如何在同一列中显示 30 天的数据，其余的将显示 90 天？

Question

几天前我问了一个问题，它使用以下查询解决了我的问题。我刚刚添加了 and t.date_of_created > DATE_SUB(NOW(), INTERVAL 90 DAY) 以仅显示 90 天的数据并且它正在运行。

现在我的问题是，我必须只显示 SUM(f_filestatus = 3) AS tcount3 30 天，剩下的将显示 90 天的数据。

有什么想法可以做到这一点吗？

SELECT
  SUM(f_filestatus = 1) AS tcount1,
  SUM(f_filestatus = 2) AS tcount2,
  SUM(f_filestatus = 3) AS tcount3
FROM (
  SELECT t.f_bankid, t.f_filestatus
  FROM tbl_fileStatus t
  WHERE t.f_id = (SELECT f_id FROM tbl_fileStatus WHERE f_bankid = t.f_bankid and t.date_of_created > DATE_SUB(NOW(), INTERVAL 90 DAY) ORDER BY f_id DESC LIMIT 1)
) t

Answer 1

我会使用 window 函数编写您当前的查询：

SELECT SUM(f_filestatus = 1) AS tcount1,
       SUM(f_filestatus = 2) AS tcount2,
       SUM(f_filestatus = 3) AS tcount3
FROM (SELECT fs.*,
             ROW_NUMBER() OVER (PARTITION BY f_bank_id ORDER BY date_of_created DESC, f_id DESC) as seqnum
      FROM tbl_fileStatus fs
      WHERE fs.date_of_created > CURDATE() - INTERVAL 90 DAY
     ) fs
WHERE seqnum = 1;

但是您的问题的答案是简单地添加一个新条件：

SELECT SUM(f_filestatus = 1) AS tcount1,
       SUM(f_filestatus = 2) AS tcount2,
       SUM(f_filestatus = 3 AND fs.date_of_created > CURDATE() - INTERVAL 30 DAY) AS tcount3
FROM (SELECT fs.*,
             ROW_NUMBER() OVER (PARTITION BY f_bank_id ORDER BY date_of_created DESC, f_id DESC) as seqnum
      FROM tbl_fileStatus fs
      WHERE fs.date_of_created > CURDATE() - INTERVAL 90 DAY
     ) fs
WHERE seqnum = 1;

编辑：

您可以在旧版本中使用相同的想法：

SELECT SUM(f_filestatus = 1) AS tcount1,
       SUM(f_filestatus = 2) AS tcount2,
       SUM(f_filestatus = 3 AND fs.date_of_created > CURDATE() - INTERVAL 30 DAY) AS tcount3
FROM tbl_fileStatus fs
WHERE fs.f_id = (SELECT fs2.f_id
                 FROM tbl_fileStatus fs2
                 WHERE fs2.bank_id = fs.bank_id AND
                       fs2.date_of_created > CURDATE() - INTERVAL 90 DAY
                 ORDER BY fs2.f_id DESC
                 LIMIT 1
                )

如何在同一列中显示 30 天的数据，其余的将显示 90 天？

How to display the one column the data for 30 days and rest will display for 90 days with same query?

mysql

sql

subquery