Okay, what have I done wrong here?
I'm trying to .POST data from a Form with JQuery $.ajax function and return json data. But every time the post data is null.
Here is the HTML/JQuery Part
<!DOCTYPE HTML>
<html>
<head>
<link rel="stylesheet" type="text/css" href="http://ajax.googleapis.com/ajax/libs/jqueryui/1/themes/redmond/jquery-ui.css">
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js"></script>
<script src="http://ajax.googleapis.com/ajax/libs/jqueryui/1/jquery-ui.min.js"></script>
</head>
<body>
<form action="testform" method="post">
OLD ID:
<input id="OldID" name="OldID" type="text"><br>
NewID:
<input id="NewID" name="NewID" type="text"><br>
<input id="do" name="do" type="button" value="Get New ID">
</form>
<script type="text/javascript">
var CID = $('#OldID').val();
$(document).ready(function(){
$('#do').click(function(){
sendValue();
});
function sendValue(CID){
$.ajax({
type: 'POST',
url: 'getNewID.php',
data: {OldID: CID},
dataType: 'json',
cache: false,
success:
function(data){
$('#NewID').val(data.NewID);
}
}) }
});
</script>
</body>
</html>
Here's the PHP page
<?php
if (isset($_POST['OldID'])){
$value = $_POST['OldID'];
}else{
$value = "ELSE VALUE";
}
echo json_encode(array("NewID"=>$value));
?>
I'm sure this is a simple mistake on my part but I'm just going in circles at this point.
Thanks in advance!
$('#do').click(function(){
sendValue();
});
should be:
$('#do').click(function(){
sendValue("VALUE TO SEND TO THE SERVER");
});
because
function sendValue(CID){
takes the scope of CID which it seems you might think should be this var CID = $('#OldID').val();
Your mistake is that you are getting the value of the OldID field in your script, and not on some event handler method. This causes when the script is loaded (when the page is being viewed), the value of the text input is kept on variable CID. and when the user clicks on the #do button, then the value of CID is still the same as the one when the page was loaded. So no value is stored in CID variable, so no value is sent via AJAX.
Define another function (Say sendOldID) that reads the value of the text input in the CID variable, and then calls the sendValue() method with CID paramter. The use that function as the event handler of the #do button clicked.
<script type="text/javascript">
$(document).ready(function(){
$('#do').click(function(){
sendOldID();
});
function sendOldID() {
var CID = $('#OldID').val();
sendValue(CID);
}
function sendValue(CID) {
$.ajax({
type: 'POST',
url: 'getNewID.php',
data: {OldID: CID},
dataType: 'json',
cache: false,
success:
function(data){
$('#NewID').val(data.NewID);
}
}) }
});
</script>
Related
I am trying to insert a name into my database. There is a form asking to enter name.There is also a submit button.on clicking the submit button no alert is show.I am new to ajax.
this is index.php.
<html>
<head>
<title>Untitled Document</title>
<script type="text/javascript" src="http://ajax.microsoft.com/ajax/jquery/jquery-1.6.2.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$("#submit").submit(function() {
var name = $('#name').val();
$.ajax({
type: "POST",
url: "ajax.php",
data: "name=" + name,
success: function() {
alert("sucess");
}
});
});
});
</script>
</head>
<body>
<form action="" method="post">
Name:
<input type="text" name="name" id="name" />
<br />
<input type="submit" name="submit" id="submit" />
</form>
</body>
</html>
This is ajax.php
<html>
<body>
<?php
$query=mysql_connect("localhost","root","root");
mysql_select_db("freeze",$query);
$name='l';//$_POST['name'];
mysql_query("insert into tt values('','$name')");
?>
</body>
</html>
Be careful with url's in ajax. Your url is 'ajax.php', this is a relative url, so, if your page is at url http://localhost.com/index.html (for example), ajax will fire a post to http://localhost.com/index.html/ajax.php (and that's probably wrong). Try use absolute url, for that, add '/' at begin of url, then ajax will be fired against http://localhost.com/ajax.php.
Your ajax code should look like code below:
$(document).ready(function() {
$("#submit").submit(function() {
var name = $('#name').val();
$.ajax({
type: "POST",
url: "/ajax.php",
data: "name=" + name,
success: function() {
alert("sucess");
}
});
});
});
The submit event is triggered on the form, not the submit button. So
$("#submit").submit(function() {
will not ever be triggered, because #submit selects the button, not the form. You need to give an ID to the form:
<form id="myform">
then use
$("#myform").submit(function() {
You also need to use event.preventDefault() or return null in the event handler, to prevent the form from being submitted normally when the function returns.
I am trying to have the output from a PHP file display in the original page containing the form, without having to refresh the page.
I am currently using AJAX to do so but I am not able to pass the value of the submitted PHP file back to the index.php page - this needs to be done without having the page refreshed.
Overall
User enters some data and submits the form
That data is passed to a PHP file through AJAX
Some pieces of data, which were manipulated through the PHP file, are then echoed out onto the original index.html file without the page refreshing
Below you will find the code that I am currently using to try and achieve this.
index.php
<form>
<input type="text" name="hi">
<input type="submit">
</form>
<script>
$(function () {
$('form').on('submit', function (e) {
e.preventDefault();
$.ajax({
type: 'post',
url: 'post.php',
data: $('form').serialize(),
});
});
});
</script>
post.php
<?php
echo "Hello";
$name = $_POST['hi'];
echo $name . "This is a test";
?>
I would appreciate any help, thanks!
How about this simple solution. Just copy and paste this code into a file called index.php and whatever text you enter in the input field will be outputted by PHP. Hope it helps!.
<?php
$data = array();
if(isset($_POST['userText']) && !empty($_SERVER['HTTP_X_REQUESTED_WITH'])){
$data = 'The data you entered is: ' . $_POST['userText'];
echo json_encode($data);
die();
}
?>
<html>
<head>
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
</head>
<body>
<form>
<input type="text" name="hi" id = "someID">
<input type="submit" id = "sendInfo">
</form>
<div id = "random"></div>
<script type = "text/javascript">
$("#sendInfo").click(function(){
var someText = document.getElementById("someID").value;
var userText = JSON.stringify(someText);
$.ajax({
url: "index.php",
method: "POST",
dataType: "json",
data: {userText: userText},
success: function (result) {
console.log(result);
$("#random").html(result);
}
});
return false;
});
</script>
</body>
</html>
Add 'success' to your object and pass in a parameter to the function. That will hold the results from the php file.
$.ajax({
type: 'post',
url: 'post.php',
data: $('form').serialize(),
success: function(results) {
console.log(results);
}
});
What this does is, on a successful ajax call, stores all of the content sent from the post.php file into the results variable.' In this case it will be whatever $name is and 'This is a test.'
First of all I'm sorry if my English isn't totally correct.
I'm learning to use json and I want to try with this simple code, where I insert name and surname and with Ajax I send them to a json struct for then show them in a table.
ajax.php
<form id="iscrizione">
Nome: <input type="text" id="nome" /><br />
Cognome: <input type="text" id="cognome" /></br >
<input type="submit" id="invia" value="ISCRIVITI" />
</form>
<table>
<tr><td>Nome: </td><td><span id="td_nome"></span></td></tr>
<tr><td>Cognome: </td><td><span id="td_cognome"></span></td></tr>
</table>
<script type="text/javascript">
$("#iscrizione").submit(function(){
var nome = $("#nome").val();
var cognome = $("#cognome").val();
$.ajax({
url: "json.php",
type: "POST",
data: {nome: nome, cognome: cognome},
dataType: "json",
success: function(msg){
$("span#td_nome").html(msg.nome);
$("span#td_cognome").html(msg.cognome);
},
error: function() {
alert ("Chiamata Fallita");
}
});
});
</script>
json.php
<?php
header("Content-Type: application/json", true);
$dati = array( 'nome'=>$_POST['nome'], 'cognome'=>$_POST['cognome'] );
echo json_encode($dati);
?>
Where are the mistakes? Because the outputs are shown for just a second and then they will disappear.
Thank you to everybody.
When you submit the form it will reload all page again, so assigned html visible for short time only.
Use preventDefault() function to overcome this situation.
<script type="text/javascript">
$("#iscrizione").submit(function(event){
event.preventDefault()
var nome = $("#nome").val();
var cognome = $("#cognome").val();
$.ajax({
url: "json.php",
type: "POST",
data: {nome: nome, cognome: cognome},
dataType: "json",
success: function(msg){
$("span#td_nome").html(msg.nome);
$("span#td_cognome").html(msg.cognome);
},
error: function() {
alert ("Chiamata Fallita");
}
});
});
On submitting form, it will reload page by default. We can cancel the submit action by calling .preventDefault() on the event object or by returning false from our handler. Read more about .submit() event.
1.On using returning false after ajax call.
$("#iscrizione").submit(function(){
var nome = $("#nome").val();
var cognome = $("#cognome").val();
$.ajax({
....
....
});
return false; // add return false here
});
2.By using event.preventDefault()
$("#iscrizione").submit(function(event){
event.preventDefault();// use preventDefault() on event
var nome = $("#nome").val();
var cognome = $("#cognome").val();
$.ajax({
....
....
});
});
Either return false or preventDefault() will work, as the other two answers have described. However, it's important to understand that you don't have to submit the form at all, and indeed it isn't best practice to do so. After all, it doesn't make much sense to use HTML to submit the form and then disable the submit behavior in your code. Especially when you don't have to.
AJAX was invented to overcome the problems (such as yours) that result when you reload the page. The point about AJAX is that it gives you the ability to update your page with new data without reloading it.
So, the cleaner way to do what you want is to make your button simply a button, and call your code from the button's click event. Have a look at this:
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>
Test
</title>
<!--jQuery file-->
<script src="includes/jquery/jquery-2.2.3.js"></script>
<script type="text/javascript">
$(document).ready(function () {
function showMyPHP() {
var nome = $("#nome").val();
var cognome = $("#cognome").val();
$.ajax({
url: "json.php",
type: "POST",
data: {nome: nome, cognome: cognome},
dataType: "json",
success: function(msg){
$("#td_nome").html(msg.nome);
$("#td_cognome").html(msg.cognome);
},
error: function() {
alert ("Chiamata Fallita");
}
});
}
$('#invia').on('click', function(e, ui){
showMyPHP();
});
});
</script>
</head>
<body>
<form id="iscrizione">
Nome: <input type="text" id="nome" /><br />
Cognome: <input type="text" id="cognome" /></br >
<button type="button" id="invia">"ISCRIVITI"</button>
</form>
<table>
<tr><td>Nome: </td><td><span id="td_nome"></span></td></tr>
<tr><td>Cognome: </td><td><span id="td_cognome"></span></td></tr>
</table>
</body>
</html>
You can see that we're not submitting the form at all here. Rather, we're calling the AJAX through the button's click event. You'll see that you get the results you're looking for, without resorting to using HTML to submit the form and then disabling the default submit behavior in code.
I am having a problem is retrieving data sent to a php file through .ajax() via jquery
Following is my html:
<!-- Jquery tute no 94 onwards -->
<html lang="en">
<head>
<meta charse="utf-8">
<title> jquery4 </title>
<link rel="stylesheet" type="text/css" href="jquery4.css"/>
</head>
<body>
<input id="lo" type="text"> </input>
<input id="ton" type="button" value="Load"> </input>
<div id="content"> </div>
<script type="text/javascript" src="jquery.js"> </script>
<script type="text/javascript" src="jquery4.js"> </script>
</body>
</html>
My jquery4.js is:
$(document).ready(function()
{
$('#ton').click(function()
{
var nm= $('#lo').val();
$.ajax({url: 'page.php', data1: 'name='+nm, success: function(data2)
{
$('#content').html(data2);
}
});
});
});
My page.php is:
<?php
if(isset($_GET['data1']))
{
echo $namer= $_GET['data1'];
}
?>
All the above files are in the same folder, and I have xampp installed.
I guess the error is somewhere in the jquery file where I call the
ajax() function
jQuery ajax doesn't take a data1 parameter. It takes a data parameter, which should be an object of name-value pairs.
$.ajax({
url: 'page.php',
data: {
data1: 'name=' + nm,
},
success: function(data2) {
$('#content').html(data2);
}
});
$.ajax({
type: "GET",
url: "page.php",
data: {
data1: 'name=' + nm,
}
,
success: function(data2) {
$('#content').html(data2);
}
});
Try this:
$(document).ready(function() {
$('#ton').click(function() {
var nm= $('#lo').val();
$.ajax({
url: 'page.php?name=' +nm,
success: function(data2) {
$('#content').html(data2);
}
});
});
});
You don't have to tell jQuery to use GET, as it defaults to that, if nothing else is specified.
So the ajax function does not take an argument called data1, but 'data', this is mostly used for other methods as POST, PUT and DELETE.
I prefer also sending GET requests with a normal query string, like the above example.
You can then check for get GET parameter with PHP, using $_GET['name']
I am new to ajax and jQuery. I am trying get html form value using ajax and jQuery. I am getting the value but i can not pass that value to another php file. I don't know what i am missing.. can someone help me please..
Here is my form.php file code:
<!DOCTYPE HTML>
<html lang="en-US">
<head>
<meta charset="UTF-8">
<title>Form</title>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script>
$(document).ready(function(){
$('#submit').click(function(){
var srt = $("#input").serialize();
// alert is working perfect
alert(srt);
$.ajax({
type: 'POST',
url: 'database.php',
data: srt,
success: function(d) {
$("#someElement").html(d);
}
});
});
});
</script>
</head>
<body>
<form id="input" method="post" action="">
First Name:<input type="text" name="firstName" id="firstName">
Last Name: <input type="text" name="lastName" id="lastName">
<input type="submit" id="submit" value="submit " name="submit">
</form>
<div id="someElement"></div>
</body>
</html>
Here is my database.php file code:
<?php
if(isset($_POST['submit']))
{
$firstName = $_POST['firstName'];
$lastName = $_POST['lastName'];
echo $firstName;
echo $lastName;
}
You need to add:
return false;
to the end of your click handler. Otherwise, the default submit button action takes place, and the page is refreshed.
Also, remove the line:
if (isset($_POST['submit']))
Submit buttons aren't included when you call .serialize() on a form, they're only sent when the submit button is performing normal browser submission (in case there are multiple submit buttons, this allows the server to know which was used).
You need to do the following
$('#submit').click(function(){
var srt = $("#input").serialize();
// alert is working perfect
alert(srt);
$.ajax({
type: 'POST',
url: 'database.php',
data: srt,
success: function(d) {
$("#someElement").html(d);
}
return false;
});
Form default event will fire if you dont use return false.
You can use as below
$("#input").submit(function(event) {
var srt = $("#input").serialize();
// alert is working perfect
alert(srt);
$.ajax({
type: 'POST',
url: 'database.php',
data: srt,
success: function(d) {
$("#someElement").html(d);
}
})
return false;
});
You need to prevent the default action of click event or the form will be processed using the default form action and your ajax call will never be complete.
$("#input").submit(function(event) {
event.preventDefault(); // prevent default action
var srt = $("#input").serialize();
// alert is working perfect
alert(srt);
$.ajax({
type: 'POST',
url: 'database.php',
data: srt,
success: function(d) {
$("#someElement").html(d);
}
});
});