你能在两个表之间做一个“INSERT SELECT WHERE”吗？

Question

So I've been trying to create a simple friend system. When you register, you get randomized numbers and chars of 8 in length. I save this number in a column to the user. I have been trying to insert the currently sessioned user(PHP), $SessionUser together with the friends' username, uidUsers using an "INSERT SELECT WHERE" statement, but something goes wrong. Heres something I have tried:

$sql = mysqli_query($conn, "INSERT into friends (uid1, uid2) 
    values($sessionUser, (SELECT  uidUsers FROM users WHERE idFriendCode = $idFriendCode)");

Inside the table, friends , I have two columns, uid1 (the sessioned user/sender) and uid2 (the reciever, name of specified $idFriendCode ). I want to insert the $sessionUser to the uid1 and whatever username ( uidUsers ) that matches with the $idFriendCode to the uid2 . This does not seem to work and I don't know why. I imagine the problem is that I can't use a PHP variable like this.

I know that I don't use prepared statements. I have tried to implement it, but I think it's much harder than just using a basic mysqli_query().

Answer 1

You may phrase your insert as an INSERT INTO ... SELECT :

INSERT into friends (uid1, uid2) 
SELECT $sessionUser, uidUsers
FROM users
WHERE idFriendCode = $idFriendCode;

Note that you should ideally be using a prepared statement here, so the above should look like:

INSERT into friends (uid1, uid2) 
SELECT ?, uidUsers
FROM users
WHERE idFriendCode = ?;

Answer 2

Try having a new variable for select and use it in the insert query

example:

$select_qr='SELECT  uidUsers FROM users WHERE idFriendCode = $idFriendCode'

$sql = mysqli_query($conn, "INSERT into friends (uid1, uid2) 
    values($sessionUser, $select_qr)");