$jquery.post() PHP and Mysql

Question

So I half got jQuery's ajax ($.post) to work. But, for some reason, I haven't been successful with finding the right online article to explain to me how PHP retrieves the ajax data that is sent. I've found some stuff on json_decode, but upon me doing that to basically decode it, it wont work (and yes, I am using json for the $.post command).

Here is my javascript code

$.post("notificationNum.php", {"user":"1"},
                function(data){
                        $(".example-number").html(data.amount);
                }, "json");

Here is my PHP code

<?php
session_start();
//link to db info here

$user_id_got = json_decode($_REQUEST['user']);

$checknoti = mysql_query("SELECT * FROM notifications WHERE notification_users = '".$user_id_got."' AND notification_viewed= '0'");

  echo json_encode(array("amount"=>mysql_num_rows($checknoti)));
?>

Mind you all, I've also tried using the $_POST command instead of the $_REQUEST. Any ideas how to send data to the PHP file so I can use it?

Is notificationNum.php getting called in the first place? Can you do something like a var_dump to see the contents of $_REQUEST? Maybe try using a tool like Fiddler (fiddler2.com/fiddler2) to see what data is flowing back and forth to the server. — mellamokb
– mellamokb, Commented Oct 4, 2011 at 18:26
notificationNum.php is getting called, I already did something to make sure it is being called, and it works. Just getting the information is the problem and retrieving it in PHP. — UXerUIer
– UXerUIer, Commented Oct 4, 2011 at 18:28
Your code is assuming that the jquery is actually sending over a JSON string. You're not, you're just constructing a regular POST system, so passing that data through json_decode in PHP will return null (check json_last_error() to confirm). — Marc B
– Marc B, Commented Oct 4, 2011 at 18:33

user187291 · Accepted Answer · 2011-10-04 18:26:21Z

4

"json" in your jQuery call is how your php should write its output, not how jQuery sends it. Use normal $_REQUEST in your php:

 $user_id_got = $_REQUEST['user'];

answered Oct 4, 2011 at 18:26

user187291

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5 Comments

UXerUIer Over a year ago

Actually, wait, this did work, but there was a problem with my code. However, is there a way to get a session in php and pass it through jQuery's ajax call?

UXerUIer Over a year ago

I have a $_SESSION[''] in place in the very same file the $.post is being executed. Can I send the information in the $_SESSION through $.post? Such as.... $.post("notificationNum.php", {"user":<?php $_SESSION['here']; ?>}, function(data){ $(".example-number").html(data.amount); }, "json");

mellamokb Over a year ago

@Majed: The session is on the server already, why would you need to pass it from the client to the server??

UXerUIer Over a year ago

Because, notificationNum.php is not accessing it for some reason, so instead, I need to pass it to the file.

UXerUIer Over a year ago

Good idea, thanks for the help, you gave me the right answer, thus you will be rewarded for that. :-)

Community · Accepted Answer · 2020-06-20 09:12:55Z

0

try this

notificationNum.php

<?php
//link to db info here

$user_id_got = intval($_POST['user']);

$checknoti = mysql_query("SELECT * FROM notifications WHERE notification_users = '".$user_id_got."' AND notification_viewed= '0'");

echo json_encode(array("amount"=>mysql_num_rows($checknoti)));
?>

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Oct 4, 2011 at 18:30

xkeshav

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