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.
                }
                ?>

Answer 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.