Oh my.. seems i didnt see the reference of $queryresult. ignore my comment
:)

but the first mysql_query is useless.

;Mats



> -----Original Message-----
> From: Mats Remman [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, May 24, 2001 12:35 PM
> To: annabelle.imray; [EMAIL PROTECTED]
> Subject: RE: [PHP-DB] Am I missing something important here ?
>
>
> Your problem here is simply that the $queryresult gets lost in
> the function.
> It is never returned.
>
> change return true;
> into   return $queryresult;
>
>
> Mats Remman
>
>
> > -----Original Message-----
> > From: annabelle.imray [mailto:[EMAIL PROTECTED]]
> > Sent: Thursday, May 24, 2001 1:02 AM
> > To: [EMAIL PROTECTED]
> > Subject: [PHP-DB] Am I missing something important here ?
> >
> >
> > I have been given an example of how to use php & mysql from a hosting
> > company.  They have the following function:
> >
> >         #====
> >         function ExecuteQuery ($linkdb, &$queryresult, $querytext) {
> >
> >          $queryresult = mysql_query($querytext, $linkdb);
> >
> >           if (($queryresult != false) and (mysql_errno() == 0)) {
> >               return true;
> >          } else {
> >               return false;
> >          }
> > }
> > #====
> >
> > And an example of it's usage:
> >
> >
> >     $query = "SELECT * FROM test";
> >
> >      if ($mysql_result = mysql_query($query, $linkdb)) {
> >           if (ExecuteQuery($linkdb, $result, $query)) {
> >                while ($row = NextRow($result)) {
> >                     print('"'.$row[1].'" : "'.$row[2].'"<br>');
> >                }
> >           }
> >       }
> >
> > Can anyone tell me why the first "if" exists in their usage
> > example, e.g. if
> > ($mysql_result = mysql_query($query, $linkdb)) {
> >
> > As the $mysql_result variable is not used after this statement
> > and the query
> > is performed again in the ExecuteQuery() function;
> >
> > As I have said, am I missing something important here?
> >
> > Thanks
> >
> >
> >
> >
> > --
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> >
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