Develop Reference

AJAX Internal 500 Error in Wordpress Children-Theme

I got this Ajax code on Js/Jq block (/buscador/menuLateral/menu-libros.php):
$.ajax({
url: '<?= get_stylesheet_directory_uri(); ?>' +
'/buscador/buscador-functions.php',
type: 'POST',
data: {
sendCheckedVal: 'yes',
tipo: 'editorial',
valor: valor_recogido,
archivo: this_file_id.toString()
},
success: function(data) {
alert('Checked value !');
}
});
Values on this file exists and got some value, I tried seeing it with a GET on a Stringify.
And this, must be getted in a file like this (/buscador/buscador-functions.php) :
<?php
if (ISSET($_POST['sendCheckedVal'])){
echo 'hi, u reached here' ;
}
?>
The values doesn't passes from js code file to next.
I get this error on console:
POST
[WP-URL-HIDDEN-ON-PURPOSE]/themes/ChildrenNewspaper/buscador/buscador-functions.php
500 (Internal Server Error)
On right side of the line-error :
jquery.min.js:2
Someone knows how to repair this on ajax working on a wordpress theme.Thanks

Issue is here : alert('hi, u reached here');. alert() is not PHP function it is javascript function. So instead of it you can use echoor return
<?php
if (isset($_POST['sendCheckedVal'])){
echo 'hi, u reached here';
}
?>

To start you off, alert is a javascript function.
In php you need to echo out values to the page;
If you want to do it with JS you need to print out js.
echo "<script type='text/javascript'>alert('hi, u reached here');</script>";
Or you can echo out the vars to the page.

How to put the ajax response/result into variable php

Can I put the ajax response into variable? So that I can echo the variable into php?
ajax.php
<script>
function random_no(){
$.ajax({
url: "test.php",
success: function(result){
$("#random_no_container").html(result);//the result was displayed on this div
}
});
}
</script>
On my sample code above, i call the query result from test.php and the result was displayed on the div but i want to put that on the variable and echo the variable. something like the code below
<?php echo $variable;?>
Can this possible? please help me. Thank you so much.

You can't PHP is executed on server and Ajax from the client so you cannot assign PHP variable an ajax response.
Once a PHP is rendered on the client side it is just HTML no PHP code.

function random_no(){
$.ajax({
url: "test.php",
success: function(result){
var result_val=result;
alert(result);
}
});
}

As #dev answered , you need to execute server and client code differently ! I answered here for your need variable only .But in developing website, should not use javascript in php tag <?php ?>
ajax.php
<div id="random_no_container">
<?php
if(isset($_GET['variable'])) {
$variable = $_GET['variable'];
echo $variable;
} else {
echo "<script>sessionStorage.setItem('stopped',0);</script>";
}
?>
</div>
<script>
var stopped = sessionStorage.getItem("stopped");
if(stopped == 0) {
random_no();
}
function random_no(){
$.ajax({
url: "test.php",
success: function(result){
sessionStorage.setItem("stopped",1);
//$("#random_no_container").html(result);//the result was displayed on this div
location.href = "ajax.php?variable="+result;
}
});
}
</script>

How to get data from one php page using ajax and pass it to another php page using ajax

I am trying to get data from one php page and pass it to another page using Ajax.
JS :
$.ajax({
url: "action.php",
success: function(data){
$.ajax({
url: "data.php?id=data"
}
});
action.php :
<?php
$test= 1;
?>
data.php :
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script type="" src="action.js"></script>
<?php
$id = $_GET['id'];
echo $id;
?>

First of all, you need to echo your data in action.php, and second, use data parameter of AJAX request to send data to data.php.
Here's the reference:
jQuery.ajax()
So the organization of pages should be like this:
JS :
$.ajax({
url: "action.php",
success: function(data){
$.ajax({
url: "data.php",
data: {id: data},
success: function(data){
// your code
// alert(data);
}
});
}
});
action.php :
<?php
$test = 1;
echo $test;
?>
data.php :
<?php
$id = $_GET['id'];
echo $id;
?>

Try to use $.get() method to get/send data :
$.get("action.php",{}, function(data){
//data here contain 1
$.get("data.php", {id: data}, function(id){
alert(id);
}
});
Just echo $test since just the data printed in page will return as responce to the query request.
action.php :
<?php
$test=1;
echo $test;
?>
Hope this helps.

For example,
Test
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script type="" src="action.js"></script>
action.js
$('.dataClass').click(function(){
var value=$(this).attr('data-value');
$.ajax({url:"Ajax_SomePage.php?value="+value,cache:false,success:function(result){
alert("success");
}});
});
Ajax_SomePage.php
<?php
$value = $_GET['value'];
echo $value;
?>

To get data as response in ajax call, you need to echo the result from your php page; action.php page.
echo $test = 1;
In your provided code
$.ajax({
url: "data.php?id=data"
} // closing bracket is missing
you are sending the string data as id to data.php page. Instead you have to append the result with the url using + symbol like shown in the below code.
$.ajax({
url: "action.php",
success: function(data){
$.ajax({
url: "data.php?id="+data
})
}
});

do not receive the POSTed data in php code ( cakephp)

I am trying to use ajax for retrieving data from my server. I use cakephp and jquery.
The ajax code is as follows:
<script>
$(document).ready(function(){
$(".viewMode").click(function(){
$.ajax({
type: "POST",
url:"viewModeSwitching",
data: {
mobileViewCookie:12,
},
success:function(result){
// window.location.href = result;
}
}
);
});
});
</script>
...
<?php
echo "<br>";
echo $this->Form->button(__('Desktop View'), array('type' => 'button','class' =>"viewMode",));
echo "<br>";
?>
This works well, in firebug I see that the POST is sent with value mobileViewCookie=12.
The point is that my cakephp controllerfunction 'viewModeSwitching' cannot retrieve that data.
I checked $this->data, $this->params, $_POST, etc but no sign of the data that had been send in the POST message.
Any suggestions?
/Antoine

javascript array in php using Ajax on submit not working

I have an array that i pass from javascript to php and in php page i am trying to put it in session to be used in the third page. The code is as below
JavaScript:
var table_row = [];
table_row[0] = [123,123,123];
table_row[1] = [124,124,124];
table_row[2] = [125,125,125];
var jsonString = JSON.stringify(table_row);
$.ajax({
type: "POST",
url: "test1.php",
dataType: "json",
data: {myJSArray: jsonString},
success: function(data) {
alert("It is Successfull");
}
});
test1.php
<?php
session_start();
$check1 = $_POST['myJSArray'];
$_SESSION['array']= $check1;
echo $check1;
?>
test2.php
<?php
session_start();
$test = $_SESSION['array'];
echo $test;
?>
on submit i call the function in javascript and the form takes me to test2.php. It is giving error on test2.php page Notice: Undefined index: array in C:\xampp\htdocs\test\test2.php on line 13
Any suggestions please do let me know.

You don't need to stringify yourself, jquery does it for you, if you stringify it, jQuery will believe you want a string instead
var table_row = [];
table_row[0] = [123,123,123];
table_row[1] = [124,124,124];
table_row[2] = [125,125,125];
$.ajax({
type: "POST",
url: "test1.php",
dataType: "json",
data: {myJSArray: table_row},
success: function(data) {
alert("It is Successfull");
}
});
However, on the php side, you still need to decode it as it is always a string when you get it from $_POST. use json_decode to do it.
$check1 = json_decode($_POST['myJSArray']);

look at your test2.php
<?php
session_start();
$test = $_SESSION['array'];
echo $test;
?>
if it's only the code in the file then the error you got C:\xampp\htdocs\test\test2.php on line 13 is mindless, because there is not line 13,
but if you have something about the code you show us, may there be something echoed before?
because session has to be started before any output,
otherwise I've tested whole script and works fine...
To check if session really started (otherwise $_SESSION will not work), try this:
if(session_id())
{
echo "Good, started";
}
else
{
echo "Magic! strangeness";
}
if problem not found in test2.php you can check test1.php echo $_SESSION['array'] after saving it, and in your javascript callback function alert data param itself,
I'm sure you can catch the problem by this way.

i got it to work, the code is below
Javascript file: in page index.php
Either you can call this function and pass parameter or use code directly
var table_row = []; //globally declared array
var table_row[0]=["123","123","123"];
var table_row[1]=["124","124","124"];
var table_row[2]=["125","125","125"];
function ajaxCode(){
var jsonArray = JSON.stringify(table_row)
$.ajax
({
url: "test1.php",
type: "POST",
dataType: 'json',
data: {source1 : jsonArray},
cache: false,
success: function (data)
{
alert("it is successfull")
}
});
}
Page: test1.php
session_start();
unset($_SESSION['array']);
$check1 = $_POST['source1'];
$_SESSION['array']= $check1;
echo json_encode(check1);
Page: test2.php //final page where i wanted value from session
if(session_id())
{
echo "Session started<br>";
$test = $_SESSION['array'];
echo "The session is".$test;
}
else
{
echo "Did not get session";
}
?>
In index page i have a form that is submitted and on submission it calls the ajax function.
Thank you for the help, really appreciate it.