CODEWITHSUNDEEP
php
Michael Kuan
Michael Kuan

Reputation: 1101

PHP image show on browser

(1)

$file = $_GET['file'];
echo '<img src="'.$file.'" />';

(2)

 $file = $_GET['file'];
 $imginfo = getimagesize($file);
 header('Content-type: '.$imginfo['mime']);
 echo file_get_contents($file);`

From these 2 code, my image can show in the browser nicely. But what is the differences of them? Which method should I prefer?

Upvotes: 3

Views: 150

Answers (3)

Rakesh Samal
Rakesh Samal

Reputation: 11

The first example you've posted is simply: output:

<img src="image.png" />

The second option actually sets the Content-Type:

It's return the content type of image/png or image/jpg like that.

I prefered go to second example.

Upvotes: 0

Matt Lo
Matt Lo

Reputation: 5731

The key difference is output:

Example 1 references from a path whereas example 2 outputs the binary and labels it an image (so HTTP clients can interpret the response correctly).

To the point... example 1 is preferred because it doesn't have to store the file contents in memory.

Upvotes: 0

Darren
Darren

Reputation: 13128

The first example you've posted is simply "including" the image file into the DOM. It would essentially output something like:

<img src="path/to/image.png" />

While the second option actually sets the Content-Type to whatever the mime of the image is. Meaning if it's a png for example, the page that runs that script will actually be served as a whole image.

If it was a png image, it'd return the content type of image/png.

Upvotes: 2

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