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Posted by totalstranger on 10/14/05 11:53
On or about 7/20/2006 9:33 AM, it came to pass that totalstranger wrote:
> 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;
Aaaargh
This error was caused by my messing up the line following this statement
from
$result = mysql_query( $query )
to
result = mysql_query( $query )
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