I need to upload files in ASP.NET MVC. the pure javascript code works (see below), but if I convert the send part to jQuery, it gives me a jquery error (line 8458).
error:
0x8000fff - JavaScript runtime error: Argument not optional
code:
8453 jQuery.param = function( a, traditional ) {
8454 var prefix,
8455 s = [],
8456 add = function( key, value ) {
8457 // If value is a function, invoke it and return its value
8458 value = jQuery.isFunction( value ) ? value() : ( value == null ? "" : value );
8459 s[ s.length ] = encodeURIComponent( key ) + "=" + encodeURIComponent( value );
8460 };
html:
<form data-bind='submit: upload'>
<input type='file' id='fileInput' />
<input type='submit' value='upload' />
</form>
js:
that.upload = function(){
var data = new FormData();
var fileInput = $('#fileInput')[0];
var file = fileInput.files[0];
data.append(file.name, file);
var url = 'blah/Upload?id=' + that.id();
// this pure js works
var xhr = new XMLHttpRequest();
xhr.open('post', url);
xhr.send(data);
// this jquery code does NOT work
$.ajax({
type: 'post',
dataType: json',
url: url,
data: data,
});
};
controller:
public JsonResult Upload(string id){
return Json(JsonConvert.SerializeObject(true), JsonRequestBehavior.DenyGet);
}
You need to add 2 additional ajax options, processData: false
and contentType: false
$.ajax({
type: 'post',
dataType: json',
url: url,
data: data,
processData: false, // add
contentType: false, // add
....
});
Side note: You should use Url.Action()
to ensure your url is correctly generated
var url = '@Url.Action("Upload", "blah")',
and add the id
value to FormData
data.append(id, that.id);
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