get data from ajax and send to mysql

Question

I have simple ajax code:

function showMe(data) {
    $("body").append();
    if(data.success == true) {
        $("body").append("<img src="+data.data.link+" height=180 /><br /><a href="+data.data.link+">"+data.data.link+"</a>");

$.ajax ({
    type: "POST",
    url: "sql.php",
    data: "y=+data.data.link+",
});

i need to get '+data.data.link+' value and send to mysql db, but it sends data.data.link and not real link. How to get real value and send to db ? Here is sql.php:

<?php
define('IN_PHPBB', true); 
$phpbb_root_path = (defined('PHPBB_ROOT_PATH')) ? PHPBB_ROOT_PATH : 'forum/'; 
$phpEx = substr(strrchr(__FILE__, '.'), 1); 
include($phpbb_root_path . 'common.' . $phpEx); 
include($phpbb_root_path . 'includes/functions_display.' . $phpEx);
include("$phpbb_root_path/includes/functions_user.php");
$user->session_begin();
$auth->acl($user->data);
$user->setup('viewtopic');

include "forum/config.php";
$link = mysql_connect("$dbhost", "$dbuser", "$dbpasswd");
$db_selected = mysql_select_db("$dbname", $link);

$y = @$_POST['y']; 

$date = date('d.m.y');
$name = $user->data['username'];

mysql_query("INSERT INTO `gallery` (name, createdate, piclink) VALUES('$name', '$date',  '".$y."')");

unlink("gallery/$imagename");
?>

Thanks for any help :)

You'll need to show the response of sql.php

Jamie Hutber
– Jamie Hutber

2013-10-21 12:32:49 +00:00
Commented Oct 21, 2013 at 12:32 — Jamie Hutber
– Jamie Hutber, Commented Oct 21, 2013 at 12:32

bipen · Accepted Answer · 2013-10-21 12:33:27Z

3

send data as object (clean and readable) and not string..

try this

 data: {'y':data.data.link},

answered Oct 21, 2013 at 12:33

bipen

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1 Comment

bipen Over a year ago

cool..welcome... glad it helped.. happy coding..looks like you haven't accepted answer from any of your question... please accetp as answer if it helped .. for your other questions too

ErnestV · Accepted Answer · 2013-10-21 12:36:52Z

0

You didn't reveal where "data.data.link" comes from, but the major problem is within the ajax code:

$.ajax ({
    type: "POST",
    url: "sql.php",
    data: "y=\""+data.data.link+"\"",
});

The reference to the variable (data.data.link) must not be in quotes to be evaluated.

answered Oct 21, 2013 at 12:36

ErnestV

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