send variable with ajax call and get in php

Question

I want to sent a variable on an ajax call when I click on a button. The button has an id and on click of the button the id should pass through the ajax call to a php file

<button class="btn-view-products" id="mangoes">View Products</button>

and my js is

var gallery = $("#product-gallery"); 
var viewproducts = $('.btn-view-products');
viewproducts.click(function(){
pid = $(this).attr('id');
gallery.load("gallery.php?id="+ pid, hideLoading);
});

and in my php

$fname = $_GET["pid"];
$images_dir = "images/products/".$fname."/showcase/";

I am not getting the id value in php. please help. thanks a ton in advance.

The URL is using id, while PHP is using pid. Might be a typo when posting here, so I wont offer it as an answer unless of course it is the problem. — David Houde
– David Houde, Commented Sep 22, 2013 at 10:18
Why don't put a link on your button and use PHP to create your link with pid? — Donovan Charpin
– Donovan Charpin, Commented Sep 22, 2013 at 10:22
$_GET["pid"] should be $_GET["id"] as you are passing "id" not "pid" ..? is that a typo..? — Sudhir Bastakoti
– Sudhir Bastakoti, Commented Sep 22, 2013 at 10:24

Anshad Vattapoyil · Accepted Answer · 2013-09-22 10:25:16Z

2

You are passing id from js and accessing pid in php. So replace your js to,

var gallery = $("#product-gallery"); 
var viewproducts = $('.btn-view-products');
viewproducts.click(function(){
  pid = $(this).attr('id');
  gallery.load("gallery.php?pid="+ pid, hideLoading);
});

answered Sep 22, 2013 at 10:25

Anshad Vattapoyil

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3 Comments

user2290403 Over a year ago

Thanks David. I didnt notice it.

Anshad Vattapoyil Over a year ago

@Sergio What will be the problem, if changing in js? Is there any performance issue?

Sergio Over a year ago

Not me downvoting btw

David Houde · Accepted Answer · 2013-09-22 10:39:04Z

1

In your JavaScript, you are sending a request to gallery.php?id= where in your PHP script, you are looking for $_GET['pid']. One of them needs to change as they don't match.

Change:

gallery.load("gallery.php?id="+ pid, hideLoading);

To:

gallery.load("gallery.php?pid="+ pid, hideLoading);

OR Change:

$fname = $_GET["pid"];

To:

$fname = $_GET["id"];

answered Sep 22, 2013 at 10:39

David Houde

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