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Posted by totalstranger on 10/14/82 11:53
On or about 7/20/2006 9:11 AM, it came to pass that totalstranger wrote:
> Can someone please tell me why this PHP statement throws the following
> PHP error:
> Parse error: parse error, unexpected '=' in
> /var/www/html/PHP/XXXXXXXX.php on line 335 (line 335 is the Where
> portion of the script below)
>
> $query = 'SELECT Venue.*, '
> . ' CONCAT_WS(\' \', Users2.Firstname, Users2.Lastname,
> Date_Format(Venue.Vnu_Add_Date, \'%b %e, %Y %T\' )) as Mnt_Add, '
> . ' CONCAT_WS(\' \', Users.Firstname, Users.Lastname,
> Date_Format(Venue.Lastmaint, \'%b %e, %Y %T\' )) as Mnt_Data '
> . ' FROM `Venue`'
> . ' LEFT JOIN Users as Users2 ON Venue.Vnu_Add_Userid = Users2.Id'
> . ' LEFT JOIN Users ON Venue.Maint_Userid = Users.Id'
> . ' WHERE Venue.Id = '.$Id_clean;
>
>
> Similar code on other scripts works without a problem and I'm unable to
> find this error.
>
> Thank you
This version also fails with same error
$query = "SELECT Venue.*, "
.. " CONCAT_WS(' ', Users2.Firstname, Users2.Lastname,
Date_Format(Venue.Vnu_Add_Date, '%b %e, %Y %T' )) as Mnt_Add, "
.. " CONCAT_WS(' ', Users.Firstname, Users.Lastname,
Date_Format(Venue.Lastmaint, '%b %e, %Y %T' )) as Mnt_Data "
.. " FROM `Venue`"
.. " LEFT JOIN Users as Users2 ON Venue.Vnu_Add_Userid = Users2.Id"
.. " LEFT JOIN Users ON Venue.Maint_Userid = Users.Id"
.. " WHERE Venue.Id = ".$Id_clean;
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