Question: I am trying to track pageviews on my site whether a user is logged in or not and so I'm capturing the sessionId in my log. I only want to show tracking results for session ids that have at some point logged in and so my flow is as follows:

```
$pagestats = $wpdb->get_results("SELECT * FROM wp_user_stats WHERE uid=".$_GET['viewstats']);
$sessionArray = array();

foreach($pagestats as $checkSession)
{
    if( !(in_array($checkSession->sessionId, $sessionArray)))
    {
        $sessionArray[] = $checkSession->sessionId;
        }
    }

```

Next, I am trying to gather all of the stats concerning any session Ids generated by this particular user ...

```
$sessions = join(',',$sessionArray);
$pagestats = $wpdb->get_results("SELECT * FROM wp_user_stats WHERE `sessionId` IN($sessions)") or die(mysql_error());

```

This is the part that throws the error. The error is:

```
Unknown column 'sjhntmqhltknks8pbhr750voe7' in 'where clause'

```

I don't understand why it's trying to find a column that matches the session Id instead of a result within that column.

Answer: 
The problem is that the session id is not numeric and needs to be wrapped in quotes. Something like the following would add proper quote to your `$sessions` string:

`$sessions = "'" . implode("', '", $sessionArray) . "'";`

You should also make sure you are escaping any user supplied input (e.g. `$_GET['viewstats']`) before using them to help avoid SQL injection attacks. It wouldn't hurt to escape `$checkSession->sessionId` as you are adding it to the `$sessionArray` array either.

If you are using a framework (it looks like you may be using WordPress) read the manual for the database component as it may provide functions to handle some of this for you.
This isn't an answer to the specific issue, but isn't the second `SELECT` statement unnecessary? If it is all stored in one table (unless there is a typo...) then `SELECT * FROM wp_user_stats WHERE uid=$_GET['view_stats']` would retrieve all sessions for that user. Perhaps you need it done for multiple users? Even at that, you could simply do a `GROUP BY` clause. 

Maybe I'm missing something though -- if so, sorry.