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);
}
?>