Is there a way to tell PHP which form has been submitted?
Form 1
<form id="factory" action="bullets.php" method="post">
<input type="submit" value="Kopen">
</form>
And form 2
<form id="localFactory" action="bullets.php" method="post">
<input type="submit" value="Kopen">
</form>
These forms are on one page.
My javascript code:
var url;
$('form').submit(function (event) {
event.preventDefault();
url = $(this).attr('action');
location.hash = url;
$.ajax ({
url: url,
method: 'POST',
data: $(this).serialize()
}).done(function (html) {
$('#content').html(html);
});
});
If i got an input i get a $_POST variable.
So i need to know which of the above forms has been submitted?
Thanks..
This will work:
var url;
$('form').submit(function (event) {
event.preventDefault();
url = $(this).attr('action');
location.hash = url;
var data = $(this).serialize();
data += "&formId=" + encodeURIComponent($(this).attr('id')); // if you have data in the form.
// do this if you don`t have data in the form:
// data = {formId: $(this).attr('id')};
$.ajax ({
url: url,
method: 'POST',
data: data
}).done(function (html) {
$('#content').html(html);
});
});
You can then get the forms Id from $_POST['formId']
Create a submit button with a name:
<form id="factory" action="bullets.php" method="post">
<button type="submit" value="factory" name="submit">Kopen</button>
</form>
This value is now posted:
if (!empty($_POST['submit']) && $_POST['submit'] == 'factory') {
}
By namespacing the input fields, you can easily identify which fields are from which form, and by extension which form was submitted.
<form id="factory" action="bullets.php" method="post">
<input type="text" name="form_1[my_input]">
<input type="submit" value="Kopen">
</form>
<form id="localFactory" action="bullets.php" method="post">
<input type="text" name="form_2[my_input]">
<input type="submit" value="Kopen">
</form>
Then it is as simple as:
if (isset($_POST['form_1'])) {
// This post variable is an array of each field.
}
If you want an html only solution, you could add a hidden input with the form id:
<form id="factory" action="bullets.php" method="post">
<input type="hidden" value="factory" name="formId"/>
<input type="submit" value="Kopen">
</form>
And then test it with:
if (isset($_POST['formId']) && $_POST['formId'] == 'factory') {
//Do what you want here
}
Related
i m trying to submit my form without any user interaction,i don't know how to do that, do you have a lead for?
Thanks by advance
Below my code
Html form
<form id="myform" method="post">
<div>
<input type="hidden" name="print_names" id="print_names" value="print_names" />
<input type="submit" name="loginBtn" id="loginBtn" value="test" />
</div>
</form>
Ajax part for submit
$(document).ready(function(){
$('#myform').submit(function(e) {
e.preventDefault();
$.ajax({
type: "POST",
url: 'form.php',
data: $(this).serialize(),
success: function(response)
{
Your form has no target action, in that caase you could use just:
document.forms.myform.submit();
When I submitted my external HIT on Mturk, the Submit button is not working. I would appreciate if someone could help me with this. The data gets stored in my server though. Here is my code:
<div id="instruction3" class="instructions" style="display:none">
survey questions here
Submit
</div>
function SaveData() {
(some code here)
d = {
"trialStruct": trialStruct,
"critStruct": critStruct
};
console.log(d)
SendToServer(curID, d);
}
<form action="https://workersandbox.mturk.com/mturk/externalSubmit" id="mturk_form" method="post" name="mturk_form">
<input id="assignmentId" name="assignmentId" type="hidden" value="" />
<p><input id="submitButton" type="submit" value="Submit" /></p>
</form>
function SendToServer(id, curData) {
$.ajax({
type : "POST",
url : "https://xxxxxxxxxxxx/turk/save.php",
data : { json : JSON.stringify(curData) },
success : function(data) {
document.forms[0].submit();
}
});
}
Edited: the flow should be participants click on the submit button and the data gets stored and sent to the externalSubmit page. These are parts of the code from Mturk that I need to implement in my code and perhaps I am not doing it right.
<!-- HTML to handle creating the HIT form -->
<form action="https://workersandbox.mturk.com/mturk/externalSubmit" id="mturk_form" method="post" name="mturk_form">
<input id="assignmentId" name="assignmentId" type="hidden" value="" />
<!-- HTML to handle submitting the HIT -->
<p><input id="submitButton" type="submit" value="Submit" /></p>
</form>
You should call the SaveData() function inside the form tag
<form action="https://workersandbox.mturk.com/mturk/externalSubmit" id="mturk_form" method="post" name="mturk_form" onSubmit="SaveData()">
So I think you should try to move the SaveData function to the onSubmit value of the form. So you would be submitting the form and the data would get saved to the server. You have extra html code above but I think that is superfluous for what you are trying to do.
function SaveData() {
(some code here)
d = {
"trialStruct": trialStruct,
"critStruct": critStruct
};
console.log(d)
SendToServer(curID, d);
}
function SendToServer(id, curData) {
$.ajax({
type : "POST",
url : "https://xxxxxxxxxxxx/turk/save.php",
data : { json : JSON.stringify(curData) },
success : function(data) {
document.forms[0].submit();
}
});
}
<form action="https://workersandbox.mturk.com/mturk/externalSubmit" id="mturk_form" method="post" name="mturk_form" onSubmit="SaveData()">
<input id="assignmentId" name="assignmentId" type="hidden" value="" />
<p><input onclick="window.location.href = https://workersandbox.mturk.com/mturk/externalSubmit';"id="submitButton" type="submit" value="Submit" /></p>
</form>
Here is the working one, I have used https://postman-echo.com/post just to make sure it works.
function SaveData() {
var d = {
"trialStruct": trialStruct,
"critStruct": critStruct
};
console.log(d);
SendToServer(curID, d);
}
function SendToServer(id, curData) {
$.ajax({
type: "POST",
url: "https://postman-echo.com/post",
data: {
json: JSON.stringify(curData)
},
success: function(data) {
$("#mturk_form").submit();
}
});
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="instruction3" class="instructions" style="display:none">
Submit
</div>
<form action="https://postman-echo.com/post" id="mturk_form" method="post" name="mturk_form">
<input id="assignmentId" name="assignmentId" type="hidden" value="" />
<p><input id="submitButton" type="submit" value="Submit" /></p>
</form>
Hello guys i don't know to submit a form without page refresh.
This is my code:
My form:
<form method="POST" id="post123" action="" enctype="multipart/form-data">
<input class="form-control input-lg" name="book_name" id="inputLarge" type="text">
<input class="form-control" name="book_price" type="number">
<textarea class="form-control" id="exampleTextarea2"
name="book_description" maxlength="100" rows="7"></textarea>
<textarea class="form-control" name="book_content" id="exampleTextarea4" maxlength="200" rows="7"></textarea>
<input type="file" name="file" id="imgInp44" >
<button type="submit" name='action' id="sendbutton21" value="post1" class="btn btn-primary">
Submit
</button>
</form>
Javascript:
$("#sendbutton21").click(function() {
var url = "http://localhost:5000/magazin_insert"; // the script where you handle the form input.
$.ajax({
type: "POST",
url: url,
data: $("#post123").serialize(), // serializes the form's elements.
success: function(data)
{
alert(data); // show response from the php script.
}
});
return false; // avoid to execute the actual submit of the form.
});
How i can modify my code to submit along with text a file towards the mysql db?
Your first issue is in this line:
<button type="submit" name="action" ... />
You have two possibilities:
change type="submit" to type="button"
use event.preventDefault() in your $("#sendbutton21").click(function() {
After, your ajax call can be changed in order to include the file:
var form = document.getElementById('post123');
var formData = new FormData(form);
$.ajax({
url: url,
data: formData,
type: 'POST',
processData: false,
success: function(data) {
alert(data);
}
});
$("#sendbutton21").click(function() {
var url = "http://localhost:5000/magazin_insert"; // the script where you handle the form input.
// for testing...
url = 'https://api.github.com/repositories';
var form = document.getElementById('post123');
var formData = new FormData(form);
$.ajax({
url: url,
data: formData,
type: 'GET',
processData: false,
success: function (data) {
console.log(data[0].id);
}
});
});
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<script src="https://code.jquery.com/jquery-2.1.1.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
<form method="POST" id="post123" action="" enctype="multipart/form-data">
<input class="form-control input-lg" name="book_name" id="inputLarge" type="text">
<input class="form-control" name="book_price" type="number">
<textarea class="form-control" id="exampleTextarea2"
name="book_description" maxlength="100" rows="7"></textarea>
<textarea class="form-control" name="book_content" id="exampleTextarea4" maxlength="200" rows="7"></textarea>
<input type="file" name="file" id="imgInp44" >
<button type="button" name='action' id="sendbutton21" value="post1" class="btn btn-primary">
Submit
</button>
</form>
Try https://jquery-form.github.io/form/ plugin to easily send form via ajax sample code:
$('#post123').ajaxForm(function() {
// SUCCESS CALLBACK
});
To submit a form without refresh, use event.preventDefault().
$("#sendbutton21").click(function(event) {
event.preventDefault();
var url = "http://localhost:5000/magazin_insert"; // the script where you handle the form input.
$.ajax({
type: "POST",
url: url,
data: $("#post123").serialize(), // serializes the form's elements.
success: function(data)
{
alert(data); // show response from the php script.
}
});
return false; // avoid to execute the actual submit of the form.
});
Instead of
$("#sendbutton21").click(function() {
use
$("#post123").submit(function() {
I have trouble, my code doesn't work, because my server script side need a name from the submit button. I'm using Ajax method, and I'm using data: serialize, when I have Click on Submit, it doesn't work. Here is my JavaScript code:
$(function(){
$('#buy_product').submit(function(e){
e.preventDefault();
var submitData = $('#buy_product').serialize();
submitData.push({ name: this.name, value: this.value });
$('.loading').html('{{HTML::image('img/ajax-loader.gif')}}');
$.ajax({
type: "POST",
data: submitData,
url: "{{URL::base()}}/products/view/{{$products->id}}/{{Str::slug($products->name_product, '_')}}",
success: function(msg){
$('#qty').val('');
$('.loading').html(msg);
}
});
});
});
If you have clue, please tell me, I'll be glad.
My button is like this:
<input name="update" id="update" type="submit" value="update">
<input name="empty" id="empty" type="submit" value="empty">
I believe you can't do that but you can use a hidden field
<input type="hidden" name="any_name" value="any_value" />
Try this:
Java Script Code:
<script type='text/javascript'>
$(function(){
$('#form').submit(function(e){
e.preventDefault();
var submitData = $('#form').serialize();
var btnName = $('#submit').attr('name');
var btnVal = $('#submit').val();
var btn = '&'+btnName+'='+btnVal;
submitData += btn;
alert(submitData);
});
});
</script>
HTML Code:
<form action="" id="form" type='post'>
<input type="text" name="name" id="name" value='Scott'/>
<input type="submit" name="submit" id="submit" value='POST'/>
</form>
or use:
var submitData = $('#buy_product').serialize();
submitData += '&btnName=' + $('#your_submitbtn_id').attr('name') + '&btnValue='+ $('##your_submitbtn_id').attr('value');
Are there any jQuery/AJAX functions that when a form (or anything for that matter) is displayed, upon pressing a button the div containing the original form is replaced by a new form? Essentially, a multi-part form without having to reload the page.
Can I use something like this?
$('form#myForm').submit(function(event) {
event.preventDefault();
$.ajax({
type: '',
url: '',
data: $(this).serialize(),
success: function(response) {
$('#divName').html(response);
//somehow repopulate div with a second form?
}
})
return false;
});
I've used this before for adding items to a list, but I've never used it to totally repopulate it with different content. How can I direct it to the second form?
edit - I got it to work, but only when I write '#form2' for the replacement. I alerted the response and I get {"formToShow":"show2"}. I tried doing response.formToShow but it's undefined.
<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js" ></script>
</head>
<div id="divName">
<form method="POST" action = "#" id="form1">
<input type="textbox" name="textbox1" value="1"/>
<input type="submit" name="submit1"/>
</form>
<form method="POST" action = "#" id="form2" style="display: none">
<input type="textbox" name="textbox2" value="2"/>
<input type="submit" name="submit2"/>
</form>
</div>
<script>
$('form#form1').submit(function(event) {
event.preventDefault();
$.ajax({
type: 'JSON',
url: 'receiving.php',
data: $(this).serialize(),
success: function(response) {
$('#form1').hide(); //hides
//$('#form2').show(); //this will show
$(response.formToShow).show(); //this does not display form 2
}
})
return false;
});
</script>
Here is receiving.php. When I view this page {"formToShow":"show2"} is displayed
<?php
echo json_encode(array("formToShow" => "#form2"));
?>
Check the JQuery Load Function
This is personal preference, but I'd never send HTML through the response and display it like that, what I'd do is:
Send a JSON array back from the server, such as { formToShow: "#form1" }
Then you can simply do this:
success: function(response) {
$('form').hide();
$(response.formToShow).show();
}
Obviously, using this method, you'd also have to have the second form in your markup like this:
<form method="POST" action = "#" id="form2" style="display: none">
<input type="textbox" name="textbox2"/>
<input type="submit" name="submit2"/>
</form>
You'd also have to change (to pickup the array):
$.ajax({
type: 'JSON'
try this
$('form#myForm').submit(function(event) {
event.preventDefault();
$.ajax({
type: '',
url: '',
data: $(this).serialize(),
success: function(response) {
$('#divName').html(response);
$('#form1').hide();
$('#form2').show();
}
})
return false;
});
<form method="POST" action = "#" id="form1">
<input type="textbox" name="textbox1"/>
<input type="submit" name="submit1"/>
</form>
<form method="POST" action = "#" id="form2" style="dispay:none;">
<input type="textbox" name="textbox2"/>
<input type="submit" name="submit2"/>
</form>