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+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3436.Find%20Valid%20Emails/README.md
+tags:
+ - 数据库
+---
+
+
+
+# [3436. Find Valid Emails](https://leetcode.cn/problems/find-valid-emails)
+
+[English Version](/solution/3400-3499/3436.Find%20Valid%20Emails/README_EN.md)
+
+## 题目描述
+
+
+
+Table: Users
+
+
++-----------------+---------+
+| Column Name | Type |
++-----------------+---------+
+| user_id | int |
+| email | varchar |
++-----------------+---------+
+(user_id) is the unique key for this table.
+Each row contains a user's unique ID and email address.
+
+
+Write a solution to find all the valid email addresses. A valid email address meets the following criteria:
+
+
+ - It contains exactly one
@ symbol.
+ - The part before the
@ symbol contains only alphanumeric characters and underscores.
+ - The part after the
@ symbol contains a domain name that starts with a letter and ends with .com.
+
+
+Return the result table ordered by user_id in ascending order.
+
+
+Example:
+
+
+
Input:
+
+
Users table:
+
+
++---------+---------------------+
+| user_id | email |
++---------+---------------------+
+| 1 | alice@example.com |
+| 2 | bob_at_example.com |
+| 3 | charlie@example.net |
+| 4 | david@domain.com |
+| 5 | eve@invalid |
++---------+---------------------+
+
+
+
Output:
+
+
++---------+-------------------+
+| user_id | email |
++---------+-------------------+
+| 1 | alice@example.com |
+| 4 | david@domain.com |
++---------+-------------------+
+
+
+
Explanation:
+
+
+ - alice@example.com is valid because it contains one
@, alice is alphanumeric, and example.com starts with a letter and ends with .com.
+ - bob_at_example.com is invalid because it contains an underscore instead of an
@.
+ - charlie@example.net is invalid because the domain does not end with
.com.
+ - david@domain.com is valid because it meets all criteria.
+ - eve@invalid is invalid because the domain does not end with
.com.
+
+
+
Result table is ordered by user_id in ascending order.
+
Table: Users
+
+
++-----------------+---------+
+| Column Name | Type |
++-----------------+---------+
+| user_id | int |
+| email | varchar |
++-----------------+---------+
+(user_id) is the unique key for this table.
+Each row contains a user's unique ID and email address.
+
+
+Write a solution to find all the valid email addresses. A valid email address meets the following criteria:
+
+
+ - It contains exactly one
@ symbol.
+ - The part before the
@ symbol contains only alphanumeric characters and underscores.
+ - The part after the
@ symbol contains a domain name that starts with a letter and ends with .com.
+
+
+Return the result table ordered by user_id in ascending order.
+
+
+Example:
+
+
+
Input:
+
+
Users table:
+
+
++---------+---------------------+
+| user_id | email |
++---------+---------------------+
+| 1 | alice@example.com |
+| 2 | bob_at_example.com |
+| 3 | charlie@example.net |
+| 4 | david@domain.com |
+| 5 | eve@invalid |
++---------+---------------------+
+
+
+
Output:
+
+
++---------+-------------------+
+| user_id | email |
++---------+-------------------+
+| 1 | alice@example.com |
+| 4 | david@domain.com |
++---------+-------------------+
+
+
+
Explanation:
+
+
+ - alice@example.com is valid because it contains one
@, alice is alphanumeric, and example.com starts with a letter and ends with .com.
+ - bob_at_example.com is invalid because it contains an underscore instead of an
@.
+ - charlie@example.net is invalid because the domain does not end with
.com.
+ - david@domain.com is valid because it meets all criteria.
+ - eve@invalid is invalid because the domain does not end with
.com.
+
+
+
Result table is ordered by user_id in ascending order.
+