尝试从数据库检索图像并旋转它后,我继续收到此错误:
Warning: imagecreatefromstring() [function.imagecreatefromstring]: Data is not in a recognized format
以下是我将 blob 数据转换回图像以进行旋转的调用:
$SQL="SELECT * FROM images WHERE id={$id}";
$rh = mysql_query($SQL);
$image=mysql_result($rh,0,"image");
$source_image=imagecreatefromstring($image);
$rotate_image = imagerotate($source_image, 90, 0);
我是不是漏掉了一步?
最佳答案
这就是我最终为它工作所做的事情:
$SQL="SELECT * FROM images WHERE id={$id}";
$rs = mysql_query($SQL);
//rotate image
$image=mysql_result($rs,0,"image");
$source_image=imagecreatefromstring($image);
$rotate_image = imagerotate($source_image, 270, 0);
ob_start();
imagejpeg($rotate_image, null, 100);
$image_bin = mysql_real_escape_string(ob_get_contents()); data.
ob_end_clean();
关于php - 无法使用 PHP 的 imagerotate 旋转图像,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/8221967/