PHP 4.0 Bug #6666 Updated: mySQL result
| From: | Bug Database | Date: | Tue, 12 Sep 2000 05:50:44 +0000 |
| Subject: | PHP 4.0 Bug #6666 Updated: mySQL result | ||
| Groups: | php.dev | ||
| Request: | Send a blank email to php-dev+get-32940@lists.php.net to get a copy of this message | ||
ID: 6666
Updated by: derick
Reported By: ldcosta@tutopia.com
Status: Closed
Bug Type: MySQL related
Assigned To:
Comments:
Hello, this list is NOT for reporting userland errors, but you can use MySQL as follows:
$query = "select * from t1";
$result = mysql_query ($query);
while ($array_r = mysql_fetch_array ($restul))
{
printf ("C1: %s", $array_r["c1"]);
}
Please read the manual more carefully, I'm sure it is in it.
Previous Comments:
---------------------------------------------------------------------------
[2000-09-12 01:40:51] ldcosta@tutopia.com
Hi, I'm form Argentina, and I've done the configuration of PHP4 --with-mysql,
--with-apxs and --enable-ftp. Now, i'm trying to work with mysql and i can't,
it says that it has an error. Here's the code:
<html>
<body>
<?
/*$conex = mysql_connect("localhost", "nobody");
mysql_select_db("hardsite", $conex);
$resultado = mysql_query("SELECT * FROM hardsite", $conex);
echo "Vendedor: ".mysql_result($resultado, 0,
"vendedor")."<BR>";
echo "e-mail: ".mysql_result($resultado, 0, "mail")."<BR>";
echo "Producto: ".mysql_result($resultado, 0,
"producto")."<BR>";
echo "Precio: ".mysql_result($resultado, 0, "precio")."<BR>";
echo "Tipo de pago: ".mysql_result($resultado, 0,
"pago")."<BR>";
echo "Descripción: ".mysql_result($resultado, 0,
"descripcion")."<BR>";
*/
$link = mysql_connect("localhost", "nobody");
mysql_select_db("hardsite", $link);
$result = mysql_query("SELECT * FROM ventas", $link);
echo "Nombre: ".mysql_result($result, 0, "vendedor")."<br>";
echo "Dirección: ".mysql_result($result, 0, "mail")."<br>";
?>
</body>
</html>
And here the result of the page
Warning: Supplied argument is not a valid MySQL result resource in /home/httpd/html/miprimeradb.php
on line 17
Nombre:
Warning: Supplied argument is not a valid MySQL result resource in /home/httpd/html/miprimeradb.php
on line 18
Dirección:
-----------------------------
I know that i'have commented the first part, but it was because i was trying,
and it didn't work. But the second part was part of an example that i've
downloaded of the net, i've changed the name of the database and the fields.
Well, i hope you can help me
Regards,
Leandro Costa
---------------------------------------------------------------------------
Full Bug description available at: http://bugs.php.net/?id=6666