✅ Simple and straightforward, 100% ✅
ExplanationSimple query using LIKE operator and functions CHAR_LENGTH and REPLACE.There are 2 parts here:
email like '%.com' matches all values ending in .com.
# Explanation
Simple query using `LIKE` operator and functions `CHAR_LENGTH` and `REPLACE`.

There are 2 parts here:
- `email like '%.com'` matches all values ending in `.com`.
- `(CHAR_LENGTH(email) - CHAR_LENGTH(REPLACE(email, '@', ''))) = 1` can be split as follows:
  - the diff between the size of the column and the size of the column, assuming all chars `@` are removed. This effectively gives us the count of `@` characters.
  - finally, we want this count to be 1, so that we only allow emails that have exactly 1 `@` character.


# Code
```postgresql []
-- Write your PostgreSQL query statement below
select * from Users where email like '%.com' and (CHAR_LENGTH(email) - CHAR_LENGTH(REPLACE(email, '@', ''))) = 1
```
['PostgreSQL']