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
I was having the same problem for a while... although, I
was using this:
for($i=0;$imysql_num_rows($result);$i++)
doStuffTo(mysql_result($result, $i, foo);
If I remember correctly... it has to do with zero-based
indexing versus 1-based indexing. Now... I think I
fixed it by using =
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
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 row_
TH.
Jayme.
-Mensagem Original-
De: Trond Erling Hundal [EMAIL PROTECTED]
Para: PHP-DB-LIST [EMAIL PROTECTED]
Enviada em: segunda-feira, 5 de maro de 2001 09:56
Assunto: [PHP-DB] mysql_fetch_array problem...!
I want to run a query to my db, fetching different fields from three
different ta
