Question

请参阅下面的SQL查询，它会计算CheckDate

之间的“是”和“否”的数量

在MatchSales字段上 - 它无法实现我想要的样子。它应该找到D.MobileNO（取决于CheckDate）与S.mobile的匹配（检查sales表中的所有记录）

SELECT D.Username, 
    SUM(CASE WHEN D.type = 'Yes' AND UNIX_TIMESTAMP(CheckDate) >= $From_timestamp AND UNIX_TIMESTAMP(CheckDate) <= $To_timestamp THEN 1 ELSE 0 END) as Yes, 
    SUM(CASE WHEN D.type = 'No' AND UNIX_TIMESTAMP(CheckDate) >= $From_timestamp AND UNIX_TIMESTAMP(CheckDate) <= $To_timestamp THEN 1 ELSE 0 END) as No, 
    SUM(CASE WHEN S.mobile IS NULL THEN 0 ELSE 1 END) as MatchSales
FROM dairy as D 
   LEFT JOIN (SELECT DISTINCT mobile FROM sales) as S on D.MobileNo = S.mobile
WHERE source = 'Company' 
GROUP BY D.Username

Answer 1

只需将CASE中的Checkdate条件添加到where子句：

SELECT D.Username,      
      SUM(CASE WHEN D.type = 'Yes' THEN 1 ELSE 0 END) AS Yes,      
      SUM(CASE WHEN D.type = 'No'  THEN 1 ELSE 0 END) AS No,      
      SUM(CASE WHEN S.mobile IS NULL THEN 0 ELSE 1 END) AS MatchSales 
 FROM dairy AS D     
 LEFT JOIN (SELECT DISTINCT mobile FROM sales) AS S ON D.MobileNo = S.mobile 
 WHERE D.source = 'Company'  
 AND   UNIX_TIMESTAMP(D.CheckDate) >= $From_timestamp 
 AND   UNIX_TIMTIMESTAMP(D.CheckDate) <= $To_timestamp 
 GROUP BY D.Username

没有它你将全面扫描乳制品

Answer 2

- 我认为你就是这样做的：

SELECT SUM(Yes) AS Yes, SUM(No) AS No
FROM (
    SELECT D.Username, 
        (CASE WHEN D.type = 'Yes' AND UNIX_TIMESTAMP(CheckDate) >= $From_timestamp AND UNIX_TIMESTAMP(CheckDate) <= $To_timestamp THEN 1 ELSE 0 END) as [Yes], 
        (CASE WHEN D.type = 'No' AND UNIX_TIMESTAMP(CheckDate) >= $From_timestamp AND UNIX_TIMESTAMP(CheckDate) <= $To_timestamp THEN 1 ELSE 0 END) as [No], 
        (CASE WHEN S.mobile IS NULL THEN 0 ELSE 1 END) as MatchSales
    FROM dairy as D 
        LEFT JOIN (SELECT DISTINCT mobile FROM sales) as S on D.MobileNo = S.mobile
    WHERE source = 'Company' 
    GROUP BY D.Username
) T

左连接问题 - 匹配数

2 个答案: