@mysql_select_db("be"); // this doesn't fail, because only the second
(UPDATE) query fails. The first query (SELECT) is done!
----- Original Message -----
From: "Marek Kilimajer" <[EMAIL PROTECTED]>
To: "Nuno Lopes" <[EMAIL PROTECTED]>
Cc: "MySQL List" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Tuesday, January 07, 2003 5:06 PM
Subject: Re: [PHP] Re: PHP and MySQL bug
> @mysql_select_db("be"); -- this failed
> do echo mysql_error(); to see what went wrong
>
>
>
> Nuno Lopes wrote:
>
> >I done a echo of Mysql_error and it returned:
> >'Nenhum banco de dados foi selecionado'
> >
> >(I have the mysql server in portuguese, but the translation is something
> >like 'no db was selected')
> >
> >
> >----- Original Message -----
> >From: "David Freeman" <[EMAIL PROTECTED]>
> >To: <[EMAIL PROTECTED]>
> >Sent: Sunday, January 05, 2003 10:29 PM
> >Subject: RE: [PHP] Re: PHP and MySQL bug
> >
> >
> >
> >
> >> > @MYSQL_QUERY("UPDATE d SET h='$h' WHERE id='$id'"); // this
> >> > query doesn't work!!!!
> >>
> >>Personally, I'd call it bad programming practice to do a database update
> >>and not check to see if it worked or not. In this case, how are you
> >>determining that the query did not work? Are you manually checking the
> >>database? You don't have anything in your code to check the status of
> >>this query.
> >>
> >>Perhaps this might get you somewhere:
> >>
> >>$qid = @mysql_query("UPDATE d SET h = '$h' WHERE id = '$id'");
> >>
> >>if (isset($qid) && mysql_affected_rows() == 1)
> >>{
> >> echo "query executed";
> >>} else {
> >> echo "query failed: " . mysql_error();
> >>}
> >>
> >>At least this way you might get some indication of where the problem is.
> >>
> >>CYA, Dave
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