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我試圖創建一個圖像出我的畫布,但我似乎無法得到一個嘗試後,後端工作上傳的PNG。有人可以看看這裏發生了什麼嗎?通過jquery上傳toDataUrl圖像
我的javascript:
stage.toDataURL({
callback: function(dataUrl) {
var imgURL = dataUrl; // keep the entire url
$.ajax({
type: "POST",
url: "http://www.xxxx.nl/pointer/upload.php",
data: ({imgData : imgURL}),
cache: false,
success: function(result){
//window.open(dataUrl); // Show result stage in a new window
alert(result); // show php error if exists
}
});
}
});
而且我很基本的.PHP現在:
$im = imagecreatefrompng($_POST['imgData']);
header('Content-Type: image/png');
imagepng($im);
imagedestroy($im);
我的錯誤日誌:
[Fri Mar 08 11:29:16 2013] [error] [client 24.132.214.139] mod_security: Access denied with code 500. Error reading request body, error code 70007: The timeout specified has expired [hostname "www.ccc.nl"] [uri "/pointer/upload.php"]
你的JavaScript代碼是完全正常的。在PHP方面,您可以使用['DataUriUpload']附帶的[PHP-FileUpload](https://github.com/delight-im/PHP-FileUpload)(https://github.com/delight- IM/PHP的文件上傳/ BLOB/023f812226673ac9e0696d8a3579bb7380606dda/src目錄/ DataUriUpload.php)。它記錄在[這裏](https://github.com/delight-im/PHP-FileUpload/tree/023f812226673ac9e0696d8a3579bb7380606dda#data-uri-uploads)。或者,您可以查看其源代碼。 – caw 2017-12-04 23:53:13