issues are still there...pls help
I am unable to load external file while using AJAX jquery. I want to use Jquery ajax to pop up form then validate, enter data in mysql. but starting from a simple ajax function. kindly let me know where i am going wrong
<link rel="stylesheet" type="text/css" media="all" href="test_style.css">
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script>
$(document).ready(function(){
$("#ajax-contact-form").submit(function(){
var str = $(this).serialize();
$.ajax(
{
type: "POST",
url:"contact.php",
data: str,
success:function(result)
{
$("#div1").html(result);
}
});
});
});
</script>
</head>
<body>
<div id="contact_form">
<form id="ajax-contact-form" name="contact" action="">
<fieldset>
<label for="name" id="name_label">Name</label>
<input type="text" name="name" id="name" size="30" value="" class="text-input" />
<label class="error" for="name" id="name_error">This field is required.</label>
<INPUT class="button" type="submit" name="submit" value="Send Message">
</fieldset>
</form>
</div>
</body>
</html>
and contact.php file is
<?php
echo "Hello";
?>
You need to return false; to prevent the form from submitting and refreshing the page and check if your $("#div1") is missing.
$(document).ready(function(){
$("#ajax-contact-form").submit(function(){
var str = $(this).serialize();
$.ajax(
{
type: "POST",
url:"contact.php",
data: str,
success:function(result)
{
$("#div1").html(result);
}
});
return false;
});
});
Simple. Since you are posting your form through ajax, you must prevent the default form submit by returning a false inside the submit method. Below is the correct version:
<script>
$(document).ready(function(){
$("#ajax-contact-form").submit(function(){
var str = $(this).serialize();
$.ajax({
type: "POST",
url:"contact.php",
data: str,
success:function(result) {
$("#div1").html(result);
}
});
return false;
});
});
</script>
You can use a more simple form of post request as follows:
$.post("url",{var1: value1, var2: value2},function(data,status){
if(status=='success')
alert(data);
});
the second argument you can pass as many using this post request. The first argument url, if ofcourse relative to the document in which this js is loaded or you can give the exact url on the server.
According to your php file, data=='Hello'.
Similar is the procedure for any GET request also.
Make sure you are missing the div1
please use
<div id="div1"><div>
Related
i have a form and i wanna use jQuery with it. but when i submit it it just doesnt use the code at all.
<form id="form">
<label>Name</label><br />
<input type="text" name="name"><br /><br />
<label>Message</label><br />
<textarea name="message"></textarea><br /><br />
<button type="submit">Submit form</button>
</form>
<script src="https://code.jquery.com/jquery-3.6.3.js" integrity="sha256-nQLuAZGRRcILA+6dMBOvcRh5Pe310sBpanc6+QBmyVM=" crossorigin="anonymous">
//have to use jQuery to get the form data
jQuery(document).ready(function($){
$('#form').submit(function(e){
e.preventDefault();
alert('hello');
var data = $(this).serialize();
$.ajax({
url: 'http://localhost:8888/wp-json/v1/contact_form/submit',
type: 'POST',
data: data,
success: function(response){
console.log(response);
}
});
});
});
</script>
i used a alert to see if it works but it just does nothing.
when i submit i get to a blank page. so the e.preventDefaults also doesnt work.
i tried using copilot but it didnt do anything usefull
You must declare your scripts in different tags.
<script src="https://code.jquery.com/jquery-3.6.3.js" integrity="sha256-nQLuAZGRRcILA+6dMBOvcRh5Pe310sBpanc6+QBmyVM=" crossorigin="anonymous"></script>
<script>
//have to use jQuery to get the form data
jQuery(document).ready(function($){
$('#form').submit(function(e){
e.preventDefault();
alert('hello');
var data = $(this).serialize();
$.ajax({
url: 'http://localhost:8888/wp-json/v1/contact_form/submit',
type: 'POST',
data: data,
success: function(response){
console.log(response);
}
});
});
});
</script>
I have a file called try.php where there is code below having all javascript, PHP and html file in itself.
<?php
if(isset($_POST["submit"])){
echo "hello";
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Submit Form Using AJAX and jQuery</title>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
</head>
<body>
<form method="POST" id="myForm">
<input name="name" type="text" placeholder="Your Name">
<input name="email" type="text" placeholder="email">
<input name="submit" type="submit" value="Submit">
<div id="display"></div>
</form>
<script>
$(document).ready(function(){
$("#myForm").submit(function(event) {
event.preventDefault(); //prevent default action
window.history.back();
var form_Data = $(this).serialize();
$.ajax({
type: "POST",
url: "try.php",
data: form_data,
cache: false,
success:function(response){
alert(response);
}
});
});
});
</script>
</body>
</html>
The target of above code is just to submit form without reloading the page simply using AJAX and the form data should be handled by php here just echo "hello". The above code works fine it submits and php handles all properly but page is reloading. What should be the change in code?
Try this as javascript code
$(document).ready(function(){
$("#myForm").click(function(event) {
event.preventDefault(); //prevent default action
var form_Data = $(this).serialize();
$.ajax({
type: "POST",
url: "try.php",
data: form_Data,
cache: false,
success:function(){
alert("hello");
}
});
});
});
Please I am new to jQuery so i just copied the code:
<div id="container">
<input type="text" id="name" placeholder="Type here and press Enter">
</div>
<div id="result"></div>
<script type="text/javascript" src="jquery.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$('#name').focus();
$('#name').keypress(function(event) {
var key = (event.keyCode ? event.keyCode : event.which);
if (key == 13) {
var info = $('#name').val();
$.ajax({
method: "POST",
url: "action.php",
data: {name: info},
success: function(status) {
$('#result').append(status);
$('#name').val('');
}
});
};
});
});
</script>
And here is the php code:
<?php
if (isset($_POST['name'])) {
echo '<h1>'.$_POST['name'];
}
?>
Its Working perfectly but now i want to have more than one input field like this:
<input type="text" id="name" >
<input type="text" id="job">
but i don't know how to run the jQuery code for the 2 input fields so that it can transfer them to the php page. Please i need help
You can pass multiple values using data param of ajax request like this.
$.ajax({
method: "POST",
url: "action.php",
data: {
name: $('#name').val(),
job: $('#job').val()
},
success: function(status) {
$('#result').append(status);
$('#name, #job').val(''); // Reset value of both fields
}
});
You need to change your code with some addition in html and JS.
Wrap your inputs in form tag. and add a preventDefault on submit.
Use jQuery .serialize() method
and event.preventDefault()
event.preventDefault() : If this method is called, the default
action of the event will not be triggered. (it will prevent page
reload / redirection) to any page.
.serialize() : Encode a set of form elements as a string for
submission.
serialized string output will be like key=value pair with & separated. :
name=john&job=developer.....
HTML
<form id="myform">
<input type="text" id="name" placeholder="Type here and press submit">
<input type="text" id="job" placeholder="Type here and press submit">
<input type="submit" name="submit" value="Submit Form">
</form>
JS
$(document).ready(function() {
$('#myform').submit(function(event) {
event.preventDefault();
var serialized = $('#myform').serialize();
$.ajax({
method: "POST",
url: "action.php",
data: serialized,
success: function(status) {
$('#result').append(status);
$('#myform').reset();
}
});
});
});
I've been racking my brains for days looking at examples and trying out different things to try and get my form to submit with Ajax without a page refresh. And Its not even sending the data now.. I don't know what I'm doing wrong..Can someone run through my ajax and form please.
Toid is the users id and newmsg is the text in which the user submits. The two values get sent to the insert.php page.
I would really appreate the help. I'm new to Ajax, and I look at some of it and don't have a clue. If I finally got it working, It may help me realize what I've done wrong. I am looking up tutorials and watching videos..but it can be very time consuming for something that would be simple to someone in the know on here. It maybe that I've got the wrong idea on the ajax and it makes no sense at all, sorry about that.
<script type="text/javascript">
$(document).ready(function(){
$("form#myform").submit(function() {
homestatus()
event.preventDefault();
var toid = $("#toid").attr("toid");
var content = $("#newmsg").attr("content");
$.ajax({
type: "POST",
url: "insert.php",
data: "toid="+content+"&newmsg="+ newmsg,
success: function(){
}
});
});
return false;
});
</script>
<form id="myform" method="POST" class="form_statusinput">
<input type="hidden" name="toid" id="toid" value="<?php echo $user1_id ?>">
<input class="input" name="newmsg" id="newmsg" placeholder="Say something" autocomplete="off">
<div id="button_block">
<input type="submit" id="button" value="Feed" onsubmit="homestatus(); return false" >
</div>
</form>
INSERT.PHP
$user1_id=$_SESSION['id'];
if(isset($_POST['toid'])){
if($_POST['toid']==""){$_POST['toid']=$_SESSION['id'];}
if(isset($_POST['newmsg'])&isset($_POST['toid'])){
if($_POST['toid']==$_SESSION['id']){
rawfeeds_user_core::create_streamitem("1",$_SESSION['id'],$_POST['newmsg'],"1",$_POST['toid']);
}else{
rawfeeds_user_core::create_streamitem("3",$_SESSION['id'],$_POST['newmsg'],"1",$_POST['toid']);
Try using firebug to identify bugs in your code. It's a really good companion for developing javascript. Nearly all of your bugs led to error messages in the firebug console.
You had several errors in your code, here is the corrected version:
$(document).ready(function(){
$("form#myform").submit(function(event) {
event.preventDefault();
var toid = $("#toid").val();
var newmsg = $("#newmsg").val();
$.ajax({
type: "POST",
url: "insert.php",
data: "toid=" + content + "&newmsg=" + newmsg,
success: function(){alert('success');}
});
});
});
And here the corrected html:
<form id="myform" method="POST" class="form_statusinput">
<input type="hidden" name="toid" id="toid" value="<?php echo $user1_id; ?>">
<input class="input" name="newmsg" id="newmsg" placeholder="Say something" autocomplete="off">
<div id="button_block">
<input type="submit" id="button" value="Feed">
</div>
</form>
Actually onsubmit event has to be used with form so instead of
<input type="submit" id="button" value="Feed" onsubmit="homestatus(); return false" >
it could be
<form id="myform" method="POST" class="form_statusinput" onsubmit="homestatus();">
and return the true or false from the function/handler, i.e.
function homestatus()
{
//...
if(condition==true) return true;
else return false;
}
Since you are using jQuery it's better to use as follows
$("form#myform").on('submit', function(event){
event.preventDefault();
var toid = $("#toid").val(); // get value
var content = $("#newmsg").val(); // get value
$.ajax({
type: "POST",
url: "insert.php",
data: "toid=" + toid + "&newmsg=" + content,
success: function(data){
// do something with data
}
});
});
In this case your form should be as follows
<form id="myform" method="POST" class="form_statusinput">
...
</form>
and input fields should have a valid type and value attribute, Html form and Input.
I think you should read more about jQuery.
Reference : jQuery val and jQuery Ajax.
change the form to this
<form id="myform" ... onsubmit="homestatus(); return false">
you don't need the onsubmit attribute on the submit button, but on the form element instead
homestatus might be out of scope
function homestatus () {
var toid = $("#toid").attr("toid");
var content = $("#newmsg").attr("content");
$.ajax({
type: "POST",
url: "insert.php",
data: "toid="+content+"&newmsg="+ newmsg,
success: function(){
}
});
}
This isn't tested, but try this (I annotated some stuff using comments)
<script type="text/javascript">
$(document).ready(function(){
$("form#myform").submit(function(event) {
// not sure what this does, so let's take it out of the equation for now, it may be causing errors
//homestatus()
// needed to declare event as a param to the callback function
event.preventDefault();
// I think you want the value of these fields
var toid = $("#toid").val();
var content = $("#newmsg").val();
$.ajax({
type: "POST",
url: "insert.php",
data: "toid="+toid +"&newmsg="+ content,
success: function(){
}
});
return false;
});
});
</script>
<form id="myform" method="POST" class="form_statusinput">
<input type="hidden" name="toid" id="toid" value="<?php echo $user1_id ?>">
<input class="input" name="newmsg" id="newmsg" placeholder="Say something" autocomplete="off">
<div id="button_block">
<input type="submit" id="button" value="Feed" / >
</div>
</form>
It's a lot easier to let .serialize() do the work of serializing the form data.
The submit handler also needs event as a formal parameter, otherwise an error will be thrown (event will be undefined).
With a few other changes, here is the whole thing:
<script type="text/javascript">
$(document).ready(function(){
$("form#myform").submit(function(event) {
event.preventDefault();
homestatus();
var formData = $(this).serialize();
$.ajax({
type: "POST",
url: "insert.php",
data: formData,
success: function(data) {
//...
}
});
});
});
</script>
<form id="myform" class="form_statusinput">
<input type="hidden" name="toid" id="toid" value="<?php echo $user1_id ?>">
<input class="input" name="newmsg" id="newmsg" placeholder="Say something" autocomplete="off">
<div id="button_block">
<input type="submit" id="button" value="Feed" >
</div>
</form>
Unless you are omitting some of your code, the problem is this line:
homestatus()
You never defined this function, so the submit throws an error.
You may want to take a look at jQuery (www.jquery.com) or another js framework.
Such frameworks do most of the stuff you normally have to do by hand.
There are also a bunch of nice helper functions for sending form data or modifying html elements
I have an insert and load record (jQuery & PHP) script working fine without using AJAX. but after the AJAX call, insert (jQuery) doesn't work.
This is my code:-
$(".insert").live("click",function() {
var boxval = $("#content").val();
var dataString = 'content='+ boxval;
if(boxval==''){
alert("Please Enter Some Text");
}
else{
$.ajax({
type: "POST",
url: "demo.php",
data: dataString,
cache: false,
success: function(html){
$("table#update tbody").prepend(html);
$("table#update tbody tr").slideDown("slow");
document.getElementById('content').value='';
}
});
}
return false;
});
$(".load").live("click",function() {
$.ajax({
type: "POST",
url: "test.php",
success: function(msg){
$("#container").ajaxComplete(function(event, request, settings){
$("#container").html(msg);
});
}
});
});
});
Definitely recommend using your browsers dev tools to examine the exact request that is submitted and see if there is a problem there first.
You might also want to change the way you pass the dataString to the ajax request.
If your boxval has a "&" in it then you'll end up with an incorrectly formatted string. So, try initialising data instead as:
var data = {};
data.content = boxval;
This will ask jQuery to escape the values for you.
I'd be curious to see your form markup and your back-end PHP code; it may provide a clue.
Often I'll have a form variable called 'action', just to tell the PHP code what I want it to do (especially if that PHP script is a controller for many different actions on an object). Something like <input type="hidden" name="action" value="insert"/> or even multiple <input type="submit" name="action"/> buttons, each with a different value. In the PHP code I'll have something like:
switch ($_POST['action']) {
case 'insert':
// insert record and send HTML
break;
// other actions
}
If you've done something like this, perhaps the PHP is looking for the presence of a variable that doesn't exist.
Without being able to look at your code, I'd highly recommend the incredibly handy jQuery Form Plugin http://jquery.malsup.com/form/ . It allows you to turn a form into an AJAX form, formats your data properly, and doesn't forget the data from any of your form elements (except <input type="submit"/> buttons that weren't clicked on, which is the same behaviour that a non-AJAX form exhibits). It works just like the standard $.ajax() method.
I solved the problem
I replaced this code
$("#container").ajaxComplete(function(event, request, settings){
$("#container").html(msg);
});
with
$("#container").html(msg);
Thank you very much for your answers
<!--
To change this template, choose Tools | Templates
and open the template in the editor.
-->
<!DOCTYPE html>
<html>
<head>
<title></title>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.0/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$('.clicker').click(function(){
var fname = $('.fname').val();
var lname = $('.lname').val();
var message=$('.message').val();
$.ajax({
type:"POST",
url: "submit.php",
cache:false,
data: "fname="+fname+"&lname="+lname+"&message="+message,
success: function(data){
$(".result").empty();
$(".result").html(data);
}
});
return(false);
});
});
</script>
</head>
<body>
<div>Data Form</div>
<form id="form1" name="form1" method="post" action="">
<input name="fname" type="text" class="fname" size="20"/><br />
<input name="lname" type="text" class="lname" size="20"/><br />
<div class="result"><textarea name="message" rows="10" cols="50" class="message"> </textarea></div>
<input type="button" value="calculate" class="clicker" />
</form>
</body>
</html>
submit.php
<?php
$con=mysql_connect("localhost","root","");
mysql_select_db("ajaxdb",$con);
$fname=$_REQUEST['fname'];
$lname=$_REQUEST['lname'];
$message=$_REQUEST['message'];
$sql="insert into person(fname,lname,message) values('$fname','$lname','$message')";
mysql_query($sql) or die(mysql_error());
echo "The data has been submitted successfully.";
?>