leetcode 197. 上升的溫度

題目


給定一個 Weather 表,編寫一個 SQL 查詢,來查找與之前(昨天的)日期相比溫度更高的所有日期的 Id。

+---------+------------------+------------------+
| Id(INT) | RecordDate(DATE) | Temperature(INT) |
+---------+------------------+------------------+
|       1 |       2015-01-01 |               10 |
|       2 |       2015-01-02 |               25 |
|       3 |       2015-01-03 |               20 |
|       4 |       2015-01-04 |               30 |
+---------+------------------+------------------+

例如,根據上述給定的 Weather 表格,返回如下 Id:

+----+
| Id |
+----+
|  2 |
|  4 |
+----+

解法


這裏我們用到SQL Server 裏面的DATEDIFF函數,此函數返回在指定的startdate和enddate之間相差的指定的日期部分邊界的計數。就是說你拿天數作比較,它會返回你的天數之差。

官方文檔

DATEDIFF ( datepart , startdate , enddate )

例子

SELECT DATEDIFF(year,        '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(quarter,     '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(month,       '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(dayofyear,   '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(day,         '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(week,        '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(hour,        '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(minute,      '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(second,      '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(millisecond, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');
SELECT DATEDIFF(microsecond, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');

代碼

/* Write your T-SQL query statement below */
SELECT
    w2.id AS 'Id'
FROM
    Weather w1, Weather w2 
WHERE
    DATEDIFF(day, w1.RecordDate, w2.RecordDate) = 1
    AND w1.Temperature < w2.Temperature;

結果

在這裏插入圖片描述

發表評論
所有評論
還沒有人評論,想成為第一個評論的人麼? 請在上方評論欄輸入並且點擊發布.
相關文章