为什么我无法旋转输出图像?

时间:2014-08-05 16:44:03

标签: php gd

<?php 

$img = imagecreatefromjpeg('simpletext.jpg');
$imgRotated = imagerotate($img, 45,-1);

imagejpeg($imgRotated,'newsimpletext.jpg',100);


?>



<img src="simpletext.jpg" />

<hr>

<img sec="newsimpletext.jpg" />

我想以旋转版本输出图像,但它有错误

警告:imagejpeg()要求参数1为资源,第6行的C:\ xampp \ htdocs \ developphp \ index.php中给出布尔值

1 个答案:

答案 0 :(得分:0)

路径中一定有错误, 测试此代码:

if(is_file('simpletext.jpg')){
   $img = imagecreatefromjpeg('simpletext.jpg');
   $imgRotated = imagerotate($img, 45,-1);

   imagejpeg($imgRotated,'newsimpletext.jpg',100);
}
else echo "error on image file or path";

echo '<img src="simpletext.jpg" /><hr><img src="newsimpletext.jpg" />';

结果是:

enter image description here