i want to to a request.JSON on a dynamic form.
is it possible to POST the complete form via a form id ?
<table>
<form id="form">
<tr>
<td>username</td>
<td><input type="text" id="username"></td>
</tr>
<tr>
<td>password</td>
<td><input type="password" id="password"/></td>
</tr>
<tr>
<td></td>
<td><input type="submit" id="go" value="go"/></td>
</tr>
</table>
and on the script side:
$('go').addEvent('click',function( event )
{
event.stop();
var myForm = document.id('form');
var request= new Request.JSON (
{
url: 'check_login.php',
method: 'post',
data:
{
form: myForm
},
onSuccess: function( response )
{
if ( response.status )
{
document.location = "home.php"
}
else
{
$('response').addClass('error');
$('response').set('html', response.message);
}
},
});
request.send();
and in the php script i want to check via the $_POST variable.
but i cannot find a method to do it like this. actually im gettinh nothing.
greetz
Sure, look at this link where it shows how to do it:
http://demos111.mootools.net/Ajax.Form
It basically selects the form via your ID, add a listener to submit event and when it comes it sends all the form the way you are looking for :-)
Related
I design a simple page were user can put name, password and image using html.
I try to sent those data using ajax to specific php page, but I cannot implement this.
how I do this thing
Html code
<?php include('connection.php'); ?>
<form id="form" enctype="multipart/form-data">
<table>
<tr>
<td>Name:</td>
<td><input type="name" id="name"></td>
</tr>
<tr>
<td>Password:</td>
<td><input type="password" id="pass"></td>
</tr>
<tr>
<td>Photos:</td>
<td><input type="file" id="photos"></td>
</tr>
<tr>
<td><input type="submit" id="go"></td>
</tr>
</table>
</form>
Jquery and ajax
<script>
$(document).ready(function(){
$('#go').click(function(){
var img_name = $('#img_name').val();
var name = $('#name').val();
var pass = $('#pass').val();
$.ajax({
type : "POST",
url : "singup_submit.php",
data : { img_name:img_name, name:name, pass:pass},
success : function(done){
alert(done);
}
});
});
});
</script>
Php code
<?php
include('connection.php');
if(isset($_POST)){
$name = $_POST['name'];
$pass = $_POST['pass'];
$img_name=$_FILES['img_name']['name'];
$qr="INSERT INTO data (name,pass,img_name) VALUES ('$name','$pass','$img_name')";
$ex=mysqli_query($con,$qr) or die(mysqli_error($con));
if($ex==1)
echo 'done';
else
echo 'not done';
}
Follow this code ..It may help you
<script>
$(document).ready(function(){
$("#go").click(function(){
var name = $("#name").val();
var pasword = $("#password").val();
var photos = $("#photos").val();
if(name==''||pass==''||photos==''){
alert("Please Fill All Fields");
}
else{
$.ajax({
type : "POST",
url : "singup_submit.php",
data : formData,
cache : false,
success: function(result){
alert(result);
}
});
}
return false;
});
});
</script>
you are not sending any file in your ajax request.
$(document).ready(function(){
$("#go").on('submit',function(e){
e.preventDefault()
$.ajax({
url: 'singup_submit.php',
type: 'POST',
data: new FormData(this),
contentType: false,
processData: false,
success: function(response){
alert(done);
},
error: function(data){
console.log("error");
console.log(data);
}
},
});
});
});
and then take data from global variables in php as you are doing now.
and please assign name to your form fields like so
<form id="form" enctype="multipart/form-data">
<table>
<tr>
<td>Name:</td>
<td><input type="name" name="name" id="name"></td>
</tr>
<tr>
<td>Password:</td>
<td><input type="password" name="password" id="pass"></td>
</tr>
<tr>
<td>Photos:</td>
<td><input type="file" name="img" id="photos"></td>
</tr>
<tr>
<td><input type="submit" name="submit" id="go"></td>
</tr>
</table>
</form>
and it should work.
I noob and get mad when submit php form, convert input value to json, and other php file get it.
html
<form action="submit.php" method="post" name="form1" id="myform">
<table width="100%" border="0" style="font-size: 65px;">
<tr>
<td>Name</td>
<td><input type="text" name="name" id="name"></td>
</tr>
<tr>
<tr>
<td></td>
<td><button id="submit">Submit</button></td>
</tr>
</table>
</form>
<script src="script.js"></script>
script.js
$('#myform').submit(function (event) {
name = $('#name').val();
var data = {
name: name
}
$.ajax({
type: "POST",
url: 'submit.php',
contentType: 'application/json',
data: JSON.stringify(data),
dataType: 'json'
});
return false
});
php file
header('Content-Type: application/json');
$name_dirty = json_decode($_POST['name']);
echo $name_dirty;
Can someone help me? submit.php got blank, I cant get the value that I submit from html page. Big Thanks
Your Html
<table width="100%" border="0" style="font-size: 65px;">
<tr>
<td>Name</td>
<td><input type="text" name="name" id="name"></td>
</tr>
<tr>
<tr>
<td></td>
<td><button id="submit">Submit</button></td>
</tr>
</table>
<script src="script.js"></script>
Your JS
$('#submit').click(function() {
name = $('#name').val()
var data = {
name: name
}
$.ajax({
type: "POST",
url: 'submit.php',
data: data
dataType: 'json'
complete: function (resultData) {
alert('data has been send')
})
})
In your php:
<?php
print_r($_POST['data'])
A few notes. Make sure you check your paths. In my answer i assumed that the problem is in the code and not in wrong paths. Also when you use form tag you can use a submit button like <input type="submit" value="Submit"> to submit your form without using Javascript. It would work in your case but it's just another way to tackle your issue.
In my answer i removed your form tags and triggered the action on button click. There will be no redirect to the page but you can set a redirect inside the js if it is important to you, on success function of the ajax that i added. At the moment i just throw an alert message when it works successfully.
I have two same name multiple input fields. I want to send all fields value from another page using jquery ajax post method but i am not getting all rows input fields value. Please review my code.
Javascript code
<script type="text/javascript">
function getValue()
{
$.post("paidamt.php",
{
paidamt : $('#paidamt').val(),
uid : $('#uid').val()
},
function( data){
/*alert(data);*/
$("#divShow").html(data);
});
}
</script>
Html Code
<div>
<form method="post">
<table border="1">
<tr>
<th>Product</th>
<th>Price</th>
<th>Paid Amount</th>
<th>Check</th>
</tr>
<?php
$sql = mysql_query("SELECT * FROM `tbldemo`");
while ($result = mysql_fetch_array($sql)) {
?>
<tr>
<td><?php echo $result['pname']; ?> </td>
<td><?php echo $result['price']; ?></td>
<td><input type="text" name="paidamt[]" id="paidamt"></td>
<td><input type="checkbox" name="uid[]" id="uid"
value="<?php echo $result['id']; ?>"></td>
</tr>
<?php }
?>
</table><br>
<input type="button" name="submit" id="submit"
onclick="getValue(1)" value="Save Amt.">
</form>
</div>
<div id="divShow">
</div>
Try this one
var paidamt = $("input[name=paidamt]").map(function(){
return $(this).val();
}).get().join(",");
var uid = $("input[name=uid]").map(function(){
return $(this).val();
}).get().join(",");
$.ajax(
{
type: "POST",
url: 'paidamt.php',
data:
{
paidamt:paidamt,
uid:uid
}
});
Firstly you have given the input elements the same id which is repeated in the loop. This will end up in your HTML being invalid, you should change the id to class:
<form method="post">
<table border="1">
<tr>
<th>Product</th>
<th>Price</th>
<th>Paid Amount</th>
<th>Check</th>
</tr>
<?php
$sql = mysql_query("SELECT * FROM `tbldemo`");
while ($result = mysql_fetch_array($sql)) { ?>
<tr>
<td><?php echo $result['pname']; ?> </td>
<td><?php echo $result['price']; ?></td>
<td><input type="text" name="paidamt[]" class="paidamt"></td>
<td><input type="checkbox" name="uid[]" class="uid" value="<?php echo $result['id']; ?>"></td>
</tr>
<?php }
?>
</table><br>
<button type="submit" name="submit" id="submit">Save Amt.</button>
</form>
To actually send the input values in the AJAX request you can simply serialize() the containing form when the form is submit:
$(function() {
$('form').submit(function(e) {
$.ajax({
url: "paidamt.php",
type: 'POST',
data: $(this).serialize(),
success: function(data) {
$("#divShow").html(data);
});
});
});
});
I suggest to add class instead of id, since identically class can be repeated but id should not.
<script type="text/javascript">
function getValue()
{
var paidamtval = [];
$('#paidamt').each(function(){
paidamtval.push($(this).val());
});
$.post("paidamt.php",
{
paidamt : paidamtval,
uid : $('#uid').val()
},
function( data){
/*alert(data);*/
$("#divShow").html(data);
});
}
</script>
Since you will have many of these, id - needs to be unique, which in your case isn't, so remove "id="paidamt"
<td><input type="text" name="paidamt[]" id="paidamt"></td>
That's your first mistake. And secondly don't use $.post, to submit this form. Either remove AJAX submit, or bind form using something like jQuery Form plugin.
You try this code
$('document').ready(function(){
$('#submit').click(function(){
jQuery.ajax({
type: "POST",
url: "paidamt.php",
data: new FormData(this),
contentType: false,
cache: false,
processData:false,
success: function(html){
try{
$("#divShow").html(data);
}catch (e){
alert(JSON.stringify(e));
}
},
error : function(e){alert("error "+JSON.stringify(e)); }
});
});
});
in you paidamt.php file
$paidamt=$_POST['paidamt'];// its can array values
print_r($paidamt);// result display
On entering the customer name in textbox it searches for customer info. I have generated successfully using JQuery by passing the entire table through Json variable, as I dont want any page refresh. Now I want to select the customer id generated from mysql db (php) through radio button, but the radio button event is not working. For testing purpose I have put a static table having the same radio button properties in that particular div(place for dynamic record generation using JQuery) and working fine. Hence I found that the data received through JQuery got some problem. Hope I am clear. Please find a way. Thanks in advance.
below is the code
abc.php
<input type="text" placeholder="full name" id="fullName" name="fullName" class="txt" style="width: 250px" />
<input type="button" id="btSelect" value="Select" class="button-crystal" />
<div id="disp"></div>
script.js
$('#btSelect').click(function () {
var form_data = {
genCustDetails: $('#fullName').val(),
is_ajax: 1
};
$.ajax({
type: "POST",
url: "xyz.php",
data: form_data,
dataType: "json",
success: function (response)
{
$('#disp').html(response);
}
});
return false;
});
xyz.php
if (isset($_POST['genCustDetails'])) {
$out="";
$result = mysql_query("select * from dbcustdetails where name like '$_POST[genCustDetails]%'");
while ($row = mysql_fetch_assoc($result)) {
$out.='
<table style="background-color:#eee; margin-bottom:5px;">
<tr>
<th class="td3">Customer ID</th>
<td class="td4">
'.$row["custID"].' <input type="radio" id="cust_ID" name="cust_ID" value="'.$row["custID"].'" />
</td>
</tr>
<tr>
<th class="td3">Name</th>
<td class="td4">'.$row["name"].'</td>
</tr>
<tr>
<th class="td3">Phone No.</th>
<td class="td4">'.$row["phno"].'</td>
</tr>
<tr>
<th class="td3">Email</th>
<td class="td4">'.$row["email"].'</td>
</tr>
<tr>
<td colspan="2" style="padding:0;">
<b>Address</b><br/>'.$row["addr"].'
</td>
</tr>
</table>';
}
echo json_encode($out);
}
Maybe You should'nt bind the event properly for the dynamic elements in the DOM.
Try Like this
$('body').on('change','.radiobuttonClassorID',function(){
//actions
});
that is because your newly generated radio button is not having any event handler assigned to it.
you have to assign an event handler after the ajax call (on ajax success).
something like
$('input[type="radio"]').unbind('click').click(function(){
//Your handler code
})
I'm trying to send simple form data from my index page to a PHP function, checklogin.php. What I'm finding is that no data, serialized or even direct input, is getting through to the PHP function. I know for a fact that the jQuery function is getting called, and I also know that the PHP file is called because I've put various post functions in.
When I submit data directly from the form using action="checklogin.php" the $_POST data is correctly received and I can process it. However when using jQuery to intercept the form submit, no data is received.
One thing to note is that the actual login form HTML is inserted from another page, to allow me to work on multiple pages instead of one big index file.
Form - login.php
<form id="loginform" name="loginform" method="post">
<td>
<table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF">
<tr>
<td colspan="3"><strong>Member Login</strong></td>
</tr>
<tr>
<td width="78">Username</td>
<td width="6">:</td>
<td width="294"><input name="username" type="text" id="username"></td>
</tr>
<tr>
<td>Password</td>
<td>:</td>
<td><input name="password" type="text" id="password"></td>
</tr>
<tr>
<td> </td>
<td> </td>
<td><input type="submit" name="Submit" value="Login"></td>
</tr>
</table>
</td>
</form>
Submit event hook - Index.php
$( document ).ready(function() {
$.post("login.php", function(data) { // retrieve login form from other page.
$("#page5").html(data);
$('#loginform').submit(function () {
event.preventDefault();
var formData = $(this).serialize();
console.log("Form Data: " + formData); // this is not blank
$.post("checklogin.php", formData, function(data) {
// post-login actions
})
});
});
})
});
checklogin.php (temporary until is problem resolved)
<?php
print_r($_POST);
?>
As I said, even direct input such as "username=test$password=blah" instead of 'formData' results in an empty $_POST array. I've also tried using the ajax equivalent post call with no luck.
JavaScript Console Output
// console.log line above $.post
Form Data: username=test&password=blah
// Result from PHP print_r
Array
(
)
Use .on()
As elements are added dynamically you can not bind events directly to them .So you have to use Event Delegation.
$(document).on('submit','#loginform',function () { //code here }
or better use
$('#page5').on('submit','#loginform',function () { //code here }
Syntax
$( elements ).on( events, selector, data, handler );