leetcode197 上升的溫度 Rising Temperature

給定一個Weather表，編寫一個SQL查詢來查找與以前(昨天的)日期相比溫度更高的全部日期的id。

 

 

建立表和數據：

 

 

-- ----------------------------
-- Table structure for `weather`
-- ----------------------------
DROP TABLE IF EXISTS `weather`;
CREATE TABLE `weather` (
 `Id` int(11) DEFAULT NULL,
 `RecordDate` date DEFAULT NULL,
 `Temperature` int(11) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;
-- ----------------------------
-- Records of weather
-- ----------------------------
INSERT INTO `weather` VALUES ('1','2015-01-01', '10');
INSERT INTO `weather` VALUES ('2','2015-01-02', '25');
INSERT INTO `weather` VALUES ('3','2015-01-03', '20');
INSERT INTO `weather` VALUES ('4','2015-01-04', '30');

 

解法：

1.思路簡單。表自鏈接，找出溫度比前一天高的行。

問題的關鍵是肯定日期的前一天。

日期函數： DATEDIFF(date1,date2) ，返回date1與date2之間相差的天數。

SELECT W1.Id
FROM weather AS W1 JOIN weather AS W2 ON (DATEDIFF(W1.RecordDate,W2.RecordDate) = 1 AND W1.Temperature > W2.Temperature)

2.一樣，函數： TIMESTAMPDIFF(unit,begin,end)，返回begin與end之間相差多少個unit。unit爲DAY時，即爲相差「天」。

SELECT W2.Id
FROM weather AS W1 JOIN weather AS W2 
ON (TIMESTAMPDIFF(DAY,W1.RecordDate,W2.RecordDate) = 1 AND W1.Temperature < W2.Temperature);

TO_DAYS函數，用來將日期換算整天數

SELECT w1.Id FROM Weather w1, Weather w2
WHERE w1.Temperature > w2.Temperature AND TO_DAYS(w1.RecordDate)=TO_DAYS(w2.RecordDate) + 1;

使用Subdate函數，來實現日期減1

 

SELECT w1.Id FROM Weather w1, Weather w2
WHERE w1.Temperature > w2.Temperature AND SUBDATE(w1.RecordDate, 1) = w2.RecordDate;