I was having the same problem for a while... although, I
was using this:
for($i=0;$i
Good luck,
Ben
Quoting BrianSander <[EMAIL PROTECTED]>:
> Greetings,
>
> I'm experiencing the strangest problem and I was
> wondering if anyone
> else has had the same problem.
>
> I have a fairly simple scr
Greetings,
I'm experiencing the strangest problem and I was wondering if anyone
else has had the same problem.
I have a fairly simple script setup that queries a mySQL database and
displays the records in a HTML table. Everything works fine except it
keeps omitting the first record. Running the
s.
HTH.
Jayme.
-Mensagem Original-
De: Trond Erling Hundal <[EMAIL PROTECTED]>
Para: PHP-DB-LIST <[EMAIL PROTECTED]>
Enviada em: segunda-feira, 5 de março de 2001 09:56
Assunto: [PHP-DB] mysql_fetch_array problem...!
> I want to run a query to my db, fetching diffe
> How can I refer to one specific row in this query..?
> What I mean is, how can i refer to result row number 4...?
>
> If I only selected rows from one table I could do something like this:
>
> $i = mysql_fetch_array($sql) ;
> echo "$i[4]" ;
Actually, no. mysql_fetch_array return the _current r
elism
eBusiness Infrastructure Technology
http://www.openlinksw.com
> -Original Message-
> From: Trond Erling Hundal [mailto:[EMAIL PROTECTED]]
> Sent: Monday, March 05, 2001 7:56 AM
> To: PHP-DB-LIST
> Subject: [PHP-DB] mysql_fetch_array problem...!
>
>
> I want to r
I want to run a query to my db, fetching different fields from three
different tables.
In order to recognise the individual fields I give them names:
select portal.portal as portal, portal.portalid as id... etc etc
How can I refer to one specific row in this query..?
What I mean is, how can i r
