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Posted by Luigi Donatello Asero on 07/12/05 15:54
"Luigi Donatello Asero" <jaggillarfotboll@telia.com> skrev i meddelandet
news:K2PAe.29003$d5.182048@newsb.telia.net...
>
> "Richard Conway" <ric@shoe.com> skrev i meddelandet
> news:NOOAe.4096$2I4.2995@fe05.ams...
> > Luigi Donatello Asero wrote:
> > > "Richard Conway" <ric@shoe.com> skrev i meddelandet
> > > news:FaOAe.2$HQ1.0@fe08.ams...
> > In that case, there is something wrong with your query. Try changing
> > the following line from:
> >
> > $result = mysql_query("SELECT * FROM name of the table", $db);
> >
> > to:
> >
> > $result = mysql_query("SELECT * FROM name of the table", $db) or
> > die(mysql_error());
> >
> > This will trap the error and output it for you so you can get some idea
> > of where you've gone wrong.
>
> After some changes:
>
> <html> <body>
> <?php $db = mysql_connect("localhost", "user",
> "password");
> mysql_select_db("databas", $db);
> $result = mysql_query("SELECT * FROM table", $db) or
That was "FROM name of the table" and not FROM "table"
--
Luigi Donatello (un italiano che vive in Svezia)
https://www.scaiecat-spa-gigi.com/
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