我正在尝试获取 referral_in
的行数和referral_out
存在于(作为单独的变量)。这是我的代码:
$username = $_SESSION['username'];
$connect = mysql_connect("xxxx", "xxxx", "xxxx!") or die("Couldnt Connect to Server");
mysql_select_db("xxxxx") or die("Couldnt find database");
$samecheck = mysql_query("SELECT `referral_in` FROM `users` WERE `username`=$username");
$same = mysql_num_rows($namecheck);
$leadcheck = mysql_query("SELECT `referral_out` FROM `users` WERE `username`=$username");
$leading = mysql_num_rows($namecheck);
echo "$leading / $same"
何时执行此操作,我收到此错误:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/content/50/8492150/html/buyarandom/member.php on line 23
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/content/50/8492150/html/buyarandom/member.php on line 25
最佳答案
您忘记了 $username
周围的引号
$samecheck = mysql_query("SELECT `referral_in` FROM `users` WERE `username`='".$username."'");
还请尝试转义$username
,因为您的代码容易受到SQL注入(inject)
$samecheck = mysql_query("SELECT `referral_in` FROM `users` WERE `username`='".mysql_real_escape_string($username)."'");
旁注:不要使用mysql_query
而是使用mysqli
This extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the MySQLi or PDO_MySQL extension should be used.
关于php - 为什么我收到错误? (PHP、MySQL),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17604901/