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 = $stmt->execute();
i get the error that id cannot be null why doesn't it get the id can anyone help me?
the error;
Fatal error: Uncaught PDOException: SQLSTATE[23000]: Integrity constraint violation: 1048 Column 'id' cannot be null in C:xampphtdocsxxxxxxx-admingalleryuploadxd.php:75 Stack trace: #0 C:xampphtdocsxxxxxxx-admingalleryuploadxd.php(75): PDOStatement->execute() #1 {main} thrown in C:xampphtdocsxxxxxxx-admingalleryuploadxd.php on line 75
It tells me I can't get the id. how can I get the id on the option thats all? I need i'm trying to categorize the uploads thats why i'm doing something like this.
Thanks for the helps!
php forms pdo
edited Nov 20 at 19:34
Funk Forty Niner
80.5k1247100
asked Nov 20 at 19:22
andrew
32
|
show 9 more comments
-1
this is my gallery.php;
<br>
<center><form action="galleryuploadxd.php" method="post" enctype="multipart/form-data">
Category
<select>
<option value="">Select...</option>
<option id_image="1" value="1">Admin Images</option>
<option id_image="2" value="2">User Images</option>
</select>
<br><br>
Select image to upload:
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="submit" value="Upload Image" name="submit">
</form></center>
So i want the option tag id_images to be written in my database. I did something like this.
galleryuploadxd.php;
<?php
session_start();
if(isset($_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 = $stmt->execute();
i get the error that id cannot be null why doesn't it get the id can anyone help me?
the error;
Fatal error: Uncaught PDOException: SQLSTATE[23000]: Integrity constraint violation: 1048 Column 'id' cannot be null in C:xampphtdocsxxxxxxx-admingalleryuploadxd.php:75 Stack trace: #0 C:xampphtdocsxxxxxxx-admingalleryuploadxd.php(75): PDOStatement->execute() #1 {main} thrown in C:xampphtdocsxxxxxxx-admingalleryuploadxd.php on line 75
It tells me I can't get the id. how can I get the id on the option thats all? I need i'm trying to categorize the uploads thats why i'm doing something like this.
Thanks for the helps!
php forms pdo
edited Nov 20 at 19:34
Funk Forty Niner
80.5k1247100
asked Nov 20 at 19:22
andrew
32
The message seems very self explanatory to me. I would say that$idhas no value
– RiggsFolly
Nov 20 at 19:25
1
You need aname="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$_POSTarray and not the$_GETarray
– 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 $_POST['user_type']` hold the ONE option selected.
– RiggsFolly
Nov 20 at 19:31
|
show 9 more comments
-1
-1
-1
1
this is my gallery.php;
<br>
<center><form action="galleryuploadxd.php" method="post" enctype="multipart/form-data">
Category
<select>
<option value="">Select...</option>
<option id_image="1" value="1">Admin Images</option>
<option id_image="2" value="2">User Images</option>
</select>
<br><br>
Select image to upload:
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="submit" value="Upload Image" name="submit">
</form></center>
So i want the option tag id_images to be written in my database. I did something like this.
galleryuploadxd.php;
<?php
session_start();
if(isset($_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 = $stmt->execute();
i get the error that id cannot be null why doesn't it get the id can anyone help me?
the error;
Fatal error: Uncaught PDOException: SQLSTATE[23000]: Integrity constraint violation: 1048 Column 'id' cannot be null in C:xampphtdocsxxxxxxx-admingalleryuploadxd.php:75 Stack trace: #0 C:xampphtdocsxxxxxxx-admingalleryuploadxd.php(75): PDOStatement->execute() #1 {main} thrown in C:xampphtdocsxxxxxxx-admingalleryuploadxd.php on line 75
It tells me I can't get the id. how can I get the id on the option thats all? I need i'm trying to categorize the uploads thats why i'm doing something like this.
Thanks for the helps!
php forms pdo
edited Nov 20 at 19:34
Funk Forty Niner
80.5k1247100
asked Nov 20 at 19:22
andrew
32
this is my gallery.php;
<br>
<center><form action="galleryuploadxd.php" method="post" enctype="multipart/form-data">
Category
<select>
<option value="">Select...</option>
<option id_image="1" value="1">Admin Images</option>
<option id_image="2" value="2">User Images</option>
</select>
<br><br>
Select image to upload:
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="submit" value="Upload Image" name="submit">
</form></center>
So i want the option tag id_images to be written in my database. I did something like this.
galleryuploadxd.php;
<?php
session_start();
if(isset($_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 = $stmt->execute();
i get the error that id cannot be null why doesn't it get the id can anyone help me?
the error;
Fatal error: Uncaught PDOException: SQLSTATE[23000]: Integrity constraint violation: 1048 Column 'id' cannot be null in C:xampphtdocsxxxxxxx-admingalleryuploadxd.php:75 Stack trace: #0 C:xampphtdocsxxxxxxx-admingalleryuploadxd.php(75): PDOStatement->execute() #1 {main} thrown in C:xampphtdocsxxxxxxx-admingalleryuploadxd.php on line 75
It tells me I can't get the id. how can I get the id on the option thats all? I need i'm trying to categorize the uploads thats why i'm doing something like this.
Thanks for the helps!
php forms pdo
php forms pdo
edited Nov 20 at 19:34
Funk Forty Niner
80.5k1247100
asked Nov 20 at 19:22
andrew
32
edited Nov 20 at 19:34
Funk Forty Niner
80.5k1247100
asked Nov 20 at 19:22
andrew
32
edited Nov 20 at 19:34
Funk Forty Niner
80.5k1247100
edited Nov 20 at 19:34
Funk Forty Niner
80.5k1247100
edited Nov 20 at 19:34
Funk Forty Niner
80.5k1247100
80.5k1247100
asked Nov 20 at 19:22
andrew
32
asked Nov 20 at 19:22
andrew
32
asked Nov 20 at 19:22
andrew
32
32
The message seems very self explanatory to me. I would say that$idhas no value
– RiggsFolly
Nov 20 at 19:25
1
You need aname="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$_POSTarray and not the$_GETarray
– 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 $_POST['user_type']` hold the ONE option selected.
– RiggsFolly
Nov 20 at 19:31
|
show 9 more comments
The message seems very self explanatory to me. I would say that$idhas no value
– RiggsFolly
Nov 20 at 19:25
1
You need aname="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$_POSTarray and not the$_GETarray
– 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
|
show 9 more comments
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'];
answered Nov 20 at 19:36
Mahfuzar Rahman
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
|
show 6 more comments
-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'];
answered Nov 20 at 19:26
m.Sarto
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 thename=""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
add a comment |
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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'];
answered Nov 20 at 19:36
Mahfuzar Rahman
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
|
show 6 more comments
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'];
answered Nov 20 at 19:36
Mahfuzar Rahman
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
|
show 6 more comments
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'];
answered Nov 20 at 19:36
Mahfuzar Rahman
14919
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'];
answered Nov 20 at 19:36
Mahfuzar Rahman
14919
answered Nov 20 at 19:36
Mahfuzar Rahman
14919
answered Nov 20 at 19:36
Mahfuzar Rahman
14919
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
|
show 6 more comments
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
|
show 6 more comments
-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'];
answered Nov 20 at 19:26
m.Sarto
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 thename=""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
add a comment |
-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'];
answered Nov 20 at 19:26
m.Sarto
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 thename=""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
add a comment |
-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'];
answered Nov 20 at 19:26
m.Sarto
395
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'];
answered Nov 20 at 19:26
m.Sarto
395
answered Nov 20 at 19:26
m.Sarto
395
answered Nov 20 at 19:26
m.Sarto
395
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 thename=""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
add a comment |
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 thename=""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
add a comment |
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The message seems very self explanatory to me. I would say that
$idhas 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
$_POSTarray and not the$_GETarray– 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