Question

尝试从我的数据库中检索图像并旋转后，我继续收到此错误：

Warning: imagecreatefromstring() [function.imagecreatefromstring]: Data is not in a recognized format

以下是我要求将blob数据转换回图像以便轮换的内容：

$SQL="SELECT * FROM images WHERE id={$id}";
$rh = mysql_query($SQL);

$image=mysql_result($rh,0,"image");
$source_image=imagecreatefromstring($image);

$rotate_image = imagerotate($source_image, 90, 0);

我错过了一步吗？

Answer 1

这就是我最终要做的工作：

$SQL="SELECT * FROM images WHERE id={$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();

Answer 2

我建议将图像数据存储在base64中。存储字符和二进制文件没问题。

然后就这么简单：

<?php
$data = 'iVBORw0KGgoAAAANSUhEUgAAABwAAAASCAMAAAB/2U7WAAAABl'
       . 'BMVEUAAAD///+l2Z/dAAAASUlEQVR4XqWQUQoAIAxC2/0vXZDr'
       . 'EX4IJTRkb7lobNUStXsB0jIXIAMSsQnWlsV+wULF4Avk9fLq2r'
       . '8a5HSE35Q3eO2XP1A1wQkZSgETvDtKdQAAAABJRU5ErkJggg==';
$data = base64_decode($data);

$im = imagecreatefromstring($data);
if ($im !== false) {
    header('Content-Type: image/png');
    imagepng($im);
    imagedestroy($im);
}
else {
    echo 'An error occurred.';
}
?>

来源：http://php.net/manual/fr/function.imagecreatefromstring.php

无法使用PHP的imagerotate旋转图像

2 个答案: