How do I do this in PHP?

I would like to use ImageMagick's identify within PHP to identify an image!

Problem (for me as I'm new to PHP), is how do I return the result(s) that are in an Array?

When I run identify /var/www/web1/web/tmp/upload/test.jpg from a SSH session I get this: JPEG 90x69 DirectClass 2kb 0.000u 0:01

All I need the PHP code to returm is the JPEG

This is what I have as PHP code till now.

Code:

<?php

exec("identify /var/www/web1/web/tmp/upload/test.jpg");

// the data returned is in an Array.
// How do I show the value(s) from the Array?

?> 

 

In your example it seems that JPEG is the first array element so you it is $result[0].

Here's another neat little function that lets you display (multi-)dimensional arrays quite easily:

So you could build a script like this:

 

Thank you... The code looks nice

I was just reading about the explode option in PHP, so I made this code;

Code:

<?php

$str=exec("identify /var/www/web1/web/tmp/upload/test.jpg");

$pieces = explode(" ", $str);
echo $pieces[0]; // 1st value
echo "<br>";
echo $pieces[1]; // 2nd value

?>

 

well, if the exec functions returns an array then there is no need for exploding ^^

 

But the array returned for: identify /var/www/web1/web/tmp/upload/test.jpg

looks like this: /var/www/web1/web/tmp/upload/test.jpg JPEG 125x164 DirectClass 8kb 0.000u 0:01

I only need the JPEG

With the explode function I'm breaking the array at every 'space' it finds in the array.

I guess I called it an Array, but it's not..?

 

well, after you explode at " " then it's the second element $result[1] ^^

 

You could as well use PHP's getimagesize function to find out about the type of an image file: http://www.php.net/manual/en/function.getimagesize.php