我有一个如下查询,它检索表中不同商店的状态并将其显示为不同的列。
SELECT a.Store_ID,b.total as order_completed,c.total as order_cancelled,d.total as order_processed,e.total as order_failed FROM ORDER_HISTORY a
-> LEFT OUTER JOIN(select Store_ID,count(*) as total from ORDER_HISTORY where Status = 57 group by Store_ID)b on a.Store_ID = b.Store_ID
-> LEFT OUTER JOIN(select Store_ID,count(*) as total from ORDER_HISTORY where Status = 53 group by Store_ID)c on a.Store_ID = c.Store_ID
-> LEFT OUTER JOIN(select Store_ID,count(*) as total from ORDER_HISTORY where Status = 52 group by Store_ID)d on a.Store_ID = d.Store_ID
-> LEFT OUTER JOIN(select Store_ID,count(*) as total from ORDER_HISTORY where Status = 62 group by Store_ID)e on a.Store_ID = e.Store_ID
-> group by a.Store_ID;
任何人都可以建议使用联接的替代方法,因为它会影响数据库操作的性能。
最佳答案
通过(Store_ID, Status)
在ORDER_HISTORY
上创建索引,那么这应该足够快。
SELECT
Store_ID,
status,
COUNT(*) as total
FROM
ORDER_HISTORY
GROUP BY
Store_ID,
status;
然后使用您的应用程序在列中显示少数结果行数据。实现起来应该不难。
<小时/>另一种方法是(与上面相同的索引):
SELECT
Store_ID,
SUM(CASE WHEN Status = 57 THEN 1 ELSE 0 END) AS order_completed,
SUM(CASE WHEN Status = 53 THEN 1 ELSE 0 END) AS order_cancelled,
SUM(CASE WHEN Status = 52 THEN 1 ELSE 0 END) AS order_processed,
SUM(CASE WHEN Status = 62 THEN 1 ELSE 0 END) AS order_processed
FROM
ORDER_HISTORY
GROUP BY
Store_ID;
根据需要替换 NULL 值。
关于mysql - mysql 中联接的替代方案,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50169662/