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copied!<p><strong>Important</strong><br> Before I answer the question, I have to make sure that you understand the insecurity of what you're doing. You REALLY need to go and read about <a href="http://en.wikipedia.org/wiki/SQL_injection" rel="nofollow">SQL injection</a> and re-evaluate how you are designing your queries. You should be escaping ALL of your values (including ones coming out of the database).</p> <p><strong>Answer</strong><br> Ok, so that said, what you <em>should</em> do, is have a <code>pal</code> table which simply links users to other users. The fields would be <code>user_id</code> and <code>pal_id</code>. Both of which are foreign keys of the <code>user_id</code> field in the user table.</p> <p>You can then write a query like this to get the pal information:</p> <pre><code>SELECT usr.* FROM pal INNER JOIN users ON pal.pal_id = users.user_id WHERE pal.user_id = XXX (User's ID here) </code></pre> <p>The array method of storing "pals" is going to get really old, really quick once you start wanting to do more elaborate things (like finding pals of pals and such).</p> <p>You <strong>can</strong> do the query as you have it now if you REALLY want by using the <code>IN</code> keyword (though I recommend changing your DB structure; that will also eliminate the headache of escaping this string). Assuming that your <code>$pal_array</code> variable has a list of user ids separated by a comma: <code>$query_pal_info = "SELECT users.user_id, user_first_name, user_last_name, username, picture_thumb_url, avatar FROM users LEFT JOIN picture ON users.user_id = picture.user_id AND picture.avatar=1 WHERE users.user_id IN ($pal_array)";</code></p>

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