You must use a loop to show each row of your query.
"mysql_fetch_array($result)" only get the current row.
while($row = mysql_fetch_array($result) ){
// do something
}
The "mysql_fetch_array" function returns an associative array where you can
use the field names to have access to its value :
$row = mysql_fetch_array($result) ;
$v1 = $row["fieldname1"] ;
$v2 = $row["fieldname2"] ;
$v3 = $row["fieldname3"] ;
You're going to use the alias names you are creating on the query.
See more details on php manual at mysql functions.
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 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 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]" ;
>
>
>
>
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