NYCPHP Meetup

[Nestor Florez]

Maybe, My eyes are old but I do not see the f table in your from
statement.

Nestor :-)

-----Original Message-----
From: Phil Powell <soazine at erols.com>
Sent: Sep 12, 2003 1:27 PM
To: NYPHP Talk <talk at lists.nyphp.org>
Subject: [nycphp-talk] had to give you guys a break

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<DIV><FONT face=Arial size=2>I'm not kidding, the only reason yesterday you 
didn't hear from me was because I wasn't coding, but today I am doing something 
quick, and yes, as always it failed.. right at the SQL statement:</font></div>
<DIV><FONT face=Arial size=2></font> </div><FONT face=Arial size=2>
<DIV><BR>   $sql = 'SELECT v.nnet_produkt_varegruppe_navn, ' 
.<BR>          
'       f.nnet_produkt_farge_navn, ' 
.<BR>          
'       s.nnet_produkt_storrelse_navn, ' 
.<BR>          
'       k.nnet_produkt_kvalitet_navn, ' 
.<BR>          
'       p.* ' 
.<BR>          'FROM nnet_produkt 
p, nnet_produkt_varegruppe v, ' 
.<BR>          
'     nnet_produkt_kvalitet k, nnet_produkt_storrelse s ' 
.<BR>          'WHERE 
p.nnet_produkt_varegruppe_code = v.nnet_produkt_varegruppe_code ' 
.<BR>          ' AND 
p.nnet_produkt_farge_code = f.nnet_produkt_farge_code ' 
.<BR>          ' AND 
p.nnet_produkt_storrelse_id (+) = s.nnet_produkt_storrelse_id ' 
.<BR>          ' AND 
p.nnet_produkt_kvalitet_id (+) = k.nnet_produkt_kvalitet_id';</div>
<DIV> </div>
<DIV>I am getting "invalid SQL near (+).  That is totally VALID SQL syntax 
for an outer join!  If mySQL doesn't allow for that, what am I supposed to 
do in lieu of the fact that I have to logically join two tables together or if 
the one is null (the case for an outer join)?</div>
<DIV> </div>
<DIV>Phil</font></div></body></html>