我是PHP的新手,所以我决定遵循this tutorial进行简单的登录.我得到了代码设置,但是当我尝试登录时出现此错误:
Warning: MysqL_num_rows(): supplied argument is not a valid MysqL result resource in (a long file path to the script) on line 27
我从教程中获得的代码是:
<?PHP
ob_start();
$host="thehost"; // Host name
$username="myusername"; // MysqL username
$password="mypass"; // MysqL password
$db_name="test"; // Database name
$tbl_name="members"; // Table name
// Connect to server and select databse.
MysqL_connect("$host", "$username", "$password")or die("cannot connect");
MysqL_select_db("$db_name")or die("cannot select DB");
// Define $myusername and $mypassword
$myusername=$_POST['myusername'];
$mypassword=$_POST['mypassword'];
// To protect MysqL injection (more detail about MysqL injection)
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$myusername = MysqL_real_escape_string($myusername);
$mypassword = MysqL_real_escape_string($mypassword);
$sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword'";
$result=MysqL_query($sql);
// MysqL_num_row is counting table row
$count=MysqL_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row
if($count==1){
// Register $myusername, $mypassword and redirect to file "login_success.PHP"
session_register("myusername");
session_register("mypassword");
header("location:login_success.PHP");
}
else {
echo "Wrong Username or Password";
}
ob_end_flush();
?>
Aslo,作者给出了PHP5版本和普通的PHP版本.我都尝试过并得到相同的错误.如果有人知道为什么会这样,将不胜感激.
解决方法:
$result=MysqL_query($sql) or die ("bad sql query: ".MysqL_error());
因此,如果sql错误;我收到一个有用的错误,通常是有罪的sql片段
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