I try to use Ajax (my first time) and i don't know what i am doing bad. I have two files: home.html and ajax.php
On first one (home.html) i have this
<html>
<head>
<script type="text/javascript" src="jquery-2.1.3.min.js"></script>
<script type="text/javascript">
jQuery(document).ready(function(){
jQuery('#Client_ID').live('change', function(event) {
$.ajax({
url : 'ajax.php',
type : 'POST',
dataType: 'json',
data : $('#myform').serialize(),
})
.done(function(data) {
for(var id in data) {
$(id).val( data[id] );
}
});
});
});
</script>
</head>
<body>
<form id='myform'>
<select name='Client_ID' id='Client_ID'>
<option value=''>Select</option>
<option value='1'>Client 1</option>
<option value='2'>Client 2</option>
</select>
<input type='text' name='address1' id='address1'>
</form>
</body>
</html>
and my AJAX.php file it's
<?php
define('WP_USE_THEMES', false);
require('../wp-blog-header.php');
global $wpdb;
$clientid = $_POST['Client_ID']; // Selected Client Id
$result = $wpdb->get_row( "SELECT * FROM wp_com_plantillas WHERE id=$clientid" );
$addr1 = $result->asunto;
$arr = array( 'input#address1' => $addr1);
echo json_encode( $arr );
?>
On database connection i use wordpress global var $wpdb.
To test ajax.php i change POST by GET and load url like ajax.php?client_id=1 and result it's ok, but when i test it on home.html, when dropdown change, input don't fill.
Thank you so much
If you are using Wordpress, consider taking a look in this article that explains how to properly declare ajax on Wordpress
Also, make usage of your Developer Tools, if you use Chrome. Pressing F12, you can see a tab named Network. Every request that your page does, it will be logged there. If you want to filter only AJAX requests, click on filter button then select XHR. Each line is a request and each request, when clicked, can provide information about the response. Check if that response is working, if response code is actually 200.
Related
I have a drpcategory dropdown in a form. I will just paste the dropdown code below;
<div class="form-group">
<label>Category</label>
<select class="form-control bg-dark btn-dark text-white" id="drpcategory" name="drpcategory" required>
<?php
$category = ''.$dir.'/template/post/category.txt';
$category = file($category, FILE_IGNORE_NEW_LINES);
foreach($category as $category)
{
echo "<option value='".$category."'>$category</option>";
}
?>
</select>
</div>
Then I AJAX post every time I make a selection in the above drpcategory dropdown as below;
<script>
$(function(){
$('#drpcategory').on('change',function()
{
$.ajax({
method: 'post',
data: $(this).serialize(),
success: function(result) {
console.log(result);
}
});
});
});
</script>
This seems to be currently working as I'm getting outputs like below in Chrome Browser > Inspect > Network tab every time I make a selection in drpcategory. Here is the screenshot;
The question is how can I capture this AJAX post data using PHP within the same page and echo it within the same page? So far I have tried;
<?php
if(isset($_POST['drpcategory']))
{
echo 'POST Received';
}
?>
I'm looking for a solution using only PHP, JQuery and AJAX combined.
This question was later updated and answered here:
AJAX POST & PHP POST In Same Page
First of all, this line -> type: $(this).attr('post') should be type: $(this).attr('method'),. So this will give the value ** type:post** and
As far as i understand, you are asking to send ajax whenever you select options from drpcategory. Why are you submitting the entire form for this. If i where you, i should have done this problem by following way
$("#drpcategory").change(function(){
e.preventDefault();
var drpcategory=$(this).val();
$.ajax({
type: 'post',
data: drpcategory,
success: function(result) {
console.log(result);
}
});
});
On you php side, you can get your data like,
echo $_POST['drpcategory'];
I recommend you read the documentation for the ajax function, I tried to replicate it and I had to fix this:
$.ajax({
// If you don't set the url
// the request will be a GET to the same page
url: 'YOU_URL',
method: 'POST', // I replaced type by method
data: $(this).serialize(),
success: function(result) {
console.log(result);
}
});
http://api.jquery.com/jquery.ajax/
OUTPUT:
First change to $value
<div class="form-group">
<label>Category</label>
<select class="form-control bg-dark btn-dark text-white" id="drpcategory" name="drpcategory" required>
<?php
$category = ''.$dir.'/template/post/category.txt';
$category2 = file($category, FILE_IGNORE_NEW_LINES);
foreach($category2 as $value)
{
echo "<option value='".$value."'>".$value."</option>";
}
?>
</select>
then add url
<script>
$(function()
{
$('#form').submit(function(e)
{
e.preventDefault();
$.ajax({
url:'folder/filename.php',
type: 'post',
data: '{ID:" . $Row[0] . "}',
success: function(result) {
console.log(result);
}
});
});
$('#drpcategory').on('change',function()
{
$("#form").submit();
});
});
try request
if(isset($_REQUEST['ID']))
The result will/should send back to the same page
Please try this code:
$.post('URL', $("#FORM_ID").serialize(), function (data)
{
alert('df);
}
I think you have an eroror syntax mistake in ajax jQuery resquest because ajax post 'http://example.com/?page=post&drpcategory=Vehicles' does not return this type url in browser Network Tab.
<?php var_dump($_POST); exit; ?> please do this statment in your php function if anything posted to php page it will dump.
Here ajax request example
$("#drpcategory").change(function(){
e.preventDefault();
var drpcategory=$(this).val();
$.ajax({
type: 'post',
data: drpcategory,
success: function(result) {
console.log(result);
}
});
});
`
It sounds like you're trying to troubleshoot several things at once. Before I can get to the immediate question, we need to set up some ground work so that you understand what needs to happen.
First, the confusion about the URL:
You are routing everything through index.php. Therefore, index.php needs to follow a structure something like this:
<?php
// cleanse any incoming post and get variables
// if all your POST requests are being routed to this page, you will need to have a hidden variable
// that identifies which page is submitting the post.
// For this example, assume a variable called calling_page.
// As per your naming, I'll assume it to be 'post'.
// Always check for submitted post variables and deal with them before doing anything else.
if($_POST['calling_page'] == 'post') {
// set header type as json if you want to use json as transport (recommended) otherwise, just print_r($_POST);
header('Content-Type: application/json');
print json_encode(array('message' => 'Your submission was received'));
// if this is from an ajax call, simply die.
// If from a regular form submission, do a redirect to /index.php?page=some_page
die;
}
// if you got here, there was no POST submission. show the view, however you're routing it from the GET variable.
?>
<html>
(snip)
<body>
<form id="form1" method="post">
<input type="hidden" name="calling_page" value="page" />
... rest of form ...
<button id="submit-button">Submit</button>
</form>
}
Now, confusion about JQuery and AJAX:
According to https://api.jquery.com/jquery.post/ you must provide an URL.
All properties except for url are optional
Your JQuery AJAX will send a post request to your index.php page. When your page executes as shown above, it will simply print {message: "Your submission was received"} and then die. The JQuery will be waiting for that response and then do whatever you tell it to do with it (in this example, print it to the console).
Update after discussion
<div class="form-group">
<label>Category</label>
<select class="form-control bg-dark btn-dark text-white" id="drpcategory" name="drpcategory" required>
<?php
$category = ''.$dir.'/template/post/category.txt';
$category = file($category, FILE_IGNORE_NEW_LINES);
foreach($category as $category)
{
echo "<option value='".$category."'>$category</option>";
}
?>
</select>
</div>
<!-- HTML to receive AJAX values -->
<div>
<label>Item</label>
<select class="" id="drpitem" name="drpitem"></select>
</div>
<script>
$(function(){
$('#drpcategory').on('change',function() {
$.ajax({
url: '/receive.php',
method: 'post',
data: $(this).serialize(),
success: function(result) {
workWithResponse(result);
}
});
});
});
function workWithResponse(result) {
// jquery automatically converts the json into an object.
// iterate through results and append to the target element
$("#drpitem option").remove();
$.each(result, function(key, value) {
$('#drpitem')
.append($("<option></option>")
.attr("value",key)
.text(value));
});
}
</script>
receive.php:
<?php
// there can be no output before this tag.
if(isset($_POST['drpcategory']))
{
// get your items from drpcategory. I will assume:
$items = array('val1' => 'option1','val2' => 'option2','val3' => 'option3');
// send this as json. you could send it as html, but this is more flexible.
header('Content-Type: application/json');
// convert array to json
$out = json_encode($items);
// simply print the output and die.
die($out);
}
Once you have everything working, you can take the code from receive.php, stick it in the top of index.php, and repoint the ajax call to index.php. Be sure that there is no output possible before this code snippet.
I have a problem concerning my code which should change content in a div onclick "More News articles" as the change will happen only once. I see in Chrome Developer mode that it fires every click a request. What goes wrong?
Output.php
<?php
require_once('../pe13f/SSI.php');
require_once ('../PE13/smf_2_api.php');
?>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script>
function MakeRequest(id)
{
$.ajax({
url : 'display.php',
data:{"id":id},
type: 'GET',
success: function(data){
$('#streaminnern').html(data);
}
});
}
</script>
<div id="stream" class="bg4 roundedcrop shadow">
<div class="ph25 pv20">
<h1>News</h1>
<input id="streamcnt" name="streamcnt" type="hidden" value="" />
</div>
<div id="streamadd"></div>
<div id="streaminnern">
<?php
$num_recent = 5;
echo $num_recent;
?>
</div>
<div onclick="MakeRequest(<?php echo $num_recent; ?>);" id="streammore">More News articles</div>
</div>
backend php display.php
<?php
$num_recent = $_GET['id']+5;
echo $num_recent;
?>
Greetings Emil
Check the source that is produced by output.php. You'll find there onclick="MakeRequest(5);". Basically - on every click you call MakeRequest(5) which always fires call display.php?id=5 (you probably see that in your dev console).
Try something like this:
<script>
var lastId = 0; // var that stores last fetched ID
function MakeRequest(id)
{
if(!lastId) // if there is no last ID use the one from initial onclick
lastId = id;
$.ajax({
url : 'display.php',
data:{"id":lastId}, // note that we are using the lastId var
type: 'GET',
success: function(data){
$('#streaminnern').html(data);
lastId = data; // save fetched ID in our global var
}
});
}
</script>
The request is always the same. Suppose $num_recent is initially set to 5.
Then as per your code MakeRequest(5) will be executed. And your ajax call updates a div with class streaminnern. So the new id has no impact on the next ajax call. For the ajax request to be sent updated value you may set
$.ajax({
url : 'display.php',
data:{"id":$('#streaminnern').html()},
.................
});
Ok I have a onchange event on a select field. It now works great. When the dropdown "networks" is changed it refreshes the second dropdown. I also want the ajax code at the top to trigger on page load as well as onchange so the second list gets populated. This is due to it being on an edit page. Here is the ajax call im using first
function get_cities(networks) {
$.ajax({
type: "POST",
url: "select.php", /* The country id will be sent to this file */
beforeSend: function () {
$("#folder").html("<option>Loading ...</option>");
},
//data: "idnetworks="+networks,
data: "idnetworks="+networks +"&doc="+ <?php echo $row_rs_doc['parentid']; ?>,
success: function(msg){
$("#folder").html(msg);
}
});
}
now the two dropdown boxes
<label for="networks"></label>
<select name="networks" id="networks" onChange='get_cities($(this).val())'>
<?php
do {
?>
<option value="<?php echo $row_rs_net['idnetworks']?>"<?php if (!(strcmp($row_rs_net['idnetworks'], $row_rs_doc['network']))) {echo "selected=\"selected\"";} ?>><?php echo $row_rs_net['netname']?></option>
<?php
} while ($row_rs_net = mysql_fetch_assoc($rs_net));
$rows = mysql_num_rows($rs_net);
if($rows > 0) {
mysql_data_seek($rs_net, 0);
$row_rs_net = mysql_fetch_assoc($rs_net);
};
?>
</select>
<select name="folder" id="folder">
</select>
You can use .trigger() to trigger a change event onto the select-box so the onchange code will be called like it would if the user just switched the option.
jQuery('#networks').trigger('change');
Just include this into the load event/function for the page.
jQuery(document).ready(function() {
jQuery('#networks').trigger('change');
});
I'm not 100% clear what you want but the standard way to do something with JQuery when the page is loaded, is to use
$(document).ready(handler)
This waits till the page is "ready" which is better.
So, in your document head you'd have something like this...
<script type="text/javascript">
$(document).ready( function(){
do_some_stuff();
});
</script>
Can't you just call get_cities($('#networks').val()) when the DOM is ready?
$(function() { // will be run by jQuery when DOM is ready
get_cities($('#networks').val());
});
I am using php/ajax to submit a form without page refresh. Here are my files-
coupon.js
jQuery(document).ready(function(){
jQuery(".appnitro").submit( function(e) {
$.ajax({
url : "http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street/sms.php",
type : "post",
dataType: "json",
data : $(this).serialize(),
success : function( data ) {
for(var id in data) {
jQuery('#' + id).html( data[id] );
}
}
});
//return false or
e.preventDefault();
});
});
sms.php
<?php
//process form
$res = "Message successfully delivered";
$arr = array( 'mess' => $res );
echo json_encode( $arr );//end sms processing
unset ($_POST);
?>
and here is code for my html page -
<form id="smsform" class="appnitro" action="http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street/sms.php" method="post">
...
</form>
<div id="mess" style="background:green;"></div>
Now when i click on submit button nothing happens and firebug shows following under console panel -
POST http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street/sms.php
404 Not Found 1.29s `jquery.min.js (line 130)`
Response
Firebug needs to POST to the server to get this information for url:
http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street/sms.php
This second POST can interfere with some sites. If you want to send the POST again, open a new tab in Firefox, use URL 'about:config', set boolean value 'extensions.firebug.allowDoublePost' to true
This value is reset every time you restart Firefox This problem will disappear when https://bugzilla.mozilla.org/show_bug.cgi?id=430155 is shipped
When i set 'extensions.firebug.allowDoublePost' to true then following results show up -
POST http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street/sms.php
404 Not Found 1.29s `jquery.min.js (line 130)`
Response -
{"mess":"Message successfully delivered"}
CaN anyone help me in fixing this firebug error of 404 not found. And why is it showing jquery.min.js (line 130) along side?
P.S -do not worry about http://174.132.194.155/~kunal17/devbuzzr/wp-content/themes/street this is my base url
You may want to try putting the e.preventDefault() statement before the $.ajax call.
EDIT:
My x.html, corresponds to your HTML page
<!DOCTYPE html>
<html>
<head>
<title>x</title>
<script
type="text/javascript"
src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js">
</script>
<script type="text/javascript" src="/so/x.js"></script>
</head>
<body>
<form id="smsform" class="appnitro" action="/so/x.php">
<input type="text" name="zz">
<input type="submit">
</form>
<div id="mess" style="background:green;"></div>
</body>
</html>
My x.js, corresponds to your coupon.js
jQuery(document).ready(function(){
jQuery(".appnitro").submit( function(e) {
e.preventDefault();
$.ajax({
url : "/so/x.php",
type : "post",
dataType: "json",
data : $(this).serialize(),
success : function( data ) {
for(var id in data) {
jQuery('#' + id).html( data[id] );
}
}
});
//return false or
//e.preventDefault();
});
});
My x.php, corresponds to your sms.php
<?php
$res = "Message successfully delivered.";
$arr = array('mess'=>$res);
echo json_encode($arr);
unset($_POST);
?>
This actually works in my environment, although I do not have the rest of the HTML markup or the additional PHP form processing code. The "Message successfully delivered." shows up in green directly below the input field.
When inside the Ajax call this refers to the Ajax object you need to do this
var __this = this;
Before going into the Ajax call, then it would be
data : __this.serialize()
Or look up the use of context within an Ajax call in Google. Or serialise your data into a variable before going into the Ajax call.
I have been trying to create a simple calculator. Using PHP I managed to get the values from input fields and jump menus from the POST, but of course the form refreshes upon submit.
Using Javascript i tried using
function changeText(){
document.getElementById('result').innerHTML = '<?php echo "$result";?>'
but this would keep giving an answer of "0" after clicking the button because it could not get values from POST as the form had not been submitted.
So I am trying to work out either the Easiest Way to do it via ajax or something similar
or to get the selected values on the jump menu's with JavaScript.
I have read some of the ajax examples online but they are quite confusing (not familiar with the language)
Use jQuery + JSON combination to submit a form something like this:
test.php:
<script type="text/javascript" src="jquery-1.4.2.js"></script>
<script type="text/javascript" src="jsFile.js"></script>
<form action='_test.php' method='post' class='ajaxform'>
<input type='text' name='txt' value='Test Text'>
<input type='submit' value='submit'>
</form>
<div id='testDiv'>Result comes here..</div>
_test.php:
<?php
$arr = array( 'testDiv' => $_POST['txt'] );
echo json_encode( $arr );
?>
jsFile.js
jQuery(document).ready(function(){
jQuery('.ajaxform').submit( function() {
$.ajax({
url : $(this).attr('action'),
type : $(this).attr('method'),
dataType: 'json',
data : $(this).serialize(),
success : function( data ) {
for(var id in data) {
jQuery('#' + id).html( data[id] );
}
}
});
return false;
});
});
The best way to do this is with Ajax and jQuery
after you have include your jQuery library in your head, use something like the following
$('#someForm').submit(function(){
var form = $(this);
var serialized = form.serialize();
$.post('ajax/register.php',{payload:serialized},function(response){
//response is the result from the server.
if(response)
{
//Place the response after the form and remove the form.
form.after(response).remove();
}
});
//Return false to prevent the page from changing.
return false;
});
Your php would be like so.
<?php
if($_POST)
{
/*
Process data...
*/
if($registration_ok)
{
echo '<div class="success">Thankyou</a>';
die();
}
}
?>
I use a new window. On saving I open a new window which handles the saving and closes onload.
window.open('save.php?value=' + document.editor.edit1.value, 'Saving...','status,width=200,height=200');
The php file would contain a bodytag with onload="window.close();" and before that, the PHP script to save the contents of my editor.
Its probably not very secure, but its simple as you requested. The editor gets to keep its undo-information etc.