Ajax submit one page PHP - php

I have code with single page PHP method
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js">
</script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/bootstrap-datepicker/1.6.4/js/bootstrap-datepicker.js"></script>
<?php
if(isset($_POST['submit'])){
$date=$_POST['date'];
}else{
$date=date('Y-m-d');
}
?>
<form action="" method="post" id="formId">
<input type="text" name="date" id="datepicker">
<input type="submit" name="submit">
</form>
<?php echo $date; ?>
<script>
$(function() {
$('#datepicker').datepicker({
autoclose: true
})
});
</script>
how, if i want to use ajax to keep the page not to reload? please help me

ok so here is quick tutorial. ajax helps you to submit your form without page loading. You have to use action="somepage.php". when you click it it will pass the values there right away. so here is basic coding
<form id="formId">
<input type="text" name="dte" id="dte">
<button type="button" id="buttonsubmit">submit</button>
</form>
and ajax should be look like. In my coding id="buttonsubmit" is button id so when user will click it it run the function and takes the values
var dte;
$("#buttonsubmit").click(function(){
dte = $("#dte").val();
$.ajax({
url:"somepage.php",
method:"POST",
data:{dte: dte},
success: function (data){
if(data == "done"){
window.location='http://somedomain.com/';
}
}
});
});
in somepage.php you will have
<?php
if(isset($_POST['dte'])){
$date=$_POST['dte'];
echo "done";
}else{
$date=date('Y-m-d');
}
?>
note: i am writing echo "done"; in success method it will come and
according to code when success will see it will redirect the page

Related

Post form with ajax and update div

I want to post simple form with ajax and update content of div (id result), but I get redirected to server.php file.
index.php:
<!DOCTYPE HTML>
<html>
<head>
<meta charset="UTF-8">
</head>
<body>
<script src="js/jquery.js"></script>
<script src="js/ajax.js" type="text/javascript"></script>
<form id="wbForm" action="server.php" method="POST">
Date1: <input type="text" name="date1" value="2000-01-21"><br>
Date2: <input type="text" name="date2" value="2000-01-02"><br>
<input type="submit" name="submit" value="Submit">
<div id="result"></div>
</form>
</body>
</html>
ajax.js:
$(document).ready(function showHint(form) {
$.ajax({
type:'POST',
url: 'server.php',
data:$('#wbForm').serialize(),
success: function(response) {
$('#wbForm').find('.result').html(response);
}});
});
server.php:
<?php
$input=$_POST;
//... compute something
echo "result";
?>
String "result" should appear in div with id=result, but I get redirected to /server.php where I can see "result", why?
HTML
<!DOCTYPE HTML>
<html>
<head>
<meta charset="UTF-8">
</head>
<body>
<script src="js/jquery.js"></script>
<script src="js/ajax.js" type="text/javascript"></script>
<form id="wbForm" action="server.php" method="POST">
Date1: <input type="text" name="date1" value="2000-01-21"><br>
Date2: <input type="text" name="date2" value="2000-01-02"><br>
<input type="submit" name="submit" value="Submit">
<div id="result"></div>
</form>
</body>
</html>
ajax.php
$(document).on("ready", function(){
//Form action
$("#wbForm").on("submit", function(event){
// Stop submit event
event.preventDefault();
$.ajax({
type:'POST',
url: 'server.php',
data:$('#wbForm').serialize(),
success: function(response) {
$('#wbForm').find('.result').html(response);
}});
});
});
server.php
<?php
$input = $_POST;
print_r( $input );
?>
Happy Codding!!
There are several issues. First, you are getting redirected to server.php when you hit submit because of your form action "server.php". If you want the AJAX call to happen when clicking the button you should put the AJAX call in a JavaScript function and call that function onclick()
The reason the jQuery .AJAX call isn't triggering success is because it's expecting JSON. Try:
<?php
header("content-type:application/json");
$input=$_POST;
//... compute something
echo json_encode("result");
?>
Hope this helps.

no data received from in AJAX request

I test my first AJAX form, but when I submit my form the alert message just shows '2'. I search for sending ajax data to the same page and I found, that I can do that, but URL is optional.
my PHP code:
<html>
<body>
<form action="index.php" method="post">
<input name="name" type="text" />
<input type="submit" value="submit" name="submit">
</form>
<?php
if(isset($_POST["name"]))
{
$data="test string";
echo json_encode($data);
}
?>
<script src="jquery-2.1.0.min.js"></script>
<script src="ajax.js"></script>
</body>
</html>
my AJAX code:
$(document).ready(function() {
$('form').submit(function(event) {
$.ajax({
type : 'POST',
url : 'index.php',
data : $(this).serialize(),
dataType : 'json',
encode : true
})
.done(function(data) {
alert(1);
})
.fail(function(data) {
alert(2);
});
event.preventDefault();
});
});
I don't know where I go wrong?
It's better to delegate page generation and json-response generation to different pages. At least you should isolate it, because the following part of page also ends up in ajax-response:
<html>
<body>
<form action="index.php" method="post">
<input name="name" type="text" />
<input type="submit" value="submit" name="submit">
</form>
...
<script src="jquery-2.1.0.min.js"></script>
<script src="ajax.js"></script>
</body>
</html>
Modifiyng your script, you can do something like that:
<?
if(isset($_POST["name"]))
{
// And don't forget to specify content type!
header("Content-type: application/json");
$data="test string";
echo json_encode($data);
} else {
?>
<html>
<body>
<form action="index.php" method="post">
<input name="name" type="text" />
<input type="submit" value="submit" name="submit">
</form>
<script src="jquery-2.1.0.min.js"></script>
<script src="ajax.js"></script>
</body>
</html>
<? } ?>
And, for future, please, post the exact request and response information in your questions, which you can get on Network page for developer tools of chrome, for example.

submit normally newly created ajax form

I want to submit my form. In my form using first button I create a textbox through ajax and after that I use second button to submit the form normally. When I want to get the value of newly created textbox using $_POST it gives error. How can i get value of ajax created button on submission. my php code is :
<?php`enter code here`
session_start();
ob_start();
require_once "config.php";
if ($_SERVER['REQUEST_METHOD']=="POST")
{
if (isset($_POST['subtest']))
{
print $_GET['tt'];
}
}
?>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<script type="text/javascript" src="jquery-ui-1.8.21.custom/js/jquery-1.7.2.min.js"></script>
<script type="text/javascript" src="jquery-ui-1.8.21.custom/js/jquery-ui-1.8.21.custom.min.js"> </script>
<script type="text/javascript">
jQuery(document).ready(function(){
jQuery("#subt").click(function(){
jQuery("#divload").show();
jQuery.ajax({url:"adp.php", type:"post", success:function(result){
jQuery("#disp").html(result);
jQuery("#divload").hide();
}});
return false;
});
});
</script>
<script type="text/javascript">
jQuery(document).ready(function(){
jQuery("#subtest").click( function() {
alert(jQuery("#tt").val());
});
});
</script>
</head>
<body id="#bdy">
<form id="FrmMain" method="post" action="">
<div id="Shd" style="font: 10%; margin: 50px; background-repeat:repeat-y; padding- left:120px;" >
<input type="text" id="txtFrom" name="txtFrom" />
<input type="text" id="txtUpto" name="txtUpto" />
<input type="text" id="txtOpt" name="txtOpt" />
<input type="submit" id="subt" name="subt" />
<input type="submit" id="subtest" name="subtest" />
<div id="RuBox" style="font-weight:bold;"><input type="checkbox" id="chkaccept" name="chkaccept" /> I accept terms and conditions.</div>
</div>
</form>
<div style="color:#F00; font-size:18px; display:none;" id="divload">Please wait loaidng... </div>
<div id="disp"></div>
</body>
</html>
Code of my adp.php file :
<?php
sleep(5);
?>
<div style="color:#30F; font-size:36px;">
Application testing............
<input type="text" id="tt" name="tt" />
</div>
I am not getting value of textbox named "tt" in temp form
Thanks
You're pretty much not posting anything from here:
jQuery(document).ready(function(){
jQuery("#subt").click(function(){
jQuery("#divload").show();
jQuery.ajax({url:"adp.php", type:"post", success:function(result){
jQuery("#disp").html(result);
jQuery("#divload").hide();
}});
return false;
});
});
So I would assume that you only want to generate a text field dynamically. And you can do it without the use of ajax:
var form = $('#FrmMain');
$('<input>').attr({'type' : 'text', 'id' : 'tt', 'name' : 'tt'}).appendTo(form);
Plus you're also trying to print out $_GET['tt'] while the form method is POST
if (isset($_POST['subtest']))
{
print $_GET['tt'];
}
This should also be:
echo $_POST['tt'];

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>

Passing two values for ajaxForm(), using jQuery and PHP

I want to pass both the values using ajaxForm, displaying both the values separately in test.php
-------test.php-----------------------------------------
<script language="javascript" type="text/javascript">
$('#test_form').ajaxForm({
target:'#result',
success:function() {
$('#result').show();
}
});
</script>
<form id="test_form" method="" action="test1.php">
<input type="submit" id="sub" value="sub_value">
</form>
<div id="result"></div>
-------test1.php---------------------------------------
<?
$t="test value";
$u="test value 1";
?>
Thanks
Jean
Hy Jean
Welcome to Stackoverflow
Here is the fixed code:
-------test.php-----------------------------------------
<script language="javascript" type="text/javascript">
$.post("test.php", {testval: val}, function(data){
if (data.length>0){
$('#result').show();
$("#result").html(data);
}
})
</script>
<form id="test_form" method="" action="test1.php">
<input type="submit" id="sub" value="sub_value" name="testval">
</form>
<div id="result"></div>
-------test1.php---------------------------------------
<?
// now coding in PHP is easiyer
if(isset($POST["testval"]))
{
// this is now for your Learning purposes that you see how things are
$testval = strip_tags(mysql_escape_string($POST["testval"]));
echo $testval;
}
// thats it have fun :-)
?>

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