Question

$filename = '01.jpg';
$degrees = 90;
// Content type
header('Content-type: image/jpeg');

// Load
$source = imagecreatefromjpeg($filename);

// Rotate
$rotate = imagerotate($source, $degrees, 0);

// Output
$rotate = imagejpeg($rotate);

我正在使用此请解释如何更新这些图像

Answer 1

imagejpeg - 将图像输出到浏览器或文件。将要保存文件的图像路径作为第二个参数传递。

参考：http://php.net/manual/en/function.imagejpeg.php

$filename = '01.jpg';
$degrees = 90;
$source = imagecreatefromjpeg($filename);
$rotate = imagerotate($source, $degrees, 0);
imagejpeg($rotate, $filename);

Answer 2

请使用以下代码

function RotateImg($filename = '',$angle = 0,$savename = false)
    {
       $original   =   imagecreatefromjpeg($filename);
       $rotated    =   imagerotate($original, $angle, 0);
       if($savename == false) {
                header('Content-Type: image/jpeg');
                imagejpeg($rotated);
            }
        else {
           imagejpeg($rotated,$savename);
        }
        imagedestroy($rotated);
    }

$filename   =   'http://images.all-free-download.com/images/graphiclarge/beautiful_nature_landscape_02_hd_picture_166206.jpg';

$saveto     =   $_SERVER['DOCUMENT_ROOT']."/images/test.jpg";

RotateImg($filename,90,$saveto);

如何在服务器上保存图像的旋转？

2 个答案: