I'm creating a login page and at the bottom of the pop-up form, there is another button that takes you to the registration page. The issue appears to be that when navigating to the new page it all sits under the original sign-in form which uses an ajax call to check if the user exists so when they try to submit the registration form it then calls that ajax call from the sign-in form.
Sign-in form
<div id="myForm">
<form onsubmit="return false;" id="loginForm">
<h1>Login</h1>
<label for="email"><b>Email</b></label>
<input type="email" id="email" placeholder="Enter Email" name="email" required>
<label for="psw"><b>Password</b></label>
<input type="password" id="psw" placeholder="Enter Password" name="psw" required>
<div id="message" class="alert-danger"></div>
<br />
<button type="submit" id="submit" class="btn">Login</button>
<button type="button" class="btn cancel" onclick="closeForm();">Close</button>
</form>
<div class="d-inline">
<button class="btn-info">#Html.ActionLink("User Registration", "SignUp", "SignUp_SignIn")</button>
</div>
</div>
Then the ajax call is
$(document).ready(function () {
$("form").on('submit', function (event) {
var data = {
'email': $("#email").val(),
'psw': $("#psw").val()
};
$.ajax({
type: "POST",
url: 'SignUp_SignIn/CredentialCheck',
data: data,
success: function (result) {
if (result == true) {
$("#message").text("Login attempt was successful");
}
else {
$("#message").text("Email/Password didn't match any results");
}
},
error: function () {
alert("It failed");
}
});
return false;
});
});
After looking at the comments I realized that the reason that the login form was being called from the layout.cshtml so when the ajax call was being called it was grabbing all the form tags that existed on any page that was loaded up. After changing the ajax so it was calling a specific id for the login form instead of form it allowed for proper actions to take place.
An example of what I'm refusing to
$(document).ready(function () {
$("form").on('submit', function (event) {
$.ajax({
type: "POST",
url: url,
data: data,
success: function (result) {
//Do stuff
}
});
});
});
The above will try to redirect you on the submit of any form that is loaded up, but if we go through and change the way it accesses the form like below then it will only work if the one specific form is submitted.
$(document).ready(function () {
$("#loginForm").on('submit', function (event) {
$.ajax({
type: "POST",
url: url,
data: data,
success: function (result) {
//Do stuff
}
});
});
});
Related
I am developing multi Step Form Submit without refresh. collect the data from 1st step 2nd step collect some date, 3rd step collect some date & finally submit data in the database. Can you tell me how to fix this.
My blade template.
<form id="post-form" method="post" action="javascript:void(0)">
#csrf
<div>
<input class="form-input" type="text" id="ptitle" name="ptitle" required="required"
placeholder="What do you want to achieve?">
</div>
<button type="text" id="send_form" class="btn-continue">Continue</button>
</div>
</form>
Ajax Script
$(document).ready(function() {
$("#send_form").click(function(e){
e.preventDefault();
var _token = $("input[name='_token']").val();
var ptitle = $('#ptitle').val();
$.ajax({
url: "{{route('create.setp2') }}",
method:'POST',
data: {_token:_token,ptitle:ptitle},
success: function(data) {
alert('data.success');
}
});
});
Web.php router
Route::post('/setp2', [Abedoncontroller::class, 'funcsetp1'])->name('create.setp2');
Controller method
public function funcsetp1(Request $request) {
$postdata=$request->input('ptitle');
return response()->json('themes.abedon.pages.create-step-2');
}
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 want to open a modal and get the content (the formular) form the controller. These steps work fine.
Now I have the formular inside the modal. How I can submit these formular and process it with the same controller function?
At the Moment I do the Ajax-Call in this way
$('.modaldisconnect').click(function(){
var modid = $(this).attr('modid');
$.ajax({
url: "<?php echo base_url() ?>admin/module/disconnect_form",
method: "POST",
data: {modid:modid},
// Callback function that is executed after data is successfully sent and received
success: function(data){
// Print the fetched data of the selected phone in the section called #phone_result
// within the Bootstrap modal
$('.modal-ajax-content').html(data);
// Display the Bootstrap modal
$('#disconnect_modal').modal('show');
},
error: function(error){
// Show error message
alert('error');
}
});
});
This is the form which will be generate from the controller by the call above
echo' <div class="modal-body">
<div id="modul_disc_infos">
<span class="module-headlines-title">'._l("modul_name").': <span id="disc_modulname">'._l('module_'.$result->folder.'_titel').'</span></span>
<span class="module-headlines-subtitle">'._l("modul_disconnect_deadline").': <span id="disc_enddate">'.$enddate.'</span></span>
</div>
<div id="modul_disc_check">
<div class="alert alert-danger" role="alert">'._l("modul_disconnect_disclaimer").'</div>
<hr>
<input type="checkbox" name="check_disclaimer" value="1"> '._l("modul_disconnect_disclaimer_check").'
<input type="hidden" name="modid" value="'.$this->input->post('modid').'">
<input type="hidden" name="formstep" value="1">
'.$errormsg.'
</div>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">'._l('close').'</button>
<button type="button" class="btn btn-danger" id="submit_disconnect" >'._l('modul_disconnect_now').'</button>
</div>';
By the way - is there a better way to parse the values of the formular instead one by one like in my code "var modid = $(this).attr('modid');"?
I've solved that in this way that i do 2 different ajax call, starts with a click or submit on the form or a specific button. One call generate the form, one call handle the form. For me it works and its a little bit flexibel to use the same functions for different jobs.
//Formular for disconnection
$('.modaldisconnect').click(function(){
var modid = $(this).attr('modid');
$('#ajax_modal_title').html('<?php echo _l('modul_disconnect'); ?>');
$.ajax({
url: "<?php echo base_url() ?>admin/module/disconnect_form",
method: "POST",
data: {modid:modid},
// Callback function that is executed after data is successfully sent and recieved
success: function(data){
// Print the fetched data of the selected phone in the section called #phone_result
// within the Bootstrap modal
$('.modal-ajax-content').html(data);
// Display the Bootstrap modal
$('#ajax_modal').modal('show');
},
error: function(error){
// Show error message
alert('error');
}
});
});
//Handle the disconnection
$("#ajax_form").on("submit", function (event) {
event.preventDefault();
subform=$('input[name="subform"]').val();
$.ajax({
url: "<?php echo base_url() ?>admin/module/"+subform+"_form",
type: "POST",
data: $('#ajax_form').serialize(),
success: function(result){
$('.modal-ajax-content').html(result);
},
error: function(error){
// Show error message
alert('error');
}
});
});
For some reason, when trying to login I need to have e.preventDefault (prevent page reloading) in order to save my JWT into local storage with an AJAX call. So when i have this:
handleLogin(e) {
//Without e.preventDefault, the jwt token is not save -> cannot access api
e.preventDefault();
const email = $('#email').val()
const password = $('#password').val()
const request = {"auth": {
"email": email,
"password": password
}}
$.ajax({
url: "http://localhost:5000/api/user_token",
type: "POST",
data: request,
dataType: "json",
success: function (result){
console.log(result.jwt)
localStorage.setItem("jwt", result.jwt)
}
})
}
Here is my simple form
render(){
return(
<div>
<form>
<input
name="email"
id="email"
type="email"
/>
<input
name="password"
id="password"
type="password"
/>
<button
onClick={this.handleLogin}
>Login</button>
<button
onClick={this.handleLogout}
>Logout</button>
</form>
<button onClick={this.getUsers}>Get Users</button>
{
this.state.users
}
</div>
)
}
I want my page to reload/go to a different page after submitting a successful login. On create-react-app and using a Rails API 5
In your case, you can try this:
success: function (result){
console.log(result.jwt)
localStorage.setItem("jwt", result.jwt)
//page reload
window.location.reload(true);
// or route to another page
window.location.href = 'foo'; // any route
}
But I would recommend to use react router so your app will never loose it's state.
If you have any query, you can ask.
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: