我的js代码将数据url转换为blob并发送到表单请求是
var canv=document.getElementById("mainCanvas");
// var dataURL = canv.toDataURL();
var dataURL = canv.toDataURL('image/jpg');
// .replace("image/png", "image/octet-stream");
documentData={"image":dataURLtoBlob(dataURL),"gameName":"emperor","userId":'userId',"gameId":56};
Game.post(documentData).success(function(response){
console.log(response);
});
创建blob函数
function dataURLtoBlob(dataurl) {
var arr = dataurl.split(','), mime = arr[0].match(/:(.*?);/)[1],
bstr = atob(arr[1]), n = bstr.length, u8arr = new Uint8Array(n);
while(n--){
u8arr[n] = bstr.charCodeAt(n);
}
return new Blob([u8arr], {type:mime});
}
我在angular 上的服务工厂
services.factory('Game', ['$http', function($http){
return {
get:function(){
return;
}
, post:function(documentData){
return $http(
{ method: 'POST',
url: 'public/Game/store',
headers: { 'content-type': 'multipart/form-data'},
data: documentData
});
}
,
delete:function(){
return ;
}
};
}]);
laravel后端
$image=$request->file('image');
$image->move("game/".$request->gameName."/play/" ,$request->userId.$request->gameId.".jpg");
return Response(["success"=>true]);
我反复收到这个错误,说
无法在非对象上调用move
我对将dataurl转换为blob的代码没有信心我试图做的是将dataurl转化为图像文件以形成数据请求
您可以提交到php,只需画布dataURL
,然后用file_put_contents()
保存,就像这个一样
/*JS - add this code in js in laravel view*/
// take data from canvas
var dataURL = canvas.toDataURL();
/*JS - add this code in js in laravel view*/
// preparing data without any dataURLtoBlob conversion
documentData={"image":dataURL,"gameName":"emperor","userId":'userId',"gameId":56}
/*JS - add this code in js in laravel view*/
// you can also remove headers: {'content-type': 'multipart/form-data'},
// from your Ajax
$http({
method: 'POST',
url: 'public/Game/store',
data: documentData
});
/*PHP - laravel backend save form controller*/
//then simply save it as image
file_put_contents(
$request->userId.$request->gameId.".jpg",
base64_decode($request->image)
);
重要信息:是在保存之前对图像数据调用base64_decode()