Обсуждение: novice, question

anyone have an idea why this won't work? when i print out "i" it says 893,
but it should be 41. and the values are null for the array $userinfo.

thanks in advance,

andy arledge

if(!($result= pg_exec($link, "SELECT * from signup, info1, info2 where
        (signup.username='$uname' or
         info1.username='$uname' or
         info2.username='$uname');"))){
         DisplayErrMsg(sprintf("Error in executing line %s stmt",
$selectStmt));
         exit();
    }
    for($i = 0; $userinfo = @pg_fetch_object($result, $i); $i++);

Andy Arledge wrote:
>
> anyone have an idea why this won't work? when i print out "i" it says 893,
> but it should be 41. and the values are null for the array $userinfo.
>
> thanks in advance,
>
> andy arledge
>
> if(!($result= pg_exec($link, "SELECT * from signup, info1, info2 where
>                 (signup.username='$uname' or
>                  info1.username='$uname' or
>                  info2.username='$uname');"))){
>          DisplayErrMsg(sprintf("Error in executing line %s stmt",
> $selectStmt));
>          exit();
>     }
>         for($i = 0; $userinfo = @pg_fetch_object($result, $i); $i++);

Why is there an '@' in front of that pg_fetch_object call?

Cheers,
                    Andrew.
--
_____________________________________________________________________
           Andrew McMillan, e-mail: Andrew@catalyst.net.nz
Catalyst IT Ltd, PO Box 10-225, Level 22, 105 The Terrace, Wellington
Me: +64 (21) 635 694, Fax: +64 (4) 499 5596, Office: +64 (4) 499 2267

Hey,

What you might have to do is
$total_rows = pg_numrows($result);

do {
pg_fetch_object($result,$i);
$i++;
} while ($i < $total_rows);

>Andy Arledge wrote:
> >
> > anyone have an idea why this won't work? when i print out "i" it says 893,
> > but it should be 41. and the values are null for the array $userinfo.
> >
> > thanks in advance,
> >
> > andy arledge
> >
> > if(!($result= pg_exec($link, "SELECT * from signup, info1, info2 where
> >                 (signup.username='$uname' or
> >                  info1.username='$uname' or
> >                  info2.username='$uname');"))){
> >          DisplayErrMsg(sprintf("Error in executing line %s stmt",
> > $selectStmt));
> >          exit();
> >     }
> >         for($i = 0; $userinfo = @pg_fetch_object($result, $i); $i++);
>
>Why is there an '@' in front of that pg_fetch_object call?

So it doesn't print an error when there are no results left.


------------------------
Chris Smith
http://www.squiz.net