php在url中读写图像

我想在URL中调整图像的大小，例如

localhost/changimage.php？percent=40

。我已经用php调整了它的大小。但用户应该通过键入URL来调整其大小 这是我的PHP代码：

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);

 ?>

---

我不知道如何在URL中执行此操作。

可能类似于：

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

试试看

$percent = 0;
   if (isset($_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);

$percent=0；
如果（isset（$_GET[“percent”]））{
$percent=（$_GET[“percent”]/100）；
}
如果（$百分比或

要检查百分比是否有值，我不确定我是否正确理解了您的问题，请尝试$percent=$\u GET['percent']；@Nerfair
$percent=array\u key\u存在（'percent'，$\u GET）？$\u GET['percent']：0；
稍好一些：
$percent = 0.5;
if (isset($_GET['percent'])) {$percent = $_GET['percent']/100;}