Download link over random php image
I am displaying a number of random images from a folder, however I'm not very good with PHP (Code sourced from the internet), how would I go about having a "download" link display on top of the image?
Display random image from folder using PHP:
function random_image($directory)
{
$leading = substr($directory, 0, 1);
$trailing = substr($directory, -1, 1);
if($leading == '/')
{
$directory = substr($directory, 1);
}
if($trailing != '/')
{
$directory = $directory . '/';
}
if(empty($directory) or !is_dir($directory))
{
die('Directory: ' . $directory . ' not found.');
}
$files = scandir($directory, 1);
$make_array = array();
foreach($files AS $id => $file)
{
$info = pathinfo($dir . $file);
$image_extensions = array('jpg', 'jpeg', 'gif', 'png', 'ico');
if(!in_array($info['extension'], $image_extensions))
{
unset($file);
}
else
{
$file = str_replace(' ', '%20', $file);
$temp = array($id => $file);
array_push($make_array, $temp);
}
}
if(sizeof($make_array) == 0)
{
die('No images in ' . $directory . ' Directory');
}
$total = count($make_array) - 1;
$random_image = rand(0, $total);
return $directory . $make_array[$random_image][$random_image];
}
Markup:
echo "<img src=" . random_image('css/images/avatars') . " />";
I've tried looking around google for an answer but I can't find anything, any help would be appreciated
php html image random
asked Nov 23 '18 at 21:58
InsignianInsignian
349
add a comment |
I am displaying a number of random images from a folder, however I'm not very good with PHP (Code sourced from the internet), how would I go about having a "download" link display on top of the image?
Display random image from folder using PHP:
function random_image($directory)
{
$leading = substr($directory, 0, 1);
$trailing = substr($directory, -1, 1);
if($leading == '/')
{
$directory = substr($directory, 1);
}
if($trailing != '/')
{
$directory = $directory . '/';
}
if(empty($directory) or !is_dir($directory))
{
die('Directory: ' . $directory . ' not found.');
}
$files = scandir($directory, 1);
$make_array = array();
foreach($files AS $id => $file)
{
$info = pathinfo($dir . $file);
$image_extensions = array('jpg', 'jpeg', 'gif', 'png', 'ico');
if(!in_array($info['extension'], $image_extensions))
{
unset($file);
}
else
{
$file = str_replace(' ', '%20', $file);
$temp = array($id => $file);
array_push($make_array, $temp);
}
}
if(sizeof($make_array) == 0)
{
die('No images in ' . $directory . ' Directory');
}
$total = count($make_array) - 1;
$random_image = rand(0, $total);
return $directory . $make_array[$random_image][$random_image];
}
Markup:
echo "<img src=" . random_image('css/images/avatars') . " />";
I've tried looking around google for an answer but I can't find anything, any help would be appreciated
php html image random
asked Nov 23 '18 at 21:58
InsignianInsignian
349
You just make a link with the same URL as the image. Like,<a>instead of<img>.
– miken32
Nov 23 '18 at 22:12
add a comment |
I am displaying a number of random images from a folder, however I'm not very good with PHP (Code sourced from the internet), how would I go about having a "download" link display on top of the image?
Display random image from folder using PHP:
function random_image($directory)
{
$leading = substr($directory, 0, 1);
$trailing = substr($directory, -1, 1);
if($leading == '/')
{
$directory = substr($directory, 1);
}
if($trailing != '/')
{
$directory = $directory . '/';
}
if(empty($directory) or !is_dir($directory))
{
die('Directory: ' . $directory . ' not found.');
}
$files = scandir($directory, 1);
$make_array = array();
foreach($files AS $id => $file)
{
$info = pathinfo($dir . $file);
$image_extensions = array('jpg', 'jpeg', 'gif', 'png', 'ico');
if(!in_array($info['extension'], $image_extensions))
{
unset($file);
}
else
{
$file = str_replace(' ', '%20', $file);
$temp = array($id => $file);
array_push($make_array, $temp);
}
}
if(sizeof($make_array) == 0)
{
die('No images in ' . $directory . ' Directory');
}
$total = count($make_array) - 1;
$random_image = rand(0, $total);
return $directory . $make_array[$random_image][$random_image];
}
Markup:
echo "<img src=" . random_image('css/images/avatars') . " />";
I've tried looking around google for an answer but I can't find anything, any help would be appreciated
php html image random
asked Nov 23 '18 at 21:58
InsignianInsignian
349
I am displaying a number of random images from a folder, however I'm not very good with PHP (Code sourced from the internet), how would I go about having a "download" link display on top of the image?
Display random image from folder using PHP:
function random_image($directory)
{
$leading = substr($directory, 0, 1);
$trailing = substr($directory, -1, 1);
if($leading == '/')
{
$directory = substr($directory, 1);
}
if($trailing != '/')
{
$directory = $directory . '/';
}
if(empty($directory) or !is_dir($directory))
{
die('Directory: ' . $directory . ' not found.');
}
$files = scandir($directory, 1);
$make_array = array();
foreach($files AS $id => $file)
{
$info = pathinfo($dir . $file);
$image_extensions = array('jpg', 'jpeg', 'gif', 'png', 'ico');
if(!in_array($info['extension'], $image_extensions))
{
unset($file);
}
else
{
$file = str_replace(' ', '%20', $file);
$temp = array($id => $file);
array_push($make_array, $temp);
}
}
if(sizeof($make_array) == 0)
{
die('No images in ' . $directory . ' Directory');
}
$total = count($make_array) - 1;
$random_image = rand(0, $total);
return $directory . $make_array[$random_image][$random_image];
}
Markup:
echo "<img src=" . random_image('css/images/avatars') . " />";
I've tried looking around google for an answer but I can't find anything, any help would be appreciated
php html image random
php html image random
asked Nov 23 '18 at 21:58
InsignianInsignian
349
asked Nov 23 '18 at 21:58
InsignianInsignian
349
edited Nov 23 '18 at 22:15
Insignian
asked Nov 23 '18 at 21:58
InsignianInsignian
349
asked Nov 23 '18 at 21:58
InsignianInsignian
349
asked Nov 23 '18 at 21:58
InsignianInsignian
349
349
You just make a link with the same URL as the image. Like,<a>instead of<img>.
– miken32
Nov 23 '18 at 22:12
add a comment |
You just make a link with the same URL as the image. Like,<a>instead of<img>.
– miken32
Nov 23 '18 at 22:12
You just make a link with the same URL as the image. Like,
<a> instead of <img>.– miken32
Nov 23 '18 at 22:12
You just make a link with the same URL as the image. Like,
<a> instead of <img>.– miken32
Nov 23 '18 at 22:12
add a comment |
2 Answers
2
active
oldest
votes
You should save the image location in a variable then use it to create a link, plus display it.
$imageUrl = random_image('css/images/avatars');
echo "<a href=" . $imageUrl . ">";
echo "<img src=" . $imageUrl . " />";
echo "</a>";
or if you want to show the text link above, seperately then
$imageUrl = random_image('css/images/avatars');
echo "<a href=" . $imageUrl . ">Click Here</a><br />";
echo "<img src=" . $imageUrl . " />";
answered Nov 23 '18 at 22:22
Nawed KhanNawed Khan
2,8141618
1
Awesome, your second example was what I was looking for, thank you.
– Insignian
Nov 23 '18 at 22:40
add a comment |
you could use simple javascript to do so, like onclick event for example :
just add this to img tag onclick='window.open('". random_image('css/images/avatars') ."')'
echo "<img onclick='window.open('". random_image('css/images/avatars') ."')' src='" . random_image('css/images/avatars') . "' />";
answered Nov 23 '18 at 22:23
M0ns1fM0ns1f
2,4912821
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53453330%2fdownload-link-over-random-php-image%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You should save the image location in a variable then use it to create a link, plus display it.
$imageUrl = random_image('css/images/avatars');
echo "<a href=" . $imageUrl . ">";
echo "<img src=" . $imageUrl . " />";
echo "</a>";
or if you want to show the text link above, seperately then
$imageUrl = random_image('css/images/avatars');
echo "<a href=" . $imageUrl . ">Click Here</a><br />";
echo "<img src=" . $imageUrl . " />";
answered Nov 23 '18 at 22:22
Nawed KhanNawed Khan
2,8141618
1
Awesome, your second example was what I was looking for, thank you.
– Insignian
Nov 23 '18 at 22:40
add a comment |
You should save the image location in a variable then use it to create a link, plus display it.
$imageUrl = random_image('css/images/avatars');
echo "<a href=" . $imageUrl . ">";
echo "<img src=" . $imageUrl . " />";
echo "</a>";
or if you want to show the text link above, seperately then
$imageUrl = random_image('css/images/avatars');
echo "<a href=" . $imageUrl . ">Click Here</a><br />";
echo "<img src=" . $imageUrl . " />";
answered Nov 23 '18 at 22:22
Nawed KhanNawed Khan
2,8141618
1
Awesome, your second example was what I was looking for, thank you.
– Insignian
Nov 23 '18 at 22:40
add a comment |
You should save the image location in a variable then use it to create a link, plus display it.
$imageUrl = random_image('css/images/avatars');
echo "<a href=" . $imageUrl . ">";
echo "<img src=" . $imageUrl . " />";
echo "</a>";
or if you want to show the text link above, seperately then
$imageUrl = random_image('css/images/avatars');
echo "<a href=" . $imageUrl . ">Click Here</a><br />";
echo "<img src=" . $imageUrl . " />";
answered Nov 23 '18 at 22:22
Nawed KhanNawed Khan
2,8141618
You should save the image location in a variable then use it to create a link, plus display it.
$imageUrl = random_image('css/images/avatars');
echo "<a href=" . $imageUrl . ">";
echo "<img src=" . $imageUrl . " />";
echo "</a>";
or if you want to show the text link above, seperately then
$imageUrl = random_image('css/images/avatars');
echo "<a href=" . $imageUrl . ">Click Here</a><br />";
echo "<img src=" . $imageUrl . " />";
answered Nov 23 '18 at 22:22
Nawed KhanNawed Khan
2,8141618
answered Nov 23 '18 at 22:22
Nawed KhanNawed Khan
2,8141618
answered Nov 23 '18 at 22:22
Nawed KhanNawed Khan
2,8141618
answered Nov 23 '18 at 22:22
Nawed KhanNawed Khan
2,8141618
2,8141618
1
Awesome, your second example was what I was looking for, thank you.
– Insignian
Nov 23 '18 at 22:40
add a comment |
1
Awesome, your second example was what I was looking for, thank you.
– Insignian
Nov 23 '18 at 22:40
1
1
Awesome, your second example was what I was looking for, thank you.
– Insignian
Nov 23 '18 at 22:40
Awesome, your second example was what I was looking for, thank you.
– Insignian
Nov 23 '18 at 22:40
add a comment |
you could use simple javascript to do so, like onclick event for example :
just add this to img tag onclick='window.open('". random_image('css/images/avatars') ."')'
echo "<img onclick='window.open('". random_image('css/images/avatars') ."')' src='" . random_image('css/images/avatars') . "' />";
answered Nov 23 '18 at 22:23
M0ns1fM0ns1f
2,4912821
add a comment |
you could use simple javascript to do so, like onclick event for example :
just add this to img tag onclick='window.open('". random_image('css/images/avatars') ."')'
echo "<img onclick='window.open('". random_image('css/images/avatars') ."')' src='" . random_image('css/images/avatars') . "' />";
answered Nov 23 '18 at 22:23
M0ns1fM0ns1f
2,4912821
add a comment |
you could use simple javascript to do so, like onclick event for example :
just add this to img tag onclick='window.open('". random_image('css/images/avatars') ."')'
echo "<img onclick='window.open('". random_image('css/images/avatars') ."')' src='" . random_image('css/images/avatars') . "' />";
answered Nov 23 '18 at 22:23
M0ns1fM0ns1f
2,4912821
you could use simple javascript to do so, like onclick event for example :
just add this to img tag onclick='window.open('". random_image('css/images/avatars') ."')'
echo "<img onclick='window.open('". random_image('css/images/avatars') ."')' src='" . random_image('css/images/avatars') . "' />";
answered Nov 23 '18 at 22:23
M0ns1fM0ns1f
2,4912821
answered Nov 23 '18 at 22:23
M0ns1fM0ns1f
2,4912821
answered Nov 23 '18 at 22:23
M0ns1fM0ns1f
2,4912821
answered Nov 23 '18 at 22:23
M0ns1fM0ns1f
2,4912821
2,4912821
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53453330%2fdownload-link-over-random-php-image%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
You just make a link with the same URL as the image. Like,
<a>instead of<img>.– miken32
Nov 23 '18 at 22:12