Answers for "javascript formdata"

6

formdata js

var form = $('form')[0]; // You need to use standard javascript object here
var formData = new FormData(form);

or specify exact data for FormData();

var formData = new FormData();
formData.append('section', 'general');
formData.append('action', 'previewImg');
// Attach file
formData.append('image', $('input[type=file]')[0].files[0]); 

Sending form

Ajax request with jquery will looks like this:

$.ajax({
    url: 'Your url here',
    data: formData,
    type: 'POST',
    contentType: false, // NEEDED, DON'T OMIT THIS (requires jQuery 1.6+)
    processData: false, // NEEDED, DON'T OMIT THIS
    // ... Other options like success and etc
});
After this it will send ajax request like you submit regular form
with enctype="multipart/form-data"

Update: This request cannot work without type:"POST" in options since all
files must be sent via POST request.
Posted by: Guest on July-02-2020
1

add formdata javascript

var formData = new FormData();

formData.append("username", "Groucho");
formData.append("accountnum", 123456); // number 123456 is immediately converted to a string "123456"

// HTML file input, chosen by user
formData.append("userfile", fileInputElement.files[0]);

// JavaScript file-like object
var content = '<a id="a"><b id="b">hey!</b></a>'; // the body of the new file...
var blob = new Blob([content], { type: "text/xml"});

formData.append("webmasterfile", blob);

var request = new XMLHttpRequest();
request.open("POST", "http://foo.com/submitform.php");
request.send(formData);
Posted by: Guest on August-15-2020
-1

jquery form data

var fd = new FormData();    
fd.append( 'file', input.files[0] );

$.ajax({
  url: 'http://example.com/script.php',
  data: fd,
  processData: false,
  contentType: false,
  type: 'POST',
  success: function(data){
    alert(data);
  }
});
Posted by: Guest on September-24-2020
-2

javascript form data

var formData = new FormData();
fromData.append('name','value');
Posted by: Guest on September-03-2020

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