我想做一個PHP腳本來裁剪dataURL並返回另一個dataURL,使用GD庫,但我總是得到錯誤,我該如何解決?PHP作物DataURL並返回DataURL
if(isset($_GET['render'])) {
if((isset($_GET['render_x'])) && (isset($_GET['render_y']))) {
if(isset($_GET['dataURL'])) {
$image = $_GET['dataURL']; // the image to crop
$image = substr($image,22);
$img = imagecreatetruecolor('200','150');
$org_img = imagecreatefromstring(base64_decode($image));
imagecopy($img,$org_img, 0, 0, 20, 20, 200, 150);
ob_start();
imagepng($img);
$image_data = ob_get_contents();
ob_end_clean();
$image_data_base64 = base64_encode($image_data);
imagedestroy($img);
echo '<img src="data:image/png;base64,'.$image_data_base64.'" ><p>';
}
}
}
我該如何解決這個問題?我得到了這些錯誤:
Warning: imagecreatefromstring(): gd warning: one parameter to a memory allocation multiplication is negative or zero, failing operation gracefully in xxx on line 29
Warning: imagecreatefromstring(): Passed data is not in 'WBMP' format in xxx on line 29
Warning: imagecreatefromstring(): Couldn't create GD Image Stream out of Data in xxx on line 29
Warning: imagecopy() expects parameter 2 to be resource, boolean given in
和黑色畫面
編輯:$ _GET [ 'dataURL' 值:data:image/png;base64,AAAFBfj42Pj4
我該如何解決呢?謝謝
'$ _GET ['dataURL']'的值是多少?爲什麼'substr'應用於圖像?請發佈樣例'$ _GET ['dataURL']'。 – Samir
@Samir post編輯 – doge45
爲什麼你需要substr對圖像數據的URL?嘗試評論它'// $ image = substr($ image,22);' – Samir