高频题（0）
1. 某个月份内的每一天
| 求某个月份内的每一天，这个月up to date的total sales
delivery_id int
order_created_at timestamp
predicted_store_pickup_time timestamp
actual_store_pickup_at timestamp
requested_delivery_time timestamp
dasher_id int
restaurant_id int
consumer_id int
Table 2:
delivery_id int
sales int (dollar mount)
delivery_id int
order_created_at timestamp
predicted_store_pickup_time timestamp
actual_store_pickup_at timestamp
requested_delivery_time timestamp
dasher_id int
restaurant_id int
consumer_id int
Table 2:
delivery_id int
sales int (dollar mount)
2. LeetCode 175
Table: Person
+-------------+---------+ | Column Name | Type | +-------------+---------+ | personId | int | | lastName | varchar | | firstName | varchar | +-------------+---------+ personId is the primary key column for this table. This table contains information about the ID of some persons and their first and last names.
Table: Address
+-------------+---------+ | Column Name | Type | +-------------+---------+ | addressId | int | | personId | int | | city | varchar | | state | varchar | +-------------+---------+ addressId is the primary key column for this table. Each row of this table contains information about the city and state of one person with ID = PersonId.
Write an SQL query to report the first name, last name, city, and state of each person in the Person table. If the address of a personId is not present in the Address table, report null instead.
Return the result table in any order.
The query result format is in the following example.
Example 1:
Input: Person table: +----------+----------+-----------+ | personId | lastName | firstName | +----------+----------+-----------+ | 1 | Wang | Allen | | 2 | Alice | Bob | +----------+----------+-----------+ Address table: +-----------+----------+---------------+------------+ | addressId | personId | city | state | +-----------+----------+---------------+------------+ | 1 | 2 | New York City | New York | | 2 | 3 | Leetcode | California | +-----------+----------+---------------+------------+ Output: +-----------+----------+---------------+----------+ | firstName | lastName | city | state | +-----------+----------+---------------+----------+ | Allen | Wang | Null | Null | | Bob | Alice | New York City | New York | +-----------+----------+---------------+----------+ Explanation: There is no address in the address table for the personId = 1 so we return null in their city and state. addressId = 1 contains information about the address of personId = 2.
3. LeetCode 620
Table: Cinema
+----------------+----------+ | Column Name | Type | +----------------+----------+ | id | int | | movie | varchar | | description | varchar | | rating | float | +----------------+----------+ id is the primary key for this table. Each row contains information about the name of a movie, its genre, and its rating. rating is a 2 decimal places float in the range [0, 10]
Write an SQL query to report the movies with an odd-numbered ID and a description that is not "boring".
Return the result table ordered by rating in descending order.
The query result format is in the following example.
Example 1:
Input: Cinema table: +----+------------+-------------+--------+ | id | movie | description | rating | +----+------------+-------------+--------+ | 1 | War | great 3D | 8.9 | | 2 | Science | fiction | 8.5 | | 3 | irish | boring | 6.2 | | 4 | Ice song | Fantacy | 8.6 | | 5 | House card | Interesting | 9.1 | +----+------------+-------------+--------+ Output: +----+------------+-------------+--------+ | id | movie | description | rating | +----+------------+-------------+--------+ | 5 | House card | Interesting | 9.1 | | 1 | War | great 3D | 8.9 | +----+------------+-------------+--------+ Explanation: We have three movies with odd-numbered IDs: 1, 3, and 5. The movie with ID = 3 is boring so we do not include it in the answer.
4. LeetCode 196
Table: Person
+-------------+---------+ | Column Name | Type | +-------------+---------+ | id | int | | email | varchar | +-------------+---------+ id is the primary key column for this table. Each row of this table contains an email. The emails will not contain uppercase letters.
Write an SQL query to delete all the duplicate emails, keeping only one unique email with the smallest id. Note that you are supposed to write a DELETE statement and not a SELECT one.
After running your script, the answer shown is the Person table. The driver will first compile and run your piece of code and then show the Person table. The final order of the Person table does not matter.
The query result format is in the following example.
Example 1:
Input: Person table: +----+------------------+ | id | email | +----+------------------+ | 1 | john@example.com | | 2 | bob@example.com | | 3 | john@example.com | +----+------------------+ Output: +----+------------------+ | id | email | +----+------------------+ | 1 | john@example.com | | 2 | bob@example.com | +----+------------------+ Explanation: john@example.com is repeated two times. We keep the row with the smallest Id = 1.
5. LeetCode 197
Table: Weather
+---------------+---------+ | Column Name | Type | +---------------+---------+ | id | int | | recordDate | date | | temperature | int | +---------------+---------+ id is the primary key for this table. This table contains information about the temperature on a certain day.
Write an SQL query to find all dates' Id with higher temperatures compared to its previous dates (yesterday).
Return the result table in any order.
The query result format is in the following example.
Example 1:
Input: Weather table: +----+------------+-------------+ | id | recordDate | temperature | +----+------------+-------------+ | 1 | 2015-01-01 | 10 | | 2 | 2015-01-02 | 25 | | 3 | 2015-01-03 | 20 | | 4 | 2015-01-04 | 30 | +----+------------+-------------+ Output: +----+ | id | +----+ | 2 | | 4 | +----+ Explanation: In 2015-01-02, the temperature was higher than the previous day (10 -> 25). In 2015-01-04, the temperature was higher than the previous day (20 -> 30).
