MySQL 从基于 email 的表中获取 ID 然后使用 ID 和更多值在另一个表中创建一行

Question

I have two tables users and features :

users
+----+-----------------+
| id | email           |
+----+-----------------+
|  1 | test1@gmail.com |
|  2 | test2@gmail.com |
|  3 | test3@gmail.com |
|  4 | test4@gmail.com |
|  5 | test5@gmail.com |
+----+-----------------+

features
+------------+---------+---------------------+------------+
| feature_id | user_id | feature_name        | can_access |
+------------+---------+---------------------+------------+
|          1 |       1 | automated-investing |          1 |
|          2 |       1 | crypto              |          0 |
|          3 |       2 | crypto              |          0 |
|          4 |       3 | automated-investing |          0 |
|          5 |       4 | automated-investing |          1 |
|          7 |       1 | financial-tracking  |          1 |
|          8 |       2 | financial-tracking  |          0 |
|          9 |       3 | financial-tracking  |          1 |
|         10 |       4 | financial-tracking  |          0 |
+------------+---------+---------------------+------------+

I am trying to get the id from users based on email and then create a new row in features using that id , feature_name and can_access ( feature_name and can_access are from a request).

I am able to do it but it is split into two parts and I am looking to do it with a single query.

Current query:

// Get ID from email
SELECT id FROM users WHERE email="test5@gmail.com";
// Insert values into features 
INSERT INTO features(user_id, feature_name, can_access) VALUES(5,"crypto", 1);

Expected output:

+------------+---------+---------------------+------------+
| feature_id | user_id | feature_name        | can_access |
+------------+---------+---------------------+------------+
|          1 |       1 | automated-investing |          1 |
|          2 |       1 | crypto              |          0 |
|          3 |       2 | crypto              |          0 |
|          4 |       3 | automated-investing |          0 |
|          5 |       4 | automated-investing |          1 |
|          7 |       1 | financial-tracking  |          1 |
|          8 |       2 | financial-tracking  |          0 |
|          9 |       3 | financial-tracking  |          1 |
|         10 |       4 | financial-tracking  |          0 |
|         11 |       5 | crypto              |          1 |
+------------+---------+---------------------+------------+

Answer 1

INSERT INTO features (user_id, feature_name, can_access) 
   SELECT id, "crypto", 1 FROM users WHERE email="test5@gmail.com";

So instead of specifying the values, you use the select statement.