Question

我要做的是使用用户名将一行添加到Guest表（此部分工作正常），并从Guest和Room中获取GuestID和HostID，并使用这些值创建一个新的Room行。我不断从SELECT中收到错误，我不知道为什么。

<html>
<h1><center>Join the Party!<center></h1>
<body>
<center>
<form method="POST">
<input type="text" placeholder="Username" name="username"<br><br>
<input type="text" placeholder="Room ID" name="room"<br><br>
<input type="submit" name="update">
</form>
</center>
</body>
</html>

<?php
    if(isset($_POST['update'])){
        $hostname="localhost";
        $uname="ethanruo_WP5HV";
        $pwd="";
        $dbname="ethanruo_WP5HV";
        $connect=mysqli_connect($hostname,$uname,$pwd,$dbname);
        $username=$_POST['username'];
        $room=$_POST['room'];
        $check1=mysqli_query($connect,"IF EXISTS(SEARCH RoomID FROM Room WHERE RoomID = '$room');");
        if($check1){
            $query1="INSERT INTO Guests VALUES (NULL,'$username');";
            $result=mysqli_query($connect,$query1);
            $gidgrab="SELECT GuestID FROM Guests WHERE Username='$username';"; 
            $gid=mysqli_query($connect,$idgrab);
            $gidgrab="SELECT HostID FROM Room WHERE RoomID='$room';"; 
            $gid=mysqli_query($connect,$idgrab);
            $query2="INSERT INTO Room VALUES ('$room','$hid',$gid');";
            $result2=mysqli_query($connect,$query2);
            if($result && $result2){
                echo "You have been added to our system!";
            }
            else{
                echo "Something happened! :(";
            }
        }
        else{
               echo "That room doesn't exist! :(";
        }
        mysqli_close($connect);
    }
    ?>

Answer 1

表名称不是来宾，而是来宾。 您的代码可能会受到sql注入的影响。始终尝试使用准备好的语句。

无法在PHP mysqli中选择行

1 个答案: