Seeing if variable is in array

Question

I am using a MySQL database. I am completely sure that the ID does actually exist in the database. Why is it going to the last else (where is says //incorrect id) ?

<?php
            //Localise user id.
            $userid = $_SESSION['userid'];

            //Get content of the article.
            $sql = "SELECT * FROM articles WHERE creatorid = '$userid'";
            $result = mysql_query($sql) or die(mysql_error()); //Execute. If fails, show error.
            $array = mysql_fetch_array($result);

            if(in_array($articleid, $array)) //If the URL id exists in the database (array)
            {
                //The article does actually exist for that user. They requested it.
                $sql = "SELECT * FROM articles WHERE id = '$articleid'";                
                $result = mysql_query($sql) or die(mysql_error()); //Execute. If fails, show error. 
                $array = mysql_fetch_array($result);

                        $content = $array['content'];

                        if($content != '') //If the article has actually been written.
                        {
                            include($_SERVER['DOCUMENT_ROOT'] . '/includes/renderimage.php');
                        }   else
                            {
                                //Article actually hasn't been written.
                            }
            }   else
                {
                    //Incorrect ID.
                }
                ?>

where is $articleid? is it initialised? if not then else part will execute. — Yogesh Suthar
– Yogesh Suthar, Commented Oct 13, 2012 at 6:32
If I'm not wrong, you can't use in_array in that context. I have a gut feeling it doesn't iterate multiple dimensions. — BudwiseЯ
– BudwiseЯ, Commented Oct 13, 2012 at 6:32
Please don't use mysql_* functions in new code. They were removed from PHP 7.0.0 in 2015. Instead, use prepared statements via PDO or MySQLi. See Why shouldn't I use mysql_* functions in PHP? for more information. — vascowhite
– vascowhite, Commented Oct 13, 2012 at 6:35
$articleid is initialised in the script. $articleid = $_GET['ref']; — KriiV
– KriiV, Commented Oct 13, 2012 at 6:41

Barmar · Accepted Answer · 2012-10-13 06:42:51Z

1

You're only looking in the first row that's returned. You need to call mysql_fetch_array in a loop to get each row. Also, you shouldn't use in_array(), since the article ID might appear in some other column (what if you're checking for article #3 and user #3?).

But if you just want to see if the article was created by this user, you can use a different query:

SELECT * FROM articles WHERE creatorid = '$userid' AND articleid = '$articleid';

This should return either 0 or 1 row depending on whether the user created the article. You can then use mysql_num_rows() to test for this.

answered Oct 13, 2012 at 6:42

Barmar

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Comments

Coldstar · Accepted Answer · 2012-10-13 06:36:44Z

0

It appears you are accessing the array incorrectly. On top of that you are returning multiple articles if the creator posted more than one so your in_array() is totally invalid. Change the limit on your query to one record (LIMIT 0,1) and access the creator id by calling:

$result[0]->creatorid or $result['creatorid']

depending on how your resource is queried

answered Oct 13, 2012 at 6:36

Coldstar

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