仅当下一行与上一行不同时才计数

Question

我有一个带有 12 个寄存器的 table。如果列 "AREA_OPERATIVA" 与上一行不同（按日期升序排序），我想计算该行。

例如，从第 1 行到第 2 行，它不应该计算任何东西，因为它们具有相同的面积 'CROSS' 但在第 2 行和第 3 行之间，它应该计算（或总和 1，我不在乎) 因为 'CROSS' 和 'UTRDANIOS' 是不同的。所以整个table.

的最终计数应该是3

是否可以通过查询来做到这一点，或者我是否需要为此目的制作一个带有光标的脚本？

我试过这个：

SELECT  a.creclama, 
sum (CASE WHEN b.area_operativa  NOT LIKE  a.area_operativa THEN 1 ELSE 0 END) AS increment
FROM TR_ASGAREOPE a
INNER JOIN TR_ASGAREOPE b ON a.creclama = b.creclama 
                          and a.cdistribuidora = b.cdistribuidora 
                          and a.secuencia = b.secuencia
WHERE a.creclama = 10008354
group by a.creclama;

但是正在计算完整的 12 行。

编辑：

最后我可以通过下一个查询解决这个问题：

select sum (
    CASE WHEN (comparacion.area_operativa  not like  comparacion.siguiente_fila THEN 1 ELSE 0 END) AS incremento 
from (    
    select creclama,
           area_operativa, 
           lead(area_operativa) over (order by fmodifica) as siguiente_fila
    from TR_ASGAREOPE
    where creclama = 10008354
    order by fmodifica
);

希望对以后的人有用，真的卡了我一天。谢谢大家。

Answer 1

您可以尝试使用超前或滞后等分析函数，例如

    SELECT CRECLAMA,
           CASE WHEN AREA_OPERATIVA <> NEXTROW THEN 1 ELSE 0 END AS INCREMENT
      FROM (
            SELECT CRECLAMA,
                   AREA_OPERATIVA,
                   LEAD(AREA_OPERATIVA) OVER (PARTITION BY 1 ORDER BY CRECLAMA) AS NEXTROW
              FROM TR_ASGAREOPE
            )

Answer 2

可以使用lag()解析函数:

with t as
(
 select a.*,
        lag(a.area_operativa,1,a.area_operativa) over (order by a."date") as lg
   from asgareope a
  where a.creclama = 10008354
)   
select t.creclama, sum(case when lg = area_operativa then 0 else 1 end) as "increment"
  from t     
 group by t.creclama

Answer 3

我想你可以简单地使用 COUNT(DISTINCT ...) 来解决你的问题，如果 AREA_OPERATIVA 不能 return 到以前使用的值：

SELECT CRECLAMA, COUNT(DISTINCT AREA_OPERATIVA)
  FROM TR_ASGAREOPE
 GROUP BY CRECLAMA

Answer 4

这是一种使用 LEAD 的方法：

WITH TR_ASGAREOPE(CRECLAMA, AREA_OPERATIVA, DATE_FIELD) AS
                 (SELECT 10008354, 'CROSS', DATE '2019-01-01' FROM DUAL UNION ALL
                  SELECT 10008354, 'CROSS', DATE '2019-01-02' FROM DUAL UNION ALL      -- 1
                  SELECT 10008354, 'UTRDANIOS', DATE '2019-01-03' FROM DUAL UNION ALL  -- 2
                  SELECT 10008354, 'EXP263', DATE '2019-01-04' FROM DUAL UNION ALL     -- 3
                  SELECT 10008354, 'EXP6', DATE '2019-01-05' FROM DUAL UNION ALL
                  SELECT 10008354, 'EXP6', DATE '2019-01-06' FROM DUAL UNION ALL
                  SELECT 10008354, 'EXP6', DATE '2019-01-07' FROM DUAL UNION ALL
                  SELECT 10008354, 'EXP6', DATE '2019-01-08' FROM DUAL UNION ALL
                  SELECT 10008354, 'EXP6', DATE '2019-01-09' FROM DUAL UNION ALL
                  SELECT 10008354, 'EXP6', DATE '2019-01-10' FROM DUAL UNION ALL
                  SELECT 10008354, 'EXP6', DATE '2019-01-11' FROM DUAL UNION ALL
                  SELECT 10008354, 'EXP6', DATE '2019-01-12' FROM DUAL UNION ALL
                  SELECT 12345678, 'AREA49', DATE '2019-02-01' FROM DUAL UNION ALL
                  SELECT 12345678, 'AREA49', DATE '2019-02-02' FROM DUAL UNION ALL  -- 1
                  SELECT 12345678, 'AREA50', DATE '2019-02-03' FROM DUAL UNION ALL
                  SELECT 12345678, 'AREA50', DATE '2019-02-04' FROM DUAL UNION ALL  -- 2
                  SELECT 12345678, 'AREA52', DATE '2019-02-05' FROM DUAL UNION ALL
                  SELECT 12345678, 'AREA52', DATE '2019-02-06' FROM DUAL UNION ALL
                  SELECT 12345678, 'AREA52', DATE '2019-02-07' FROM DUAL UNION ALL  -- 3
                  SELECT 12345678, 'AREA53', DATE '2019-02-08' FROM DUAL UNION ALL  -- 4
                  SELECT 12345678, 'AREA52', DATE '2019-02-09' FROM DUAL UNION ALL  -- 5
                  SELECT 12345678, 'AREA53', DATE '2019-02-10' FROM DUAL),
     cteData AS (SELECT CRECLAMA,
                         LEAD(CRECLAMA) OVER (ORDER BY DATE_FIELD) AS NEXT_CRECLAMA,
                         AREA_OPERATIVA,
                         LEAD(AREA_OPERATIVA) OVER (ORDER BY DATE_FIELD) AS NEXT_AREA_OPERATIVA
                    FROM TR_ASGAREOPE)
SELECT CRECLAMA, COUNT(*)
  FROM cteData
  WHERE CRECLAMA = NEXT_CRECLAMA AND
        AREA_OPERATIVA <> NEXT_AREA_OPERATIVA
  GROUP BY CRECLAMA
  ORDER BY CRECLAMA;

我为另一个 CRECLAMA 值添加了数据以显示它是如何工作的。

结果：

CRECLAMA    COUNT(*)
10008354    3
12345678    5

dbfiddle here

仅当下一行与上一行不同时才计数

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