Question

我想创建登录页面。首先，我从login.html获取用户名和密码，并将它们发送到login.php以检查其是否可用。但它总是会出错，我无法解决这个问题

的login.html

        <form action="login.php" method="post">
           <div class="containerLogin">
            <label><b>Username</b></label>
            <input type="text" placeholder="Enter Username"name="username" required>

            <label><b>Password</b></label>
            <input type="password" placeholder="Enter Password" name="password" required>
          </div>

          <div class="containerLogin" style="background-color:#f1f1f1">
            <input type="submit" name="submit" value="submit" class="btn btn-primary">
            <span class="password">Forgot <a href="#">password?</a></span>
          </div>
        </form>

的login.php

<?php
  include_once "connection.php";
  if (isset($_POST['submit'])) { 
    session_start();
      if($_POST['username'] && $_POST['password']) {
        $username  =  $_POST['username'];
        $password  =  $_POST['password'];
        $query = mysqli_query("SELECT * FROM studenttable WHERE username='$username' and password='$password'");
        $res = mysqli_query($con, $query);
        $count_user = mysqli_num_rows($res);
        if($count_user==1)
        {
            $row = mysqli_fetch_array($res);
            $_SESSION['username'] = $row['username'];
            $_SESSION['password'] = $row['password'];
            header("location:dashboard.php?success=1");
        }else{
                $error = 'Incorrect Username, Password and Branch.';
            }
      }
}
?>

错误 login.php

Answer 1

$query变量不应为mysqli_query，而应为字符串，因为您不能将查询作为参数传递给另一个查询。将该行（24）替换为此：

$query = "SELECT * FROM studenttable WHERE username='$username' and password='$password'";

php mysqli_query（）错误

1 个答案: