我有下面的代码,但我不确定我的试捕块是否真的正确。有人能指点我吗?
var upload = new FormData($("#video_upload")[0]);
try {
$.ajax({
url: "files/video_upload",
enctype: 'multipart/form-data',
type: 'POST',
data: upload,
async: false,
success: function (data) {
$('.filesToUpload').empty();
},
cache: false,
contentType: false,
processData: false
});
} catch (Exception e) {
alert (‘Upload Failed.');
}发布于 2022-08-27 09:24:26
而不是在ajax中使用try-catch。当请求失败时,可以使用ajax请求返回的错误函数。
$.ajax({
url: "files/video_upload",
enctype: 'multipart/form-data',
type: 'POST',
data: upload,
async: false,
success: function (data) {
$('.filesToUpload').empty();
},
//Add this
error: function (error) {
console.log(error)
},
cache: false,
contentType: false,
processData: false
});https://stackoverflow.com/questions/73509749
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