 when i choose admin images to be id in my table that's all
    – andrew
    Nov 20 at 19:43










  • did u changed the value of $id
    – Mahfuzar Rahman
    Nov 20 at 19:45










  • i need the id to upon choosing from the options if they choose admin images the value will be 1 and user images needs to be 2
    – andrew
    Nov 20 at 19:49










  • change $id = $_GET['id_image']; to $id = $_POST['id_image'];
    – Mahfuzar Rahman
    Nov 20 at 19:51










  • is your categories table have only (id,image_link)
    – Mahfuzar Rahman
    Nov 20 at 19:52














0












0








0






why you don't set the name of your select input



<select name="id_image">
<option value="">Select...</option>
<option value="1">Admin Images</option>
<option value="2">User Images</option>
</select>


use this as $id = $_POST['id_image'];






share|improve this answer












why you don't set the name of your select input



<select name="id_image">
<option value="">Select...</option>
<option value="1">Admin Images</option>
<option value="2">User Images</option>
</select>


use this as $id = $_POST['id_image'];







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 20 at 19:36









Mahfuzar Rahman

14919




14919












  • No error this time but i choose admin images and upload an image i don't see it in my table i have id and image_link i need the value 1 when i choose admin images to be id in my table that's all
    – andrew
    Nov 20 at 19:43










  • did u changed the value of $id
    – Mahfuzar Rahman
    Nov 20 at 19:45










  • i need the id to upon choosing from the options if they choose admin images the value will be 1 and user images needs to be 2
    – andrew
    Nov 20 at 19:49










  • change $id = $_GET['id_image']; to $id = $_POST['id_image'];
    – Mahfuzar Rahman
    Nov 20 at 19:51










  • is your categories table have only (id,image_link)
    – Mahfuzar Rahman
    Nov 20 at 19:52


















  • No error this time but i choose admin images and upload an image i don't see it in my table i have id and image_link i need the value 1 when i choose admin images to be id in my table that's all
    – andrew
    Nov 20 at 19:43










  • did u changed the value of $id
    – Mahfuzar Rahman
    Nov 20 at 19:45










  • i need the id to upon choosing from the options if they choose admin images the value will be 1 and user images needs to be 2
    – andrew
    Nov 20 at 19:49










  • change $id = $_GET['id_image']; to $id = $_POST['id_image'];
    – Mahfuzar Rahman
    Nov 20 at 19:51










  • is your categories table have only (id,image_link)
    – Mahfuzar Rahman
    Nov 20 at 19:52
















No error this time but i choose admin images and upload an image i don't see it in my table i have id and image_link i need the value 1 when i choose admin images to be id in my table that's all
– andrew
Nov 20 at 19:43




No error this time but i choose admin images and upload an image i don't see it in my table i have id and image_link i need the value 1 when i choose admin images to be id in my table that's all
– andrew
Nov 20 at 19:43












did u changed the value of $id
– Mahfuzar Rahman
Nov 20 at 19:45




did u changed the value of $id
– Mahfuzar Rahman
Nov 20 at 19:45












i need the id to upon choosing from the options if they choose admin images the value will be 1 and user images needs to be 2
– andrew
Nov 20 at 19:49




i need the id to upon choosing from the options if they choose admin images the value will be 1 and user images needs to be 2
– andrew
Nov 20 at 19:49












change $id = $_GET['id_image']; to $id = $_POST['id_image'];
– Mahfuzar Rahman
Nov 20 at 19:51




change $id = $_GET['id_image']; to $id = $_POST['id_image'];
– Mahfuzar Rahman
Nov 20 at 19:51












is your categories table have only (id,image_link)
– Mahfuzar Rahman
Nov 20 at 19:52




is your categories table have only (id,image_link)
– Mahfuzar Rahman
Nov 20 at 19:52













-1














In your form the method is POST and you try to retreive the records with



$id = $_GET['id_image'];


you should do:



$id = $_POST['id_image'];





share|improve this answer





















  • i tried that too same fatal error.
    – andrew
    Nov 20 at 19:28










  • Would be useful to add what I said in my comment about where the name="" attribute belongs to make this a complete answer
    – RiggsFolly
    Nov 20 at 19:28












  • But beware the code vampire, code this bad will normally require a complete rewrite and the OP wont credit you until you have fixed all their errors
    – RiggsFolly
    Nov 20 at 19:29


















-1














In your form the method is POST and you try to retreive the records with



$id = $_GET['id_image'];


you should do:



$id = $_POST['id_image'];





share|improve this answer





















  • i tried that too same fatal error.
    – andrew
    Nov 20 at 19:28










  • Would be useful to add what I said in my comment about where the name="" attribute belongs to make this a complete answer
    – RiggsFolly
    Nov 20 at 19:28












  • But beware the code vampire, code this bad will normally require a complete rewrite and the OP wont credit you until you have fixed all their errors
    – RiggsFolly
    Nov 20 at 19:29
















-1












-1








-1






In your form the method is POST and you try to retreive the records with



$id = $_GET['id_image'];


you should do:



$id = $_POST['id_image'];





share|improve this answer












In your form the method is POST and you try to retreive the records with



$id = $_GET['id_image'];


you should do:



$id = $_POST['id_image'];






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 20 at 19:26









m.Sarto

395




395












  • i tried that too same fatal error.
    – andrew
    Nov 20 at 19:28










  • Would be useful to add what I said in my comment about where the name="" attribute belongs to make this a complete answer
    – RiggsFolly
    Nov 20 at 19:28












  • But beware the code vampire, code this bad will normally require a complete rewrite and the OP wont credit you until you have fixed all their errors
    – RiggsFolly
    Nov 20 at 19:29




















  • i tried that too same fatal error.
    – andrew
    Nov 20 at 19:28










  • Would be useful to add what I said in my comment about where the name="" attribute belongs to make this a complete answer
    – RiggsFolly
    Nov 20 at 19:28












  • But beware the code vampire, code this bad will normally require a complete rewrite and the OP wont credit you until you have fixed all their errors
    – RiggsFolly
    Nov 20 at 19:29


















i tried that too same fatal error.
– andrew
Nov 20 at 19:28




i tried that too same fatal error.
– andrew
Nov 20 at 19:28












Would be useful to add what I said in my comment about where the name="" attribute belongs to make this a complete answer
– RiggsFolly
Nov 20 at 19:28






Would be useful to add what I said in my comment about where the name="" attribute belongs to make this a complete answer
– RiggsFolly
Nov 20 at 19:28














But beware the code vampire, code this bad will normally require a complete rewrite and the OP wont credit you until you have fixed all their errors
– RiggsFolly
Nov 20 at 19:29






But beware the code vampire, code this bad will normally require a complete rewrite and the OP wont credit you until you have fixed all their errors
– RiggsFolly
Nov 20 at 19:29




















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