html
<body>
<form method="post" action="" id="dataForm" enctype="multipart/form-data">
<input type="text" id="textSelector"/>
<input type="file" id="fileSelector"/>
<button name="sub-comfirm" class="btn-selection-content" type="submit" id="sub-comfirm">Send</button>
</form>
</body>
js/ajax
var fileVar = null;// global var to store file info
$(document).ready(function(){
$('#dataForm').on('submit', function(e) {
e.preventDefault();
SendData();
});
$('#fileSelector').on('change',function(ev){
fileVar = null;
fileVar = new FormData();
$.each(ev.target.files, function(key, value)
{
fileVar.append(key, value);
});
});
});
function SendData(){
var formData = new FormData($("#dataForm")[0]);
// should i add the filerVar like this ?
// formData.append("Image", fileVar);
$.ajax({
type: "POST",
url: "checkinput.php",//you need to add this '?files part' to URL
data: formData,// use fileVar here now
cache: false,
processData: false,
// contentType: false,
success:function(data)
{
console.log(data);
console.log("success");
},
error: function(jqXHR, textStatus, errorThrown)
{
console.log("failure");
}
});
}
php
print_r($_POST);
print_r($_FILES);
my intention is to send input(file) and input(text) in one ajax , i can get the input file value if i add ajax data with fileVar , but i cant get my input text value i have no idea why , can anyone tell me what i did wrong ?
var formData = new FormData($("#dataForm")[0]) is the way to get both to one ajax but i cant get any input text value.
anyone can teach me how to make this work ?
I think you need to specify input name attributes:
<body>
<form method="post" action="" id="dataForm" enctype="multipart/form-data">
<input type="text" id="textSelector" name="textSelector"/>
<input type="file" id="fileSelector" name="fileSelector"/>
<button name="sub-comfirm" class="btn-selection-content" type="submit" id="sub-comfirm">Send</button>
</form>
</body>
Hope that helps.
Related
I have a form in a modal window that allows a user to resend a confirmation email. When the form is submitted, the confirmation email is sent twice, instead of once. Everything else is working exactly as it should.
Form is pretty standard:
<form enctype="multipart/form-data" id="ReMailf" name="ReMailf" role="form" data-toggle="validator">
<fieldset>
<div class="row">
<p>You may enter a different email than your original if you wish. However, the original email will remain as the main contact on your application.</p>
<label class="desc" for="prim_email"> Email </label>
<input id="prim_email" name="prim_email" type="email" class="form-control" value="<?php echo $tE['prim_email']; ?>" data-error="Please Enter A Valid Email Address" required/>
<div class="help-block with-errors"></div>
</div>
<div class="row">
<input id="submitForm" name="submitForm" class="btn btn-success" type="submit" value="Resend Conformation "/>
<input name="uniqid" type="hidden" value="<?php echo $tE['unqID']; ?>"/>
<input name="ReMAIL" type="hidden" value="ReMAIL"/>
</div>
</fieldset>
</form>
… and here's the handler:
$(document).ready(function () {
$("#ReMailf").on("submit", function(e) {
var postData = $(this).serializeArray();
// var formURL = $(this).attr("action");
$.ajax({
url: '_remail.php',
type: "POST",
data: postData,
success: function(data, textStatus, jqXHR) {
$('#myModal .modal-header .modal-title').html("YOUR EMAIL HAS BEEN RESENT");
$('#myModal .modal-body').html(data);
// $("#ReMailf").remove();
},
error: function(jqXHR, status, error) {
console.log(status + ": " + error);
}
});
e.preventDefault();
});
$("#submitForm").on('click', function() {
$("#ReMailf").submit();
});
});
I've read a number of other post about this, and tried some of the suggestions, but nothing is working. It either doesn't submit at all, or submits twice.
This is the only form on the page...
Suggestions please?
It is because you are using a button or submit to trigger the ajax event. Use this instead:
$(document).ready(function() {
$("#ReMailf").on("submit", function(e) {
e.preventDefault(); //add this line
var postData = $(this).serializeArray();
// var formURL = $(this).attr("action");
$.ajax({
url: '_remail.php',
type: "POST",
data: postData,
success: function(data, textStatus, jqXHR) {
$('#myModal .modal-header .modal-title').html("YOUR EMAIL HAS BEEN RESENT");
$('#myModal .modal-body').html(data);
// $("#ReMailf").remove();
},
error: function(jqXHR, status, error) {
console.log(status + ": " + error);
}
or you can just use a simple form with action and method. It will do the job
I am trying to send some data from a modal dialog to my controller with Ajax. But my modelfields are always null, but I enter my actionmethod in the controller.
This is a shortend version of my cshtml-file.
#model anmespace.MyModel
<form method="post" id="formID">
...
<div class="row">
<div class="col-md-5">#Resource.GetResource("MyModal", "Firstname")</div>
<div class="col-md-7"><input type="text" class="form-control" id="firstname" value="#Html.DisplayFor(model => model.FirstName)"></div>
</div>
...
<input type="submit" class="btn btn-primary" value="Submit" />
</form>
<script>
$("#formID").on("submit", function (event) {
var $this = $(this);
var frmValues = $this.serialize();
$.ajax({
cache: false,
async: true,
type: "POST",
url: "#Url.Action("ActionName", "Controller")",
data: frmValues,
success: function (data) {
alert(data);
}
});
});
</script>
Sorry MVC/Ajax are really new for me.
If you want to bind the form data to model then, the names of HTML elements should match with Model properties.
Note: name attribute value of html input field should match to the property of a model.
When you use form and submit button then it will try to reload the page by posting data to the server. You need to prevent this action. You can do this by returning false on onSubmit event in the Form element.
When you use jquery, do not forget to keep the ajax call/events inside the $(document).ready(function(){}) function.
I have written a simple code which takes First Name as input and makes an ajax call on clicking on submit button.
Html & Jquery Code:
<script>
$(document).ready(function() {
$("#formID").on("submit", function(event) {
var $this = $(this);
var frmValues = $this.serialize();
$.ajax({
cache: false,
async: true,
type: "POST",
url: "#Url.Action("PostData", "Home")",
data: frmValues,
success: function(data) {
alert(data.FirstName);
}
});
});
});
</script>
<div>
<form method="post" id="formID" onsubmit="return false;">
<input id="FirstName" name="FirstName"/>
<input type="submit" value="submit" />
</form>
</div>
My Model :
public class Person
{
public string FirstName { get; set; }
}
Action Method:
public ActionResult PostData(Person person)
{
return Json(new { Success = true, FirstName = person.FirstName });
}
Output:
I am trying to post File using JQuery. Below is my code.
<script language="javascript">
$(document).ready(function() {
$('#frmUpdateProfile').on("submit", function(event) {
event.stopPropagation(); // Stop stuff happening
event.preventDefault(); // Totally stop stuff happening
var data = {
"FileName" : event.target.FileName.files,
"_token" : "{!! csrf_token() !!}"
};
$.ajax({
url: '{{URL::route("UpdateProfile")}}',
method: "POST",
async: true,
data: JSON.stringify(data),
processData: false,
contentType: "application/json; charset=utf-8",
success: function (msg) {
SuccessCallback(msg);
},
error: function (jqXHR) {
ErrorCallback(jqXHR);
}
});
});
});
</script>
I tried processData: false,. While debugging in Js, you can check that image is coming in the data. Below is the screenshot.
But when I print the request data in Laravel, it show blank array.
Html form is here
<form method="POST"
action="http://localhost:1234/AS6-Code/public/UpdateProfile"
accept-charset="UTF-8"
enctype="multipart/form-data"
id="frmUpdateProfile">
<input name="_token" type="hidden" value="26KWkWdNqe5iOFE8VRBf1dRnL5xKxwN25jg3tAFW">
<input type="hidden" name="_token" value="26KWkWdNqe5iOFE8VRBf1dRnL5xKxwN25jg3tAFW">
<input multiple="1" name="FileName" type="file">
<input class="btn btn-info" type="submit" value="Update">
</form>
Am I doing something wrong?
Try sending your request with FormData instead:
var data = new FormData($('#frmUpdateProfile')[0]);
Also set contentType to false:
contentType: false
Also Update
event.target.FileName.files
to
event.target.FileName.files[0]
event.target.FileName.files is a FileList object. I believe you need event.target.FileName.files[0] instead.
I'm creating a JSP/Servlet web application and I'd like to upload a file to a servlet via Ajax. How would I go about doing this? I'm using jQuery.
I've done so far:
<form class="upload-box">
<input type="file" id="file" name="file1" />
<span id="upload-error" class="error" />
<input type="submit" id="upload-button" value="upload" />
</form>
With this jQuery:
$(document).on("#upload-button", "click", function() {
$.ajax({
type: "POST",
url: "/Upload",
async: true,
data: $(".upload-box").serialize(),
contentType: "multipart/form-data",
processData: false,
success: function(msg) {
alert("File has been uploaded successfully");
},
error:function(msg) {
$("#upload-error").html("Couldn't upload file");
}
});
});
However, it doesn't appear to send the file contents.
To the point, as of the current XMLHttpRequest version 1 as used by jQuery, it is not possible to upload files using JavaScript through XMLHttpRequest. The common workaround is to let JavaScript create a hidden <iframe> and submit the form to it instead so that the impression is created that it happens asynchronously. That's also exactly what the majority of the jQuery file upload plugins are doing, such as the jQuery Form plugin (an example).
Assuming that your JSP with the HTML form is rewritten in such way so that it's not broken when the client has JavaScript disabled (as you have now...), like below:
<form id="upload-form" class="upload-box" action="/Upload" method="post" enctype="multipart/form-data">
<input type="file" id="file" name="file1" />
<span id="upload-error" class="error">${uploadError}</span>
<input type="submit" id="upload-button" value="upload" />
</form>
Then it's, with the help of the jQuery Form plugin, just a matter of
<script src="jquery.js"></script>
<script src="jquery.form.js"></script>
<script>
$(function() {
$('#upload-form').ajaxForm({
success: function(msg) {
alert("File has been uploaded successfully");
},
error: function(msg) {
$("#upload-error").text("Couldn't upload file");
}
});
});
</script>
As to the servlet side, no special stuff needs to be done here. Just implement it exactly the same way as you would do when not using Ajax: How can I upload files to a server using JSP/Servlet?
You'll only need an additional check in the servlet if the X-Requested-With header equals XMLHttpRequest or not, so that you know how what kind of response to return for the case that the client has JavaScript disabled (as of now, it is mostly the older mobile browsers which have JavaScript disabled).
if ("XMLHttpRequest".equals(request.getHeader("X-Requested-With"))) {
// Return an Ajax response (e.g. write JSON or XML).
} else {
// Return a regular response (e.g. forward to JSP).
}
Note that the relatively new XMLHttpRequest version 2 is capable of sending a selected file using the new File and FormData APIs. See also HTML5 drag and drop file upload to Java Servlet and Send a file as multipart through XMLHttpRequest.
Monsif's code works well if the form has only file type inputs. If there are some other inputs other than the file type, then they get lost. So, instead of copying each form data and appending them to FormData object, the original form itself can be given to the constructor.
<script type="text/javascript">
var files = null; // when files input changes this will be initialised.
$(function() {
$('#form2Submit').on('submit', uploadFile);
});
function uploadFile(event) {
event.stopPropagation();
event.preventDefault();
//var files = files;
var form = document.getElementById('form2Submit');
var data = new FormData(form);
postFilesData(data);
}
function postFilesData(data) {
$.ajax({
url : 'yourUrl',
type : 'POST',
data : data,
cache : false,
dataType : 'json',
processData : false,
contentType : false,
success : function(data, textStatus, jqXHR) {
alert(data);
},
error : function(jqXHR, textStatus, errorThrown) {
alert('ERRORS: ' + textStatus);
}
});
}
</script>
The HTML code can be something like following:
<form id ="form2Submit" action="yourUrl">
First name:<br>
<input type="text" name="firstname" value="Mickey">
<br>
Last name:<br>
<input type="text" name="lastname" value="Mouse">
<br>
<input id="fileSelect" name="fileSelect[]" type="file" multiple accept=".xml,txt">
<br>
<input type="submit" value="Submit">
</form>
$('#fileUploader').on('change', uploadFile);
function uploadFile(event)
{
event.stopPropagation();
event.preventDefault();
var files = event.target.files;
var data = new FormData();
$.each(files, function(key, value)
{
data.append(key, value);
});
postFilesData(data);
}
function postFilesData(data)
{
$.ajax({
url: 'yourUrl',
type: 'POST',
data: data,
cache: false,
dataType: 'json',
processData: false,
contentType: false,
success: function(data, textStatus, jqXHR)
{
//success
},
error: function(jqXHR, textStatus, errorThrown)
{
console.log('ERRORS: ' + textStatus);
}
});
}
<form method="POST" enctype="multipart/form-data">
<input type="file" name="file" id="fileUploader"/>
</form>
This code works for me.
I used Commons IO's io.jar, Commons file upload.jar, and the jQuery form plugin:
<script>
$(function() {
$('#upload-form').ajaxForm({
success: function(msg) {
alert("File has been uploaded successfully");
},
error: function(msg) {
$("#upload-error").text("Couldn't upload file");
}
});
});
</script>
<form id="upload-form" class="upload-box" action="upload" method="POST" enctype="multipart/form-data">
<input type="file" id="file" name="file1" />
<span id="upload-error" class="error">${uploadError}</span>
<input type="submit" id="upload-button" value="upload" />
</form>
boolean isMultipart = ServletFileUpload.isMultipartContent(request);
if (isMultipart) {
// Create a factory for disk-based file items
FileItemFactory factory = new DiskFileItemFactory();
// Create a new file upload handler
ServletFileUpload upload = new ServletFileUpload(factory);
try {
// Parse the request
List items = upload.parseRequest(request);
Iterator iterator = items.iterator();
while (iterator.hasNext()) {
FileItem item = (FileItem) iterator.next();
if (!item.isFormField()) {
String fileName = item.getName();
String root = getServletContext().getRealPath("/");
File path = new File(root + "../../web/Images/uploads");
if (!path.exists()) {
boolean status = path.mkdirs();
}
File uploadedFile = new File(path + "/" + fileName);
System.out.println(uploadedFile.getAbsolutePath());
item.write(uploadedFile);
}
}
} catch (FileUploadException e) {
e.printStackTrace();
} catch (Exception e) {
e.printStackTrace();
}
}
I'm new to MooTools and trying to send Ajax request with form content to the url.
<form method="post" enctype="multipart/form-data" action="<?= $PHP_SELF; ?>" name="fn" id="fn">
<input name="user_name" id="user_name">
<input name="user_mail" id="user_name">
<input name="user_text" id="user_name">
<input type="file" name="attach">
<input id="button" type="button" value="Submit">
</form>
<script type="text/javascript">
$('button').addEvent('click', function(event) {
var req = new Request({
method: 'post',
url: 'url.php',
onSuccess: function(response) {
alert(response);
});
});
</script>
When I click on button, nothing happens. How right transferring data from form?
Your code looks good, you had a } missing but aside from that you just forgot to add a .send();
Like req.send();, and you can actually pass the data as a argument to that method.
Test that and check here about the .send() method.
Suggention to how your code could look like:
<script type="text/javascript">
var req = new Request({
method: 'post',
url: 'url.php',
onSuccess: function(response) {
alert(response);
} // < --- You actually missed this one
});
$('button').addEvent('click', function(event) {
req.send();
});
</script>