当Oracle中的字段值更改时增加行号答案

【问题标题】：increment row number when value of field changes in Oracle当Oracle中的字段值更改时增加行号
【发布时间】：2013-05-09 22:31:11
【问题描述】：

在 Oracle 中为以下数据编写查询时，我需要帮助。数据按人员和日期字段排序。

    Person     Day   Flag
    ------     ---   ---- 
    person1    day1   Y
    person1    day2   Y
    person1    day3   Y
    person1    day4   N
    person1    day5   N
    person1    day6   Y
    person1    day7   Y
    person1    day8   Y

我需要一个 Group_Number 列，该列在标志值更改时递增。我的结果应该如下所示

    Person     Day   Flag  Group_Number
    ------     ---   ----  ------------
    person1    day1   Y    1
    person1    day2   Y    1
    person1    day3   Y    1
    person1    day4   N    2
    person1    day5   N    2
    person1    day6   Y    3
    person1    day7   Y    3
    person1    day8   Y    3

我认为有办法使用诸如 ROW_NUMBER、LEAD 等分析函数来获得上述结果。

【问题讨论】：

标签： oracle field increment row-number

【解决方案1】：

您可以将analytic functionsSUM（用作累计）和LAG结合起来：

SQL> WITH data AS (
  2            SELECT 'person1' person, 'day1' day, 'Y' flag FROM dual
  3  UNION ALL SELECT 'person1' person, 'day2' day, 'Y' flag FROM dual
  4  UNION ALL SELECT 'person1' person, 'day3' day, 'Y' flag FROM dual
  5  UNION ALL SELECT 'person1' person, 'day4' day, 'N' flag FROM dual
  6  UNION ALL SELECT 'person1' person, 'day5' day, 'N' flag FROM dual
  7  UNION ALL SELECT 'person1' person, 'day6' day, 'Y' flag FROM dual
  8  UNION ALL SELECT 'person1' person, 'day7' day, 'Y' flag FROM dual
  9  UNION ALL SELECT 'person1' person, 'day8' day, 'Y' flag FROM dual
 10  )
 11  SELECT person, DAY, flag, SUM(gap) over (PARTITION BY person
 12                                           ORDER BY DAY) grp
 13    FROM (SELECT person, DAY, flag,
 14                 CASE WHEN flag = lag(flag) over (PARTITION BY person
 15                                                  ORDER BY DAY)
 16                      THEN 0
 17                      ELSE 1
 18                 END gap
 19            FROM DATA);

PERSON  DAY  FLAG        GRP
------- ---- ---- ----------
person1 day1 Y             1
person1 day2 Y             1
person1 day3 Y             1
person1 day4 N             2
person1 day5 N             2
person1 day6 Y             3
person1 day7 Y             3
person1 day8 Y             3

【讨论】：