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mysqli_query():空查询mysqli_error()只需要1个参数,给定[重复] 0

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我正在创建一个注册页面 . 表格包含电子邮件,密码,确认密码,学生证,名字,第二名,课程,性别和DoB . 我希望页面收集该信息并将其存储在数据库中 .

2个错误:

mysqli_query():空查询

mysqli_error()预计只有1个参数,给定0

$sql = mysqli_query($connection, "INSERT INTO tblaccounts (Email, Password, Student_ID, ID, FirstName, SecondName, Course, Gender, DoB) VALUES ('".$email."','".$password."','".$Student_ID."','".$id."','".$FN."','".$SN."','".$course."','".$gender."','".$dob."')"); 
        $result = mysqli_query($connection, $sql) or die("Database Connection Failed" . mysqli_error());  
        $count = mysqli_num_rows($result);






    <?php
    require_once 'connect.php';
    require_once 'logincheck.php';

        if (($_COOKIE['userID']) == null){
              //true
              //show sign in
              ?>
                <li><p><center><a class="btn btn-primary btn-lg" href="login.php" role="button">Sign In</a><center></p></li>
              <?php
              //show register button
              ?>
                <li><p><center><a class="btn btn-primary btn-lg" href="register.php" role="button">Register</a><center></p></li>
              <?php
        } else {
        //false
        //show 'logged in as'
              ?>
        <li><a href="#">Logged in as: <?php echo ($_COOKIE['user']) ?></a></li>
        <?php//show 'my profile'?>
        <li><a href="#">My Profile</a></li>
        <?php//show 'settings'.?>
        <li><a href="#">Settings</a></li>
        <li><p><center><a class="btn btn-primary btn-lg" href="logout.php" role="button">Sign Out</a><center></p></li>
            <!--
<a href ="login.php">Go back to the login screen.</a>--> <!--logged in menu--> <!--<li><a href="#">User ID: <php echo ($_COOKIE['userID']) ?></a></li>--> </ul> <?php } ?>

1 回答

  • 0

    您正在执行一次查询,然后再尝试使用重新调整对象执行查询 . 将您的代码更改为:

    $sql = "INSERT INTO tblaccounts (Email, Password, Student_ID, ID, FirstName, SecondName, Course, Gender, DoB) VALUES ('".$email."','".$password."','".$Student_ID."','".$id."','".$FN."','".$SN."','".$course."','".$gender."','".$dob."')"; 
            $result = mysqli_query($connection, $sql) or die("Database Connection Failed" . mysqli_error($connection));  
            $count = mysqli_num_rows($result);
    

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