[PHP-ES] de nuevo OT de mysql

From: Xytras ( pollo .en. mzt.megared.net.mx)
Date: Wed Mar 24 2004 - 03:30:28 CET

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el dia de ayer envie un mail con un problema similar, tengo este codigo:

$resultado=mysql_query("SELECT asegurado, numref, nomase, ajustador FROM
datos WHERE '$tipo' LIKE '%$busca%' ORDER BY numref", $link);
if($row=mysql_fetch_array($resultado))

y me arroja este error: *Warning*: mysql_fetch_array(): supplied
argument is not a valid MySQL result resource in
*/srv/www/htdocs/sistema/terminados/buscabaja.php* on line *11

ayer me dijeron que pusiera entre ' ' las variables de PHP, y funciono,
hoy hice lo propio y volvio a hacer lo mismo, alguna idea?

Gracias

Xytras
*

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