Question

我有类似的桌子

+---------------------+-------+
| date_in             | title |
+---------------------+-------+
| 2018-01-01 00:00:00 | foo2  |
| 2018-01-02 00:00:00 | bar   |
| 2018-01-03 00:00:00 | man   |
| 2018-01-04 01:00:00 | foo2  |
| 2018-01-04 02:00:00 | foo   |
| 2018-01-04 03:00:00 | foo1  |
| 2018-01-04 04:00:00 | foo2  |
| 2018-01-05 00:00:00 | test  |
+---------------------+-------+

我想要的结果是

+---------------------+-------+
| date_in             | title |
+---------------------+-------+
| 2018-01-05 00:00:00 | test  |
| 2018-01-04 04:00:00 | foo2  | -- see this
| 2018-01-04 01:00:00 | foo2  | -- see this
| 2018-01-04 03:00:00 | foo1  |
| 2018-01-04 02:00:00 | foo   |
| 2018-01-03 00:00:00 | man   |
| 2018-01-02 00:00:00 | bar   |
| 2018-01-01 00:00:00 | foo2  |
+---------------------+-------+

我在下面的查询和其他几个查询中都尝试过，但是不起作用

SELECT
    detail.* 
FROM
    test detail
    JOIN ( SELECT min( date_in ) firstdate, title FROM test GROUP BY title ) firstdate ON detail.title = firstdate.title 
ORDER BY
    firstdate.firstdate,
    detail.title,
    detail.date_in

但是此查询显示的是相同标题的行，彼此上下，无论日期如何

我希望行仅在同一天显示在彼此的上方

在这里http://sqlfiddle.com/#!9/039aa/6 SQLFiddle与模式。

Answer 1

似乎您需要按日期（date_in）和标题desc排序

financialYear: "2019-2020"
holidayDate: "2019-06-07"
holidayDescription: "dfghdfh"
holidayType: "RH"

Answer 2

我认为您实际上是想在每个日期的最晚时间订购标题。在MySQL 8+中，您可以执行以下操作：

 SELECT d.* 
 FROM test d
 ORDER BY date(d.date_in),
          MAX(d.date_in) OVER (PARTITION BY date(d.date_in), d.title),
          d.title desc,
          d.date_in ;

在早期版本中，可以使用子查询。这是一种方法：

 SELECT d.* 
 FROM test d
 ORDER BY date(d.date_in),
          (SELECT MAX(d2.date_in) FROM test d2 WHERE d2.title = d.title AND DATE(d2.date_in) = DATE(d.date_in)),
          d.title desc,
          d.date_in;

或者，您可以使用显式的JOIN：

SELECT d.* 
FROM test d JOIN
     (SELECT d2.title, DATE(d2.date_in) as date, MAX(d2.date_in) as max_date_in
      FROM test d2
      GROUP BY d2.title, DATE(d2.date_in) 
     ) d2
     ON d2.title = d.title AND d2.date = DATE(d.date_in)
ORDER BY date(d.date_in),
         d2.max_date_in,
         d.title desc,
         d.date_in

Answer 3

尝试一下：

SELECT
    detail.* 
FROM
    test detail
    JOIN ( SELECT min( date_in ) firstdate, title FROM test GROUP BY title ) firstdate ON detail.title = firstdate.title 
ORDER BY
    DATE(detail.date_in),
    detail.title;

从firstdate.firstdate中删除了DATE(detail.date_in)并编辑了ORDER BY。

当且仅当在同一天发布时，才显示彼此相邻的行

3 个答案: