Problems about keep in the page with ajax - php

I have a form. My objective is send and insert the values of the form to my database. Then
Clear the input of the form
Show the successful message.
Like this:
My problems:When I press the "save" button, I am redirected to another page.
This is the operating error:
and change the page
¿What is the problem on my code?
html
<html>
<head><title>Insert Data Into MySQL: jQuery + AJAX + PHP</title></head>
<body>
<form id="myForm" action="userInfo.php" method="post">
Name: <input type="text" name="name" /><br />
Age : <input type="text" name="age" /><br />
<button type="submit" id="sub">Save</button>
</form>
<span id="result"></span>
<script src="script/jquery-1.8.1.min.js" type="text/javascript"></script>
<script src="script/elscript.js" type="text/javascript"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.8.1/jquery.min.js">
</script>
</body>
</html
This is my script
$("#myForm").submit( function(e) {
// Prevent the normal form submission event
e.preventDefault();
// Create an object of the form
var form = $(this);
// Make an AJAX request
$.post(this.action, form.serializeArray(), function(info) {
// Clear the form
form[0].reset();
// Display message
$("#result").html(info);
});
});
Finally, my code to insert
<?php
$conn = mysql_connect('localhost', 'root', '');
$db = mysql_select_db('practicas');
$name = $_POST['name'];
$age = $_POST['age'];
if(mysql_query("INSERT INTO ajaxtabla VALUES('$name', '$age')"))
echo "Successfully Inserted";
else
echo "Insertion Failed";
?>
Thanks for help me

You can simplify the code by having just one event handler for when you submit the form. Additionally, you need to include e.preventDefault() which prevents the normal function of the submit button.
$("#myForm").submit( function(e) {
// Prevent the normal form submission event
e.preventDefault();
// Create an object of the form
var form = $(this);
// Make an AJAX request
$.post(this.action, form.serializeArray(), function(info) {
// Clear the form
form[0].reset();
// Display message
$("#result").html(info);
});
});
Give your button a type attribute as well to ensure your form submits upon clicking.
<button type="submit" id="sub">Save</button>
Demo
$(function() {
$("#myForm").submit(function(e) {
// Prevent the normal form submission event
e.preventDefault();
console.log('onsubmit event handler triggered');
// Create an object of the form
var form = $(this);
// Make an AJAX request
/*$.post(this.action, form.serializeArray(), function(info) {
// Clear the form
form[0].reset();
// Display message
$("#result").html(info);
});*/
// fake AJAX request
setTimeout(function(){
// Clear the form
form[0].reset();
// Display message
$("#result").html('Successfully Inserted');
}, 3000);
});
});
<form id="myForm" action="userInfo.php" method="post">
Name: <input type="text" name="name" /><br /> Age : <input type="text" name="age" /><br />
<button type="submit" id="sub">Save</button>
</form>
<span id="result"></span>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.8.1/jquery.min.js"></script>

In ajax you are posting the form with its action so it will be redirected confirm
If you don't want to redirect then send ajax request like this
$.ajax({
method:post,
url:"File name you want to send request"
}).done(function(response){
});

Related

jQuery form plugin , how to submit only visible fields

Using the jQuery form plugin, I just want to submit the visible fields (not the hidden ones ) of the form.
HTML:
<div class="result"></div>
<form id="myForm" action="comment.php" method="post">
Name: <input type="text" name="name" />
Comment: <textarea name="comment"></textarea>
<div style="display:none;">
<input type="text" value="" name="name_1" />
</div>
<input type="submit" value="Submit Comment" />
</form>
I cannot find a way to submit only the visible fields using any of the methods below:
ajaxForm:
// wait for the DOM to be loaded
$(document).ready(function() {
// bind 'myForm' and provide a simple callback function
$('#myForm').ajaxForm(function() {
alert("Thank you for your comment!");
});
});
ajaxSubmit:
$('#myForm').ajaxSubmit({
target: '.result',
success: function(response) {
alert("Thank you for your comment!");
}
});
There is another method formSerialize but found no way to use it with the 2 methods mentioned above (usable with $.ajax however).
How to submit only the visible fields using any of the two methods ?
$("#myForm").on("submit", function() {
var visibleData = $('#myForm input:visible,textarea:visible,select:visible').fieldSerialize();
$.post(this.action, visibleData, function(result) {
alert('Thank you for your comment!');
});
// this is needed to prevent a non-ajax submit
return false;
});

Submitting forms thru in PHP on jQuery

I have a registration form that is currently in a popup modal window coded in jQuery. I have a PHP submit button on the bottom and I have added jQuery code that stops the button from submitting. This is because it will stop my modal window from closing when I submit the page. My issue now is that submitting the form would be impossible. Is there a way to submit my form over all this crowded pop-ups and jQuery? Say is it possible to use AJAX or jQuery to submit the form and allow my PHP to handle it.
Since I am writing in PHP, there is quite a bit of server side validation going on, so the point of this is to allow my viewers to fix their validation mistakes before the modal window closes.
Here is my jQuery, I didnt bother to mess with that anymore as it does what I need.
$(document).ready(function() {
$('a.modal-window').click(function() {
//Getting the variable's value from a link
var loginBox = $(this).attr('href');
$(loginBox).fadeIn(300);
var popMargTop = ($(loginBox).height() + 24) / 2;
var popMargLeft = ($(loginBox).width() + 24) / 2;
$(loginBox).css({
'margin-top' : -popMargTop,
'margin-left' : -popMargLeft
});
// Add the mask to body
$('body').append('<div id="mask"></div>');
$('#mask').fadeIn(300);
return false;
});
// When clicking on the button close or the mask layer the popup closed
$('a.close, #mask').live('click', function() {
$('#mask , .login-popup').fadeOut(300 , function() {
$('#mask').remove();
});
return false;
});
});
Here is the code I used to stop the form from submitting:
$(function () {
$(':submit').click(function (event) {
event.preventDefault();
// submit the form dynamically
});
});
and below is my form, it might not matter although its there for the viewing.
<form method="post" id="loginform" action="<?php echo $_SERVER['PHP_SELF']?>">
<table style="color: white;">
<tr><th style="float:left;">Register a new account with us.</th></tr>
<tr><td>Username</td><td><input type="text" name="txtUser"/></td></tr>
<tr><td>Password</td><td><input type="text" name="txtPass"/></td></tr>
<tr><td>Email</td><td><input type="text" name="txtEmail"/></td></tr>
<tr><td>Confirm Email</td><td><input type="text" name="txtEmail2"/></td></tr>
<tr><td>First Name</td><td><input type="text" name="txtFname"/></td></tr>
<tr><td>Last Name</td><td><input type="text" name="txtLname"/></td></tr>
<tr><td>Address</td><td><input type="text" name="txtAddress"/></td></tr>
<tr><td>City</td><td><input type="text" name="txtCity"/></td></tr>
<tr><td>Postal Code</td><td><input type="text" name="txtPostal"/></td></tr>
<tr><td>Birth Year</td><td><input type="text" name="txtBirth"/></td></tr>
<tr><td>Gender</td><td><input type="radio" id="radio-1-1" name="radicalSex" class="regular-radio" value="m" selected="true" /><label for="radio-1-1"></label> Male</td></tr>
<tr><td></td><td><input type="radio" id="radio-1-2" name="radicalSex" class="regular-radio" value="f"/><label for="radio-1-2"></label> Female</td></tr>
<tr><td colspan='2' style="color: #FF6600;float:left;font-size:70%;"><?php echo $Error;?></td></tr>
<tr><td colspan="2"><input type="submit" name="btnRegister" ID="btnBlueTemp" value="Submit Registration" /></td></tr>
<tr><td colspan='2' style="float:left; font-size:70%;">Address information is optional</td></tr>
</table>
</form>
Let me give you an example of how you can do that .
<html>
<head>
<title></title>
<script src="js/jquery-1.7.2.min.js" type="text/javascript"></script>
<script>
$(document).ready(function(){
function validate(name, addr){
if(name=="") {
alert('Name is Blank');
return false;
} else if(addr=="") {
alert('Address is Blank');
return false;
} else {
return true;
}
}
$("#save").click(function(event){
event.preventDefault();
var name = $("#name").val();
var addr = $("#addr").val();
if(validate(name,addr)){
$.ajax({
type:'POST',
data:'name='+name+'&addr='+addr,
url:'test2.php',
success:function(data) {
alert(data);
}
})
}
});
});
</script>
</head>
<body>
<form name="frm" method="POST" action="">
<input type="text" name="name" id="name" value=""/><br>
<input type="text" name="addr" id="addr" value="" /><br>
<input type="submit" name="save" id="save" value="Save"/>
</form>
</body>
</html>
Now in test2.php You can do your php codes
<?php
if(isset($_POST['name'])) {
echo $_POST['name'];
}
?>
Hope this gives you an Idea.
You need to serialize the form data before posting it to PHP.
<script type="text/javascript">
var frm = $('#loginform');
frm.submit(function () {
$.ajax({
type: frm.attr('method'),
url: frm.attr('action'),
data: frm.serialize(),
success: function (data) {
alert('submitted');
}
});
return false;//stop actual form submit
});
</script>
Then, submit your form via ajax
Jquery AJAX
On AJAX URL on which the request is sent, you can write necessary codes for validation and return accordingly. For eg. if some one the form element doesn't meet the validation, you can throw the flag accordingly as json value.
Its possible, why not.
Once you have done all the input validation at client side, just submit the form...
$("#loginform").submit();
Then you will have your server do the rest of the validation.
If you want to stay in the page and show the validation output from server, the. You should submit using Ajax.
It will send your form data to server, then you can do server validation, and output any errors. You will get this in your Ajax complete handler, which you can use to show error messages to user.
To stop the form from reloading the page you needn't call any prevent methods as a simple script request would do the trick.
For instance,
$('#loginForm').submit(function() {
// Do the relevant tasks needed here, form is already prevented from being submitted
});
Check out this demo for more information on what I am referring to

How to have two buttons in a same form to do different actions in ajax?

I have a form, which take name from form and it sends to javascript codes and show in php by Ajax. these actions are done with clicking by submit button, I need to have another button, as review in my main page. how can I address to ajax that in process.php page have "if isset(submit)" or "if isset(review)"?
I need to do different sql action when each of buttons are clicked.
how can I add another button and be able to do different action on php part in process.php page?
<script type="text/javascript">
$(document).ready(function(){
$("#myform").validate({
debug: false,
submitHandler: function(form) {
$.post('process.php', $("#myform").serialize(), function(data) {
$('#results').html(data);
});
}
});
});
</script>
<body>
<form name="myform" id="myform" action="" method="POST">
<label for="name" id="name_label">Name</label>
<input type="text" name="name" id="name" size="30" value=""/>
<br>
<input type="submit" name="submit" value="Submit">
</form>
<div id="results"><div>
</body>
process.php:
<?php
print "<br>Your name is <b>".$_POST['name']."</b> ";
?>
You just need to add a button and an onclick handler for it.
Html:
<input type="button" id="review" value="Review"/>
Js:
$("#review").click(function(){
var myData = $("#myform").serialize() + "&review=review";
$.post('process.php', myData , function(data) {
$('#results').html(data);
});
}
);
Since you have set a variable review here, you can use it to know that is call has come by clicking the review button.
Bind the event handlers to the buttons' click events instead of the form's submit event.
Use the different event handler functions to add different pieces of extra data to the data object you pass to the ajax method.

submit form using Jquery Ajax Form Plugin and php?

this a simple example in how to submit form using the Jquery form plugins and retrieving data using html format
html Code
<html>
<head>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7/jquery.js"></script>
<script src="http://malsup.github.com/jquery.form.js"></script>
<script>
// prepare the form when the DOM is ready
$(document).ready(function() {
// bind form using ajaxForm
$('#htmlForm').ajaxForm({
// target identifies the element(s) to update with the server response
target: '#htmlExampleTarget',
// success identifies the function to invoke when the server response
// has been received; here we apply a fade-in effect to the new content
success: function() {
$('#htmlExampleTarget').fadeIn('slow');
}
});
});
</script>
</head>
<body>
<form id="htmlForm" action="post.php" method="post">
Message: <input type="text" name="message" value="Hello HTML" />
<input type="submit" value="Echo as HTML" />
</form>
<div id="htmlExampleTarget"></div>
</body>
</html>
PHP Code
<?php
echo '<div style="background-color:#ffa; padding:20px">' . $_POST['message'] . '</div>';
?>
this just work fine
what i need to know if what if i need to Serialize the form fields so how to pass this option through the JS function
also i want show a loading message while form processed
how should i do that too
thank you
To serailize and post that to a php page, you need only jQuery in your page. no other plugin needed
$("#htmlForm").submit(function(){
var serializedData= $("#htmlForm").serialize();
$.post("post.php", { dat: serializedData}, function(data) {
//do whatever with the response here
});
});
If you want to show a loading message, you can do that before you start the post call.
Assuming you have div with id "divProgress" present in your page
HTML
<div id="divProgress" style="display:none;"></div>
Script
$(function(){
$("#htmlForm").submit(function(){
$("#divProgress").html("Please wait...").fadeIn(400,function(){
var serializedData= $("#htmlForm").serialize();
$.post("post.php", { dat: serializedData},function(data) {
//do whatever with the response here
});
});
});
});
The answer posted by Shyju should work just fine. I think the 'dat' should be given in quotes.
$.post("post.php", { 'dat': serializedData},function(data) {
...
}
OR simply,
$.post("post.php", serializedData, function(data) {
...
}
and access the data using $_POST in PHP.
NOTE: Sorry, I have not tested the code, but it should work.
Phery library does this behind the scenes for you, just create the form with and it will submit your inputs in form automatically. http://phery-php-ajax.net/
<?php
Phery::instance()->set(array(
'remote-function' => function($data){
return PheryResponse::factory('#htmlExampleTarget')->fadeIn('slow');
}
))->process();
?>
<?php echo Phery::form_for('remote-function', 'post.php', array('id' => ''); ?> //outputs <form data-remote="remote-function">
Message: <input type="text" name="message" value="Hello HTML" />
<input type="submit" value="Echo as HTML" />
</form>
<div id="htmlExampleTarget"></div>
</body>
</html>

Prevent Default for Ajax created form

I am getting the form on click from a file called test.php which contains the following:
<form method="post" class="adminTM">
<input type="hidden" name="execID" value="<?=$_POST['exec']?>" />
<input type="hidden" name="fromTM" value="<?=$_POST['TM']?>" />
<input type="text" name="toTM" value="<?=$_POST['TM']?>" />
<input type="hidden" name="symbol" value="<?=$_POST['symbol']?>" />
<button class="submitTM">SUBMIT</button>
</form>
The javascript looks like so:
$(function(){
$('.adminTMClick').live('click', function(e){
$(this).data('TM', this.innerHTML);
$.post('test.php', $(this).data(), function(data){
$(data).dialog({
modal: true,
beforeClose: function(){
$(this).remove();
}
});
console.log($('.adminTM'));
console.log($('.submitTM'));
});
});
$('.submitTM').live('click', function(e){
//originally had .adminTM with submit which failed
e.preventdefault();
alert('i am here');
return false;
});
});
How do i make it so that the form DOES NOT do the default submit action when the submit button is clicked?
Here is a fiddle that demonstrates basically what I am doing (i had to change it a bit because of the way jsfiddle works): http://jsfiddle.net/maniator/tQVnV/show/
You should use the submit() event on the form instead of the click() on the submit, since pressing enter will still submit the form (bypassing the submit button).
This should properly prevent the form from doing the default submit:
$('.adminTM').live('submit', function(e) {
// execute custom code
console.log("submit event fired");
// prevent default submit
return false;
});
jsFiddle: http://jsfiddle.net/bjorn/tQVnV/11/

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