I want to send POST data without submit button, when one of the checkboxes are ticked - USING AJAX.
I have this code, but I think, that when I tick checkbox, I should get an alert with posted data (to know, that its working), but I dont get any reaction at all. I've added jquery so it's some kind of error in code function..
<form method="post" name="form1">
<input type="checkbox" name="checkboxList[]" value="1">value1</input>
<input type="checkbox" name="checkboxList[]" value="2">value2</input>
<input type="checkbox" name="checkboxList[]" value="3">value3</input>
</form>
<script>
jQuery(document).ready(function($){
// jQuery code is in here
$("input[type=checkbox]").on("click", function() {
var data = $(this).val();
console.log(data); // 1, 2 or 3 (value)
$.ajax({
url: "<?php echo JURI::getInstance(); ?>", // URL should be correct!
type: "POST",
async: true,
cache: false,
data: {data: data},
dataType: 'text',
success: function(data){
alert(data);
}
});
});
});
</script>
PROBLEM: Getting HTML content of page, not value on success method
Change your Html as below 1. remove </input> and change 2. change form name to id
<form method="post" id="form1">
<input type="checkbox" name="checkboxList[]" value="1">value1
<input type="checkbox" name="checkboxList[]" value="2">value2
<input type="checkbox" name="checkboxList[]" value="3">value3
</form>
Change in JS
var data = $("#form1").serialize();
Your form was not selecting and in post there was blank data
Try below code.
<form method="post" name="form1">
<input type="checkbox" name="checkboxList[]" value="1">value1</input>
<input type="checkbox" name="checkboxList[]" value="2">value2</input>
<input type="checkbox" name="checkboxList[]" value="3">value3</input>
</form>
<script>
$("input[type=checkbox]").on("click", function() {
var data = $("form").serialize();
$.ajax({
url: "../../x.php", // URL should be correct!
type: "POST",
async: true,
cache: false,
data: {data: data},
dataType: 'json',
success: function(response_data){
alert(response_data);
}
});
});
</script>
Related
I am struggling to have multiple AJAX submitted forms on the same php page. I can get it to work with one, but when I add more, it just uses the first form all the time.
I have tried to make it unique but not sure exactly how to do this.
My code is below, if someone could please assist that would be fantastic.
I have looked at similar questions and answers on this site but can't seem to fathom how to make it work for mine.
The form should submit when the checkbox is checked or unchecked.
Thanks very much in advance for all your help.
Martyn.
HTML form
<form id="search_form" method="post">
<input readonly type="text" name="id" value=<?php echo $row1[0];?>>
<input readonly type="text" name="student" value=<?php echo $row1[1];?>>
<input readonly type="text" name="addedby" value=<?php echo $_SESSION['username'];?>>
<input readonly type="text" name="register" value="monday_morning">
<input <?php if($row1[4] == "Yes"){echo "checked";}?> type="checkbox" name="box" value="Yes">
</form>
Script
<script type="text/javascript">
$(document).on("change", '[type="checkbox"]', function () {
var url = "/register_update.php";
$.ajax({
dataType: "html",
type: "POST",
url: url,
data: $("#search_form").serialize(),
success: function (data) {
}
});
return false;
});
</script>
Try this
<script type="text/javascript">
$(document).on("change", '[type="checkbox"]', function (event) {
var url = "/register_update.php";
$.ajax({
dataType: "html",
type: "POST",
url: url,
data: $($(event.target).parent()).serialize(),
success: function (data) {
}
});
return false;
});
</script>
event.target returns the clicked checkbox.
parent() function returns the form you need to submit. This way you don't need to have unique form id.
I am trying to submit data to the database using AJAX. I have one array and I have to pass the value of the array to PHP using AJAX to display all the related records.
<form id="search-form" method="POST">
<input value="4869" name="compare_id[]" type="hidden">
<input value="4884" name="compare_id[]" type="hidden">
<input value="5010" name="compare_id[]" type="hidden">
<input type="button" id="search-button" name="search-button" value="search">
</form>
<div id="response"></div>
AJAX
<script>
$(document).ready(function(){
$('#search-button').click(function(){
$.ajax( {
type: 'POST',
url: 'response.php',
data: $('#search-form').serialize(),
dataType: 'json',
success: function(response) {
$('#response').html(response);
//alert(response);
}
});
});
});
</script>
PHP
$sql='SELECT Name, Email FROM request WHERE Id IN (' .( is_array( $_POST['compare_id'] ) ? implode( ',', $_POST['compare_id']) : $_POST['compare_id'] ).')';
$records = array();
$query=$conn->query($sql);
if ($query->num_rows > 0) {
while($row=$query->fetch_assoc()){
$records[]=$row;
}
}
echo json_encode($records);exit();
HTML
<form id="search-form" method="POST">
<input value="4869" name="compare_id[]" type="hidden">
<input value="4884" name="compare_id[]" type="hidden">
<input value="5010" name="compare_id[]" type="hidden">
<input type="button" id="search-button" name="search-button" value="search">
</form>
<div id="response"></div>
JS
<script>
$(document).ready(function(){
$('#search-button').click(function(){
$.ajax( {
type: 'POST',
url: 'response.php',
data: $('#search-form').serialize(),
dataType: 'json',
success: function(response) {
$('#response').html(response);
}
});
});
});
</script>
PHP
var_dump($_POST['compare_id']);
// it is already an array of ids. You can do whatever you want with it.
change your script as below. Your output is in array so you cant add it in div directly
<script>
$(document).ready(function(){
$('#search-button').click(function(){
$.ajax( {
type: 'POST',
url: 'action.php',
data: $('#search-form').serialize(),
dataType: 'json',
success: function(response) {
$('#response').html();
for(data in response) //loop over your data
{
$('#response').append(response[data].Email); //add email
}
//alert(response);
}
});
});
});
</script>
There are errors in your code. A good way to debug this is to print_r your POST value in your php script.
First $_POST["All"] does not exist. It is all. (php)
Second, you send a GET request not a POST one. (jQuery)
Third, format your date into json. A good way to do this is to create a variable right after compare_id.push, it's more readable, as so :
var json_data = {"my_array" : [1,2, "bonjour", 4]};
Your problem is mostly related to "how to debug". I think you should print what's happening along the way to figure out what's happening.
HTML
<input type="checkbox" name=options[cid]" value='1'
onChange="chkdeptCount(this.value)" class="test">
<input type="checkbox" name=options[cid]" value='2'
onChange="chkdeptCount(this.value)" class="test">
jquery:
function chkdeptCount(val){
$.ajax({ url: '../ajax/AjaxCall.php',
data: {Action:'IMPLODEARRAY',arrVal: val},
type: 'post',
success: function(output) {
alert(output);
$('.result').html(output);
}
});
}
PHP:
if($_POST['Action']=='IMPLODEARRAY'){
$arr_val[] = $_POST['arrVal'];
print_r($arr_val);
}
When I run this code does not return array value. It returns a single value WHY?
Note that, you are missing the quote here name=options[cid]".
And you are using onChange="chkdeptCount(this.value)" event with current value this, this will only return one value at once.
This is very basic Example:
HTML:
<form method="post" id="formID" action="">
<input type="hidden" name="Action" value="IMPLODEARRAY">
<input type="checkbox" name="options[cid]" value='1' class="test">
<input type="checkbox" name="options[cid]" value='2' class="test">
<input type="submit" name="submit" value="Submit" id="SubmitButton">
</form>
AJAX:
<script type="text/javascript">
$(document).ready(function(){
$("#SubmitButton").click(function(){ // when submit button press
var data = $("#formID").serialize(); // get all form input in serialize()
$.ajax({
url: YourURL, // add your url here
type: "POST", // your method
data: data, // your form data
dataType: "json", // you can use json/html type
success: function(response) {
console.log(response); // your response
},
beforeSend: function()
{
// if you want to display any loading message
}
}); // JQUERY Native Ajax End
return false;
});
});
</script>
PHP:
<?php
if(count($_POST) > 0){ // if you have some value in AJAX request
if($_POST['Action'] == 'IMPLODEARRAY'){ // your condition
print_r($_POST['options']); // get all checkbox value.
}
}
?>
Few more example will help you to understand, you can you use: Submitting HTML form using Jquery AJAX |
jQuery AJAX submit form
I'm generating tables of buttons with php
echo ' <td">
<form action="test.php" method="POST">
<input type="hidden" id="node" name="node" value="'.$fnode->{'name'}.'">
<input type="hidden" id="service" name="service" value="'.$flavor.'">
<input type="hidden" id="running" name="running" value="false">
<input type="submit" value="OFF" class="button">
</form>
</td>';
I want to send the values without reloading via jquery ajax and I'm using this code for it:
$(".button").click(function() {
$('.error').hide();
var dataString = 'node='+ document.getElementById('node').value + '&service=' + document.getElementById('service').value + '&running=' + document.getElementById('running').value;
$.ajax({
type: "POST",
url: "test.php",
data: dataString,
success: function() {
alert ("Success");
}
});
return false;
});
Code works so far - it just always sends the data from the first form. What is the best way to distinguish between all the buttons. I could use a counter in the form, but how would I exactly write the js "ifs".
Is there a more elegant way to do this. Number of forms is dynamic.
You can grab the parent form of the button clicked easily enough, but youll also probably want to have a unique ID on the form for other things. Also you need to either remove the ids on the inputs or make them unique.
echo ' <td">
<form action="test.php" method="POST" id="form_node_' . $fnode->{'name'} . '>
<input type="hidden" name="node" value="'.$fnode->{'name'}.'">
<input type="hidden" name="service" value="'.$flavor.'">
<input type="hidden" name="running" value="false">
<input type="submit" value="OFF" class="button">
</form>
</td>';
$(".button").click(function(e) {
e.preventDefault();
$('.error').hide();
var $form = $(this).closest('form'), // the closest parent form
dataString = $form.closest('form').serialize(); // serialize the values instead of manually encoding
$.ajax({
type: "POST",
url: "test.php",
data: dataString,
success: function() {
alert ("Success submitting form ID " + $form.attr('id'));
// you can now modify the form you submitted
}
});
return false;
});
The best way is to use unique IDs for form elements. Another way is to set classes to multiple elements with the same name.
However, the following approach is much preferable:
$("form").on("submit", function() {
$.ajax({
type: "POST",
url: "test.php",
data: $(this).serialize(),
success: function() {
alert ("Success");
}
});
return false;
});
(But anyway don't forget to remove duplicating id attributes from the form elements.)
You can give each submit buttons an id:
<input id="button-1" type="submit" value="OFF" class="button">
and then trigger the event on click of a specific button:
$("#button-1").click(function() { ... });
Start learning Ajax (and jQuery), I encounter the issue with post form data, which I could not resolve myself.
First, here is the simple example where I collect data, post it, and displayed it on a new page:
<!-- index.html -->
<form id="myForm" action="test.php" method="post">
<input type="checkbox" id="ch1" name="Ch1" value="1">Check Box 1 <br />
<input type="checkbox" id="ch2" name="Ch2" value="2">Check Box 2 <br />
<input type="submit" id="myButton" />
</form>
<?php
// test.php
print_r($_POST);
?>
Then I tried with Ajax and jQuery. I slightly modified the form pasted above:
<!-- index.html -->
<form id="myForm">
<input type="checkbox" id="ch1" name="Ch1" value="1">Check Box 1 <br />
<input type="checkbox" id="ch2" name="Ch2" value="2">Check Box 2 <br />
<input type="button" id="myButton" value="Submit"/>
</form>
And here is the jQuery function:
// script.js
$(document).ready(function () {
$("#myButton").click(function () {
$.ajax({
cache: false,
type: 'POST',
url: 'test.php',
data: $("#myForm").serialize()
});
});
});
My question is, how to reimplement the jQuery function to echo the result in the test.php tab? I crawled the many resources, but i didn't find any solution.
Thanks in advance for any pointer.
Best, Andrej
You're going to need a success callback, which I'll assume will be some HTML...
$(document).ready(function(){
$("#myButton").click(function() {
$.ajax({
cache: false,
type: 'POST',
url: 'test.php',
data: $("#myForm").serialize(),
success: function(response) {
$("#someElement").html(response);
}
});
});
});
Try this.
$(document).ready(function(){
$("#myButton").click(function() {
$.ajax({
cache: false,
type: 'POST',
url: 'test.php',
data: $("#myForm").serialize(),
success: function(data){
alert(data);
}
});
});
});
I believe you're looking for the success in the .ajax options parameter.
$.ajax({
...
success: function(d){
alert(d);
}
});