这是我的sqlite声明:
select code ,
case when (max(close)-min(close))/min(close)<0.1 then "grade1"
when (max(close)-min(close))/min(close) <0.2 then "grade2"
when (max(close)-min(close))/min(close) <0.3 then "grade3"
else "grade4" end as type;
(max(close)-min(close))/min(close)
使用了3次,如何使我的查询语句更简单?
最佳答案
将计算结果放入子查询中:
SELECT code,
CASE WHEN CloseRatio < 0.1 THEN 'grade1'
WHEN CloseRatio < 0.2 THEN 'grade2'
WHEN CloseRatio < 0.3 THEN 'grade3'
ELSE 'grade4'
END AS type
FROM (SELECT code,
(MAX(close) - MIN(close)) / MIN(close) AS CloseRatio
FROM MyTable);
关于sqlite - 在sqlite中使其简单,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13319143/