PHP Copy image and load in HTML

Question

I have uploaded an image to the folder:

/var/www/uploads/img.png

using the relative path

../uploads/img.png

I then load the image with the following code:

$img = copy('../uploads/9', '/tmp/profile_picture');

Which returns true

And my googling got me to this:

<img src="/tmp/profile_picture" alt="profile_picture" />

I have tried the above with and without .png ending. still not working.

I just need to get out 1 picture and display in an image tag, and i always know the exact path and filename of my image.

EDIT

After the first answer in this thread i have tried the following:

$image = '../uploads/9';

$info = getimagesize($image);

// output the image
header("Content-Disposition: filename={$image};");
header("Content-Type: {$info["mime"]}");
header('Content-Transfer-Encoding: binary');
header('Last-Modified: ' . gmdate('D, d M Y H:i:s', time()) . ' GMT');

readfile($image);

And i call the script here:

<img src="profile_picture.php" alt="profile_picture" />

and i get 404.

I've tried various paths for the $image

Any help appreciated.

Why don't you use <img src="/uploads/img.png" alt="profile_picture" />? If you start with / on a website, it means the web server root, not your local file system root. — ByteHamster
– ByteHamster, Commented Jan 4, 2016 at 19:57

Ema4rl · Accepted Answer · 2016-01-05 12:14:23Z

1

Step 1: correct your use of copy()

Use:
$img = copy('/tmp/profile_picture', '../uploads/9'); instead.

From the PHP.net docs, the format for the copy function is:

copy ( string $source , string $dest [, resource $context ] )

Where the parameters:
source » Path to the source file.

dest » The destination path. If the destination file already exists, it will be overwritten.

context » A valid context resource created with stream_context_create().

Step 2 (1): upload image to a web accessible path

You'll need to know your web root. Consider this folder structure:
/var/www » your www root

/var/www/project1 » your www root

/var/www/project1/assets/img » your publicly accessible images directory for your project

then upload your image to the above path: /var/www/project1/assets/img/img.png

and use the image like this (from index.html):

<img src="/assets/img/img.png" alt="Image">

Step 2 (2): OR wrap your image in a php file

This requires a custom php file, and it's use in a image tag.

Example:
image.php

<?php
$image = '/var/www/uploads/img.png';

$info = getimagesize($image);

// output the image
header("Content-Disposition: filename={$image};");
header("Content-Type: {$info["mime"]}");
header('Content-Transfer-Encoding: binary');
header('Last-Modified: ' . gmdate('D, d M Y H:i:s', time()) . ' GMT');

readfile($image);
?>

index.html

<img src="image.php" alt="Image">

Explanation

This option just wraps your image in php script and the script is called instead.

I use this if the image cannot be publicly accessible and/or I have to do some checks to see if the current user is worthy of accessing the image.

::Used by most file sharing web-apps

edited Jan 5, 2016 at 12:14

answered Jan 4, 2016 at 20:00

Ema4rl

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2 Comments

Kasper Sølvstrøm Over a year ago

I have copied your script and it returns 404, i have tried with the php script residing in the same folter as the view

Kasper Sølvstrøm Over a year ago

And the first part of your answer is obsolete and i think you misunderstood. Source is uploads/9 and destination is /temp/9

Rodrigo Prazim · Accepted Answer · 2017-03-14 03:19:10Z

0

You can do it this way.

Use the function sys_get_temp_dir() to upload the images directly to the temporary folder.

Then

$path = '/tmp/profile_picture';
$imageData = file_get_contents($path);
// display in view
echo sprintf('<img src="data:image/png;base64,%s" />', base64_encode($imageData));

answered Mar 14, 2017 at 3:19

Rodrigo Prazim

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PHP Copy image and load in HTML

2 Answers 2

Step 1: correct your use of copy()

Step 2 (1): upload image to a web accessible path

Step 2 (2): OR wrap your image in a php file

2 Comments

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Collectives™ on Stack Overflow

2 Answers 2

Step 1: correct your use of copy()

Step 2 (1): upload image to a web accessible path

Step 2 (2): OR wrap your image in a php file

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