Question

给出如下表格：

    +---------+------+--------+-----------+--------------+
    | Empcode | name | desig  | joinmonth | releivemonth |
    +---------+------+--------+-----------+--------------+
    |      1. | A1.  |    D1. | Jan-18.   |    null      |
    |      2. | A2.  |    D2. | Jan-18.   |    May-18    |
    |      3. | A3.  |    D3. | Jan-18.   |    null      |
    +---------+------+--------+-----------+--------------+

我想显示表格：

    +---------------+--------+--------+--------+--------+--------+
    |    Remarks    | jan-18 | feb-18 | mar-18 | apr-18 | may-18 |
    +---------------+--------+--------+--------+--------+--------+
    | Joinmonth     |      3 |      0 |      0 |      0 |      0 |
    | Releivedmonth |      0 |      0 |      0 |      0 |      1 |
    +---------------+--------+--------+--------+--------+--------+

Answer 1

您需要先取消透视，然后再重新透视：

select max(BMI), STDDEV(WEIGHT) FROM MEASUREMENTS where 
      MEASUREMENTS.patient_id='nick' AND 
      MEASUREMENTS.measure_date BETWEEN DATE_SUB("2013-12-30", INTERVAL 6 YEAR) AND "2013-12-30";

Answer 2

这是扩展评论而不是答案，请接受我在拒绝投票之前需要使用格式控制。

您似乎在注释中添加了查询，尽管语法并不完全正确。您经常使用标准括号()而不是括号[]，并且在结束IN()时缺少右括号。我相信您的查询应如下所示：

SELECT
    empname AS remarks
  , [1-1-18]
  , [1-2-18]
  , [1-3-18]
  , [1-4-18]
  , [1-5-18]
FROM (
    SELECT
        empname
      , joimonth
      , releivedmonth
    FROM emply
) AS s
PIVOT (
    COUNT(releivedmonth) 
    FOR joinmonth IN ([1-1-18], [1-2-18], [1-3-18], [1-4-18], [1-5-18])
) piv

您不应尝试在评论中添加查询，而只需编辑问题即可。

在此查询中，您所引用的值看起来像 1-1-18 ，但是在数据样本中根本没有看起来像这样的值。 [joinmonth]和[releivedmonth]列是什么数据类型？

对于这些列中的文本数据，您遇到了严重的问题。例如，如果所有这些都不同：Jan-18.，Jan 18，Jan-18，那么它们将不会按照您的需要对齐。像这样的数据变化将使这不可能。

CREATE TABLE emply(
   Empcode      NUMERIC(9,0)
  ,empname         VARCHAR(6)
  ,desig        VARCHAR(8)
  ,joinmonth    varchar(30)
  ,releivemonth varchar(30)
);
INSERT INTO emply(Empcode,empname,desig,joinmonth,releivemonth) VALUES (1.,'A1.','D1.','Jan-18.',NULL);
INSERT INTO emply(Empcode,empname,desig,joinmonth,releivemonth) VALUES (2.,'A2.','D2.','Jan-18.','May 18');
INSERT INTO emply(Empcode,empname,desig,joinmonth,releivemonth) VALUES (3.,'A3.','D3.','Jan-18.',NULL);

SELECT
    empname AS remarks
  , [Jan-18.]
  , [Feb-18.]
  , [Mar-18.]
  , [Apr-18.]
  , [May-18.]
FROM (
    SELECT
        empname
      , joinmonth
      , releivemonth
    FROM emply
) AS s
PIVOT (
    COUNT(releivemonth) 
    FOR joinmonth IN ([Jan-18.], [Feb-18.], [Mar-18.], [Apr-18.], [May-18.])
) piv

但是，输出是：

+----+---------+---------+---------+---------+---------+---------+
|    | remarks | Jan-18. | Feb-18. | Mar-18. | Apr-18. | May-18. |
+----+---------+---------+---------+---------+---------+---------+
|  1 | A1.     |       0 |       0 |       0 |       0 |       0 |
|  2 | A2.     |       1 |       0 |       0 |       0 |       0 |
|  3 | A3.     |       0 |       0 |       0 |       0 |       0 |
+----+---------+---------+---------+---------+---------+---------+

COUNT(releivemonth)只有一个非空值

如何根据日期表日期

2 个答案: