I have managed to send a FormData object like so:
var formData = new FormData();
formData.append('file', this.files[0]);
$.ajax({
url: urlUploadProductsFile,
type: 'POST',
data: formData,
cache: false,
contentType: false,
processData: false
}, 'json');
Now what I want to do is add an additional CustomerId
to send to the server. The following won't work:
var formData = new FormData();
formData.append('file', this.files[0]);
$.ajax({
url: urlUploadProductsFile,
type: 'POST',
data: { "file": formData, "CustomerId": 2 },
cache: false,
contentType: false,
processData: false
}, 'json');
And I also tried the following variations:
data: { "file": formData, "CustomerId": 2 }, processData: true
data: JSON.stringify({ "file": formData, "CustomerId": 2 })
data: { "file": JSON.stringify(formData), "CustomerId": 2 }
data: { file: formData, CustomerId: 2 }
Any help appreciated.
Borik Bobrujskov
Try:
var formData = new FormData();
formData.append('file', this.files[0]);
formData.append('CustomerId', 2);
/*
note:: appending in form Data will give "csrf token mismatch error".
so better you make a input feild of type hidden with name = CustomerId
and value = 2
*/
$.ajax({
url: urlUploadProductsFile,
type: 'POST',
data: formData,
cache: false,
contentType: false,
processData: false
}, 'json');
You need to either add it directly to formData
(just as you did with 'file'
), or alternatively use query (GET) parameters.
来源:https://stackoverflow.com/questions/36448724/send-formdata-object-and-an-additional-parameter-via-ajax