On Thursday 27 February 2003 01:00, Luis A wrote:
> $link = mysql_connect("localhost", "nobody");
> mysql_select_db("mydb", $link);
> $result = mysql_query("SELECT * FROM opina", $link);
> echo "Nombre: ".mysql_result($result, 0, "nombre")."";
[snip]
You need to use the mysql_fetch_*() function
i dont know whyyy he does not inser into the databse the values ??
take a look
aparently hje works all rigth
but when i present this var
";
echo "Pais: ".mysql_result($result, 0, "pais")."";
echo "Provincia: ".mysql_result($result, 0, "provincia")."";
echo "URL :".mysql_result($result,
the mysql server guives to me this error now
tk a look
¡Gracias! Hemos recibido sus datos.
Warning: mysql_free_result(): supplied argument is not a valid MySQL result
resource in c:\apache group\apache\htdocs\1\2\registered.php3 on line 39
Warning: mysql_close(): supplied argument is not a val
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