php read and write an image in url

Question

I would like to resize an image in URL such as localhost/changimage.php?percent=40. I have already resized it in php. But user should resize it by typing in URL here is my PHP code:changimage.php

<?php

 $filename = '/var/www/html/zahir/images/bike.jpeg';

  $percent = 0.5; 

  header('Content-type: image/jpeg');

    list($width, $height) = getimagesize($filename);

     $new_width = $width * $percent;

       $new_height = $height * $percent;

         $image_p = imagecreatetruecolor($new_width, $new_height);

  $image = imagecreatefromjpeg($filename)
;
imagecopyresampled($image_p, $image, 0, 0, 0, 0, $new_width, $new_height, $width, $height);

  imagejpeg($image_p, null, 100);

 ?>

I have no idea how to do it in URL.

Not sure i understood your question right, but try $percent = $_GET['percent']; — Littlebobbydroptables
– Littlebobbydroptables, Commented Mar 26, 2013 at 8:02
@Nerfair $percent = array_key_exists('percent', $_GET) ? $_GET['percent']:0; is slightly better : — HamZa
– HamZa, Commented Mar 26, 2013 at 8:06

zyntrax · Accepted Answer · 2013-03-26 08:42:36Z

1

OR

$percent = 0.5;
if (isset($_GET['percent'])) {$percent = $_GET['percent']/100;}

To check if percent has a value

answered Mar 26, 2013 at 8:42

zyntrax

1361 silver badge12 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Dieepak · Accepted Answer · 2013-03-26 09:58:34Z

1

Try it.

$percent = 0;
   if (isset($_GET["percent"])) {
       $percent = ($_GET["percent"] / 100);
      }
    if ($percent <= 0) {
        $percent = 1;
    }

    $filename = '/home/deepakgupta/Pictures/phone.jpg';

    header('Content-type: image/jpeg');

    list($width, $height) = getimagesize($filename);

    $new_width = $width * $percent;

    $new_height = $height * $percent;

    $image_p = imagecreatetruecolor($new_width, $new_height);

    $image = imagecreatefromjpeg($filename)
    ;
    imagecopyresampled($image_p, $image, 0, 0, 0, 0, $new_width, $new_height, $width, $height);

    imagejpeg($image_p, null, 100);

edited Mar 26, 2013 at 9:58

answered Mar 26, 2013 at 8:27

Dieepak

5462 silver badges7 bronze badges

Comments

Amir · Accepted Answer · 2013-03-26 08:26:11Z

0

maybe something like that:

$percent = 0.5;
if ($_GET['percent']) {$percent = $_GET['percent']/100;}

answered Mar 26, 2013 at 8:26

Amir

4,1314 gold badges18 silver badges28 bronze badges

Collectives™ on Stack Overflow

php read and write an image in url

3 Answers 3

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related