sur*_*ava 2 php mysql session login
我创建登录会话,它已显示错误:未识别myusername和mypassword在这些行中.我不知道为什么.
$myusername=$_POST['myusername'];
$mypassword=$_POST['mypassword'];
Run Code Online (Sandbox Code Playgroud)
这是我的checklogin.php
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$myusername=$_POST['myusername'];
$mypassword=$_POST['mypassword'];
$sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword'";
$result=mysql_query($sql);
$count=mysql_num_rows($result);
if($count==1){
session_register("myusername");
session_register("mypassword");
header("location:login_success.php");
}
else {
echo "Wrong Username or Password";
}
?>
Run Code Online (Sandbox Code Playgroud)
这是我的登录表格:
<form action='checklogin.php' method="post" >
Username
<input name='myusername' type="text">
Password
<input name='mypassword' type="password">
<input type="submit" name="Submit" value="Login">
</form>
Run Code Online (Sandbox Code Playgroud)
检查$_POST['myusername']并$_POST['mypassword']在使用前设置
if(isset($_POST['myusername']) && isset($_POST['mypassword']) && !empty(trim($_POST['myusername'])) && !empty(trim($_POST['mypassword']))) {
}
Run Code Online (Sandbox Code Playgroud)
它看起来login_true.php也应该是形式动作checklogin.php
<form action='checklogin.php' method="post" >
Run Code Online (Sandbox Code Playgroud)
因为你发送数据login_true.php不是check_login.php因为你得到错误:unidentified bla bla
好读: