点击登录按钮后,它仍保留在同一页面上,例如:如果输入错误的用户ID或密码,它应该回显一些内容,但它不会回显任何内容并保持不变 LOGIN.PHP FILE的代码是:
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>title of the document</title>
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
<form action"signin.php" method="POST">
<input type="text" name="uid" placeholder="Username"><br>
<input type="password" name="password" placeholder="PAssword"><br>
<button type="submit">SIGN IN</button><br>
</form>
<br><br><br><br>
<form action="signup1.php" method="POST">
<input type="text" name="firstname" placeholder="Firstname"><br>
<input type="text" name="lastname" placeholder="Lastname"><br>
<input type="text" name="uid" placeholder="Username"><br>
<input type="password" name="password" placeholder="PAssword"><br>
<button type="submit">SIGN UP</button><br>
</form>
</body>
</html>
signin.php文件是:
<?php
include 'dbh1.php';
$userid=$_POST['uid'];
$pwd=$_POST['password'];
$sql="select * from userlogin where uid='$userid' AND password='$pwd'";
$result = $conn->query($sql);
if (!$row = $result->fetch_assoc())
{
echo "YOU ARE NOT LOGGED IN INCORRECT CREDENTIALS!!";
}
else {
echo "SUCCESFULLY LOGGED IN!!";
}
答案 0 :(得分:0)
我试过这个并且它正在运行。这里学习准备好的陈述
<form action"signin.php" method="POST">
<input type="text" name="uid" placeholder="Username"><br>
<input type="password" name="password" placeholder="PAssword"><br>
<input type="submit" value="Sign In" />
</form>
<?php
include 'database.php';
$userid=$_POST['uid'];
$pwd=$_POST['password'];
$stmt = $conn->prepare("select * from userlogin where uid=? AND password=?");
$stmt->bind_param("ss",$userid, $pwd);
$stmt->execute();
$result = $stmt->get_result();
$row = $result->fetch_assoc();
$stmt->close();
通过准备好的陈述,你已经取得了很多成就。试着检查一下。