Question

我有一张包含此信息的表

ACCTCCODE | ACCTDESCRIPTION | ISSUEDATE
----------+-----------------+----------------
1031      | Blahdescription | 2018-03-11
1032      | Blahdescription | 2017-04-18
1033      | Blahdescription | 2018-04-15
1034      | Blahdescription | 2018-011-04

我想尝试两次获取日期？对于两个单独的列。例如

ACCTCCODE | ACCTDESCRIPTION | FIRSTCOUNT  | SECOUNDCOUNT
----------+-----------------+-------------+--------------
1031      | Blahdescription | 150            23
1032      | Blahdescription | 75             101
1033      | Blahdescription | 3              78
1034      | Blahdescription | 11             23

我试图用SELECT内的SELECT创建查询，但是对于sql来说是新手，所以在使其工作方面有些麻烦。

这是我想出的方法，它适用于第一个计数，但不适用于第二个计数。

SELECT DISTINCT 
    account AS ACCTCODE, Description AS ACCTDESCRIPTION,
    COUNT(issueDate) AS FIRSTCOUNT,
    (SELECT COUNT(issueDate) 
     FROM Table1 
     WHERE issueDate BETWEEN CONVERT(DATETIME, '2018-2-31') 
                         AND CONVERT(DATETIME, '2018-04-03') 
       AND account <> '') AS SECONDCOUNT
FROM 
    Table1
WHERE 
    issueDate BETWEEN CONVERT(DATETIME, '2017-11-31') 
                  AND CONVERT(DATETIME, '2018-02-01')
    AND account <> '' 
GROUP BY
    account, Description
ORDER BY 
    account, Description ASC

Answer 1

每个日期间隔一次将SUM（）与CASE一起使用

SELECT account AS ACCTCODE, description AS ACCTDESCRIPTION,
    SUM(CASE WHEN issuedate BETWEEN CONVERT(DATE, '20171131', 112) AND CONVERT(DATE, '20180201', 112) 
        THEN 1 ELSE 0 END) as FIRSTCOUNT,
    SUM(CASE WHEN issuedate BETWEEN CONVERT(DATE, '20180228', 112) AND CONVERT(DATE, '20180403', 112) 
        THEN 1 ELSE 0 END) as SECONDCOUNT
FROM Table1
GROUP BY account, description

Answer 2

您可以通过以下情况来尝试以下操作

select ACCTCCODE,ACCTDESCRIPTION  ,sum(case when issueDate>='2017-11-31' and issueDate<='2018-02-01' then 1 else 0 end)
          as FIRSTCOUNT ,count(*) as SECOUNDCOUNT
FROM Table1
 group by ACCTCCODE,ACCTDESCRIPTION

在同一列和表格中的两个不同日期之间进行计数

2 个答案: