这是我从教程中获得的代码,当我尝试登录时,它不工作!它说:PHP登录表单错误
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\index.php on line 13
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\index.php on line 16
<?php
include("config.php");
session_start();
if($_SERVER["REQUEST_METHOD"] == "POST") {
// username and password sent from form
$myusername = mysqli_real_escape_string($db,$_POST['username']);
$mypassword = mysqli_real_escape_string($db,$_POST['password']);
$sql = "SELECT * FROM tbl_users WHERE username = '$myusername' and password = '$mypassword'";
$result = mysqli_query($db,$sql);
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
$active = $row['active'];
$count = mysqli_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row
if($count == 1) {
session_register("myusername");
$_SESSION['login_user'] = $myusername;
header("location: welcome.php");
}else {
$error = "Your Login Name or Password is invalid";
}
}
?>
任何帮助将是非常赞赏! 在此先感谢。
用“.username。”更改用户名后...这是代码!
<?php
include("config.php");
session_start();
if($_SERVER["REQUEST_METHOD"] == "POST") {
// username and password sent from form
$myusername = mysqli_real_escape_string($db,$_POST['username']);
$mypassword = mysqli_real_escape_string($db,$_POST['password']);
$sql = "SELECT * FROM tbl_users WHERE username = '".$myusername."' and password = '".$mypassword."'";
$result = mysqli_query($db,$sql);
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
$active = $row['active'];
$count = mysqli_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row
if($count == 1) {
session_register("myusername");
$_SESSION['login_user'] = $myusername;
header("location: welcome.php");
}else {
$error = "Your Login Name or Password is invalid";
}
}
?>
<html>
<head><title>Login Page PHP Script</title></head>
<body>
<div align="center">
<div style="width:300px; border: solid 1px #006D9C; " align="left">
<?php
if(isset($errMsg)){
echo '<div style="color:#FF0000;text-align:center;font-size:12px;">'.$errMsg.'</div>';
}
?>
<div style="background-color:#006D9C; color:#FFFFFF; padding:3px;"><b>Login</b></div>
<div style="margin:30px">
<form action="" method="post">
<label>Username :</label><input type="text" name="username" class="box"/><br /><br />
<label>Password :</label><input type="password" name="password" class="box" /><br/><br />
<input type="submit" name='submit' value="Submit" class='submit'/><br />
</form>
</div>
</div>
</div>
</body>
</html>
当我这样做,我得到这个......“解析错误:语法错误,在C意外的文件结尾:\ XAMPP \ htdocs中\ login.php中上线141” –
而这个:' username ='“。$ myusername。”'AND password ='“。$ mypassword。”''? – Jer
@MichealHarker这不是一个问题现在..我想要的是一些工作代码登录 –