新手求教(这个语句总是报错)

sky_sun_wei

里建有guestbook表,里面有id,name等

$sqldo="select count(*) as title from guestbook where id<1";//

$result=mysql_query($sqldo);

$message_count=mysql_result($result,0,"title";

报错:
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in e:\my_php_app\liuyan\guestbook.php on line 12

是我的mysql_result()用错了吗?

各位请帮忙看一下.

ukalpa

指定数据库了么?

bull_bull

我遇到过这个问题，可能是mysql_result()返回的都是数组，所以必须这样写就不会报错了：
$sqldo="select count(*) as title from guestbook where id<1";//

$result=mysql_query($sqldo);

if ($result) {

$r1 = mysql_fetch_array($result);

$message_count = $r1[0];

}