How to pass mutiple values from ajax to php file - php

Question

I have 3 values stored in 3 seperate DIV tags and i want it to pass via ajax to php file. I have working code and stuck in passing all values to php file. Any ideas hoe to do it.
This is my js code:
$('#button').click(function(){
var slider_value = $('#slider_value').text();
var slider1_value = $('#slider1_value').text();
var slider2_value = $('#slider2_value').text();
$.ajax({
url:'placeDetailSend.php',
type: 'POST',
data: 'slider_value='+slider_value,
success: function(data){
$('#test').html(data);
}
});
});
and this it my php file:
<?php
if (isset($_POST['slider_value'])||($_POST['slider1_value'])){
echo $slider_value = $_POST['slider_value'];
echo $slider1_value = $_POST['slider1_value'];
}?>

Answer 1

Values are seperated by & as in a URL:
data: 'slider_value='+slider_value+'&slider1_value='+slider1_value+'&slider2_value='+slider2_value,

Answer 2

jQuery Code:
$('#button').click(function(){
var slider_value = $('#slider_value').text();
var slider1_value = $('#slider1_value').text();
var slider2_value = $('#slider2_value').text();
$.ajax({
url:'placeDetailSend.php',
type: 'POST',
data: {var1: slider_value, var2: slider1_value,var3:slider2_value },
success: function(data){
$('#test').html(data);
}
});
});
Use This php code for fetching values
<?php
echo $_POST['var1'];
echo $_POST['var2'];
echo $_POST['var3'];
?>

Answer 3

$('#button').click(function(){
var slider_value = $('#slider_value').text();
var slider1_value = $('#slider1_value').text();
var slider2_value = $('#slider2_value').text();
$.ajax({
url:'placeDetailSend.php',
type: 'POST',
data: {slider: [slider_value, slider1_value, slider2_value]},
success: function(data){
$('#test').html(data);
}
});
});
And in php file,
<?php
if (isset($_POST['slider'])){
$slider_value = $_POST['slider'];
echo '<pre>' . print_r($slider_value) . '</pre>';
}
?>

Answer 4

Try,
data: 'slider_value='+slider_value+'&slider1_value='+slider1_value+'&slider2_value='+slider2_value;
OR
data:{slider_value:slider_value,slider1_value:slider1_value,slider2_value:slider2_value}
You will get more about jquery ajax here

Answer 5

You are only passing one div value in the datastring. Pass all the values like this:-
$('#button').click(function(){
var slider_value = $('#slider_value').text();
var slider1_value = $('#slider1_value').text();
var slider2_value = $('#slider2_value').text();
$.ajax({
url:'placeDetailSend.php',
type: 'POST',
data: 'slider_value='+slider_value + '&slider1_value='+slider1_value + '&slider2_value='+slider2_value,
success: function(data){
$('#test').html(data);
}
});
});

Answer 6

$.ajax({
url:'placeDetailSend.php',
type: 'POST',
data: {
'slider_value': slider_value,
'slider_value1': slider_value1,
'slider_value2': slider_value2
},
success: function(data){
$('#test').html(data);
}
});
});

Answer 7

Firstly, you could store your data differently (in an array) like so:
var slider_values = new Array();
silder_values.push($('#slider_value').text(),$('#slider_value2').text(),$('#slider_value3').text());
And then, you can simply pass this array as a data object to the ajax request like so:
$.ajax({
url:'placeDetailSend.php',
type: 'POST',
data: {'slider_values': slider_values},
success: function(data){
// Do whatever here
}
});
});
However, if you use this method, you must be sure to loop through the $_POST['slider_values'] in PHP as it is now an array not a string. This is pretty simple:
foreach($_POST['slider_values'] as $value){
// Write the current value
echo $value
}