Hello so I have 2 submit buttons with different names (btn1, btn2) in my html form and what I am trying to do is to submit to another page without refreshing page. So what I wanted to do is if I click btn1 submit it will do something and if I click btn2 it will do another thing. My code in the html page is this
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Percentage</title>
<script type="text/javascript" src="jquery.js"></script>
<script>
$(document).ready(function(){
$('#myForm').on('submit',function(e) {
$.ajax({
url:'update.php',
data:$(this).serialize(),
type:'POST',
success:function(data){
console.log(data);
$("#success").show().fadeOut(5000);
},
error:function(data){
$("#error").show().fadeOut(5000); //===Show Error Message====
}
});
e.preventDefault();
});
});
</script>
</head>
<body>
<form method="POST" id="myForm">
Input Amount: <input type="text" name="txt_amount" required placeholder="Input number"> <br /> <br />
<span id="error" style="display:none; color:#F00">Some Error!Please Fill form Properly </span> <span id="success" style="display:none; color:#0C0">All the records are submitted!</span>
<input type="submit" name="btn1"> <input type="submit" name="btn2">
</form>
</body>
</html>
And the code in my update.php page
<?php
if(isset($_POST['btn1'])) {
//insert query
} else if(isset($_POST['btn2'])) {
//another insert query
}
?>
I actually got it working if I only have 1 submit button and no if(isset()) thing in the update.php page. What can I do to use 2 submits and with issets in another page without refreshing the main page?
$(this).serialize();
The above code statement doesn't include name of the submit button as a key value pair.
So, as people have suggested before me, you should use button instead of submit button. Something like this.
HTML and JS
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Percentage</title>
<script type="text/javascript" src="jquery.js"></script>
<script>
$(document).ready(function(){
$('#btn1, #btn2').on('click',function(e) {
var datastr = $(this).serialize() + "&button_id="+$(this).attr('id');
$.ajax({
url:'update.php',
data:datastr,
type:'POST',
success:function(data){
console.log(data);
$("#success").show().fadeOut(5000);
},
error:function(data){
$("#error").show().fadeOut(5000); //===Show Error Message====
}
});
e.preventDefault();
});
});
</script>
</head>
<body>
<form method="POST" id="myForm" action="update.php">
Input Amount: <input type="text" name="txt_amount" required placeholder="Input number"> <br /> <br />
<span id="error" style="display:none; color:#F00">Some Error!Please Fill form Properly </span> <span id="success" style="display:none; color:#0C0">All the records are submitted!</span>
<button id="btn1">Button1</button><button id="btn2">Button2</button>
</form>
</body>
</html>
AND PHP would be:
<?php
if($_POST['button_id'] == 'btn1') {
//do something
} else if($_POST['button_id'] == 'btn2') {
//do something else;
}
?>
Use this, may useful for you
try
$('#myForm').on('submit',function(e) {
e.preventDefault();
});
OR
<button type="button">
Use on click event on the button and add the value attribute to the submit button, the value of the click button will be pased to the php file
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Percentage</title>
<script type="text/javascript" src="jquery.js"></script>
<script>
$(document).ready(function(){
$('input[type="submit"]').on('click',function(e) {
e.preventDefault();
$.ajax({
url:'update.php',
data:{'txt_amount':$('input[name="txt_amount"]').val(),'btn': $('input[type="submit"]').val()}
type:'POST',
success:function(data){
console.log(data);
$("#success").show().fadeOut(5000);
},
error:function(data){
$("#error").show().fadeOut(5000); //===Show Error Message====
}
});
});
</script>
</head>
<body>
<form method="POST" id="myForm">
Input Amount: <input type="text" name="txt_amount" required placeholder="Input number"> <br /> <br />
<span id="error" style="display:none; color:#F00">Some Error!Please Fill form Properly </span> <span id="success" style="display:none; color:#0C0">All the records are submitted!</span>
<input type="submit" name="btn1" value="btn1"> <input type="submit" name="btn2" value="btn2">
</form>
</body>
</html>
php:
<?php
if($_POST['btn'] == 'btn1') {
//do something
} else if($_POST['btn'] == 'btn2') {
//do something else;
}
?>
Related
I just need to display names of the array using ajax. Following code is working but the result ( Nilantha Ruwan Nimal Shamitha Alex) is just display and disappears.
Index.php
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Document</title>
<link rel="stylesheet" href="assets/css/bootstrap.min.css">
</head>
<body>
<div class="container" style="margin-top:50px";>
<form >
<div class="form-group">
<input type="text" id="name" class="form-control" placeholder="Enter Name....">
</div>
<div class="form-group">
<button class="btn btn-success" id="btn">Enter</button>
</div>
</form>
<div class="msg"></div>
</div>
<script type="text/javascript" src="assets/js/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$("#btn").click(function(){
var name = $("#name").val();
$.post("ajax.php",{ajax_name:name},function(response){
$(".msg").html(response);
})
.fail(function(error){
alert(error.statusText);
})
})
})
</script>
</body>
</html>
ajax.php
<?php
if(isset($_POST['ajax_name'])){
$store = array("Nilantha","Ruwan","Nimal","Shamitha","Alex");
foreach($store as $names) {
echo $names,"<br>";
}
}
?>
Try the following code.
<script type="text/javascript">
function submit() {
jQuery("form").submit(function(e) {
e.preventDefault();
var name = jQuery("#name").val();
jQuery.ajax({
type: 'POST',
url: 'ajax.php',
data: {ajax_name:name},
success: function(response) {
jQuery(".msg").html(response);
jQuery('#name').val('');
},
error: function() {
console.log("Something wrong");
}
});});
}
jQuery(document).ready(function() {
submit();
});
</script>
I would suggest looking at the network tab of Chrome Devtools and ensuring that the AJAX call isn't running twice. I would think that if the query ran twice but the second one did not have any name attached then it would return blank once the first one had received its response.
I am trying to submit a child-form, inside parent-form via ajax-jquery, so that it does not refresh entire page. Code is:
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.2.2/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function (e) {
$(document).on('submit', '#form-2', function() {
var data = $(this).serialize();
$.ajax({
type : 'POST',
url : 'a2.php',
data : data,
success : function(data) {
$("#form-2").fadeOut(500).hide(function() {
$(".result").fadeIn(500).show(function() {
$(".result").html(data);
});
});
}
});
return false;
});
}) // document ready ends here;
</script>
</head>
<body>
<form action="a1.php" method="post" name="form-1" id="form-1">
<input type="text" name="f1" />
<input type="text" name="f2" />
<input type="text" name="f3" />
<input type="text" name="f4" />
<!-----form 2 ajax starts----->
<form method="post" name="form-2" id="form-2">
<input type="text" name="g1" />
<input type="submit" id="sf2">
</form><!-----form-2 ends----->
</form><!-----form-1 ends----->
</body>
</html>
But its not working, it does simply nothing. I too used - preventdefault()
Any help ? I am trying to simply submit form-2 value in database, from which some dropdown of form-1 is getting all option values.
You could use:
$(document).on('click', '#sf2', function(event) {
var g1 = $('#g1').val();
$.ajax({
type : 'POST',
url : 'a2.php',
data : {
g1: g1
},
success : function(data) {
$("#form-2").fadeOut(500).hide(function() {
$(".result").fadeIn(500).show(function() {
$(".result").html(data);
});
});
}
});
});
and use a normal button:
<input type="text" name="g1" id="g1" />
<button type="button" id="sf2">Submit</button>
This is not good style though as forms should not be nested.
I am sending state,city,zip code to ajax.php file. using jquery.
I am unable to get those values in response
<!DOCTYPE html>
<html lang="en">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<meta charset="utf-8">
<title>Jquery Ajax</title>
</head>
<body>
<!------------------------Jquery-------POST DATA----------------------------------------->
<!--<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script>
-->
<script type="text/javascript" src="jquery-1.3.2.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$('#id-submit').click(function() {
var reg_state = $('#reg_state').val();
var reg_city = $('#reg_city').val();
var reg_zip = $('#reg_zip').val();
var dataString = 'reg_state='+ reg_state;
//alert(dataString);
$.ajax
({
type: "POST",
url: "ajax.php",
data: dataString,
cache: false,
success: function(data)
{
//$("#state").html(data);
alert(data);
}
});
});
});
</script>
<!------------------------Jquery-----------POST DATA END------------------------------------->
<form id="registration_form" action="" method="post">
<div id="state"></div>
<div class="reg-id">
<label>
<input placeholder="State:" type="text" tabindex="3" name="user_state" id="reg_state" value="">
</label>
</div>
<div class="reg-id">
<label>
<input placeholder="City:" type="text" tabindex="3" name="user_city" id="reg_city" value="">
</label>
</div>
<div class="reg-id-last">
<label>
<input placeholder="Zip/Postal:" type="text" tabindex="3" name="user_zip" id="reg_zip" value="">
</label>
</div>
<input type="submit" value="submit" tabindex="3" name="reg_btn" id="id-submit">
</div>
</form>
</body>
</html>
This is the ajax.php file , from where i need to send response, and make it visible in div id="state
<?php
if($_POST['reg_state'])
{
echo $_POST['reg_state'];
}
else{
echo 'nothing';
}
?>
Try create an object
var dataString = { reg_state: reg_state, reg_city: reg_city, reg_zip : reg_zip }
I believe your problem lies in the type=submit. Replacing this with type=button gives you the result you want.
I'm unsure why this is, but I'm guessing it type=submit gives extra functionality to the input. The page seems to reload after hitting the submit button.
ps. don't forget you actually need to click the button in order for the function to trigger. Hitting enter after filling out the input fields will do nothing.
When ever we use input['type']="submit" in form
we have to use
$('#id-submit').click(function(e) {
e.preventDefault();
}
Here is the full working code
<!DOCTYPE html>
<html lang="en">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<meta charset="utf-8">
<title>Jquery Ajax</title>
</head>
<body>
<!------------------------Jquery-------POST DATA----------------------------------------->
<script type="text/javascript" src="http://code.jquery.com/jquery-1.11.1.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$('#id-submit').click(function(e) {
e.preventDefault();
var reg_state = $('#reg_state').val();
var reg_city = $('#reg_city').val();
var reg_zip = $('#reg_zip').val();
var dataString = { reg_state: reg_state, reg_city: reg_city, reg_zip : reg_zip }
//alert(dataString);
$.ajax
({
type: "POST",
url: "ajax_city.php",
data: dataString,
cache: false,
success: function(data)
{
$("#state").html(data);
// alert('hi');
}
});
});
});
</script>
<!------------------------Jquery-----------POST DATA END------------------------------------->
<form id="registration_form" action="" method="post">
<div id="state"></div>
<div class="reg-id">
<label>
<input placeholder="State:" type="text" tabindex="3" name="user_state" id="reg_state" value="">
</label>
</div>
<div class="reg-id">
<label>
<input placeholder="City:" type="text" tabindex="3" name="user_city" id="reg_city" value="">
</label>
</div>
<div class="reg-id-last">
<label>
<input placeholder="Zip/Postal:" type="text" tabindex="3" name="user_zip" id="reg_zip" value="">
</label>
</div>
<input type="submit" value="Response" tabindex="3" name="reg_btn" id="id-submit">
</div>
</form>
</body>
</html>
ajax_city.php
<?php
if($_POST['reg_state'])
{
echo 'STATE: '.$_POST['reg_state'].'<br/>';
echo 'CITY: '.$_POST['reg_city'].'<br/>';
echo 'ZIP: '.$_POST['reg_zip'].'<br/>';
}
else{
echo 'nothing to respond';
}
?>
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.
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'];