Oracle SQL将行中的一列与同一列中的前一行进行比较

如何解决Oracle SQL将行中的一列与同一列中的前一行进行比较

这是我的表（Table1）当前在Oracle数据库中的方式。

<CustomTooltip id="123"><Text>...</Text></CustomTooltip>

我试图获得一个输出，该输出将每一行的Product列进行比较，以得到如下所示的内容：在这里，我正在比较第1行和第2行，以查看第2行是否具有第1行中没有的新产品（NEW_PRODUCTS）。

似乎我可以使用LAG或LEAD函数，但由于产品之间使用ID Year_Mth Product 123 201901 1,2,3 123 201902 1,4,5 123 201903 2,3,6 123 201904 1,5,6分隔符，因此似乎很棘手。

解决方法

这是一种选择。看起来和您的数据模型一样丑陋：)查看代码中的注释。如果您不确定每个CTE的功能，建议您分步运行以下代码并查看其结果。

为了便于阅读，我将其分为几部分。

SQL> with
  2  test (id,year_mth,product) as
  3    -- your sample data (as well as some of my sample data)
  4    (select 123,201901,'1,2,3'   from dual union all
  5     select 123,201902,4,5' from dual union all
  6     select 123,201903,'2,3,6' from dual union all
  7     select 123,201904,5,6' from dual union all
  8     --
  9     select 888,'apple,banana' from dual union all
 10     select 888,banana' from dual union all
 11     select 888,lemon'  from dual
 12    ),

 13  py as
 14    (select id,15            year_mth ymp,-- "this" year_mth
 16            lead(year_mth) over (partition by id order by year_mth) ymn  -- "next" year_mth
 17     from test
 18     order by id,year_mth
 19    ),20  tabp as
 21    -- products that belong to "THIS" year_mth split to rows
 22    (select
 23        t.id,24        t.year_mth,25        p.ymp,26        p.ymn,27        regexp_substr(t.product,'[^,]+',1,c.column_value) product
 28      from test t join py p on t.id = p.id and t.year_mth = p.ymp cross join
 29        table(cast(multiset(select level from dual
 30                            connect by level <= regexp_count(product,',') + 1
 31                           ) as sys.odcinumberlist)) c
 32    ),33  tabn as
 34    -- products that belong to "NEXT" year_mth split to rows
 35    (select
 36        t.id,37        t.year_mth,38        p.ymp,39        p.ymn,40        regexp_substr(t.product,c.column_value) product
 41      from test t join py p on t.id = p.id and t.year_mth = p.ymn cross join
 42        table(cast(multiset(select level from dual
 43                            connect by level <= regexp_count(product,') + 1
 44                           ) as sys.odcinumberlist)) c
 45    ),

46  newprod as
47    -- MINUS set operator finds differences between "NEXT" and "THIS" year_mth
48    (select id,ymn,product from tabn
49     minus
50     select id,product from tabp
51    )
52  -- finally,aggregate new products (result of the previous MINUS set operation)
53  select
54    t.id,55    t.year_mth,56    t.product,57    listagg(case when t.rn = 1 then t.product else n.product end,')
58      within group (order by n.product) new_products
59  from (select a.id,60          a.year_mth,61          a.product,62          row_number() over (partition by a.id order by a.year_mth) rn
63        from test a
64       ) t left join newprod n on t.id = n.id and t.year_mth = n.ymn
65  group by t.id,t.year_mth,t.product
66  order by t.id,t.year_mth;

       ID   YEAR_MTH PRODUCT      NEW_PRODUCTS

      123     201901 1,3        1,3
      123     201902 1,5      4,5
      123     201903 2,6      3,6
      123     201904 1,6      1,5
      888     201901 apple,banana apple,banana
      888     201902 apple,banana
      888     201903 apple,lemon  lemon

已选择7行。

SQL>

在需要使用此类定界字符串的情况下，使用xml函数（例如fn：string-join（），fn：tokenize（））通常非常方便。

例如：

xmltable(
       'let $x:=tokenize($a,","),$y:=tokenize($b,")
        return fn:string-join($x[not(.=$y)],")'
       passing product as "a",prev_product as "b"
     columns New_Products varchar(100) path '.'
    ) x

此xmltable（）拆分输入参数product和prev_product，并从prev_product中不在product中的那些子字符串中返回：

函数tokenize($a,")使用逗号作为分隔符来拆分输入字符串$ a。
$x[not(.=$y)]从$ x返回$ y中不存在的那些值
函数string-join($arg1,")使用逗号作为分隔符来连接$ arg1中的值。

完整示例：

with
test (id,product) as
  -- your sample data (as well as some of my sample data)
  (select 123,3'   from dual union all
   select 123,5' from dual union all
   select 123,6' from dual union all
   select 123,6' from dual union all
   --
   select 888,banana' from dual union all
   select 888,lemon'  from dual
  )
select
    t.*,x.*
from 
    (
     select 
        t.*,lag(t.product)over(partition by id order by year_mth) prev_product
     from test t
    ) t,xmltable(
       'let $x:=tokenize($a,prev_product as "b"
     columns New_Products varchar(100) path '.'
    ) x;

我使上面的xquery很久了，只是为了使其更具可读性。在现实生活中，xquery会更短： fn:string-join(tokenize($a,")[not(.=tokenize($b,"))],")

with
test (id,xmltable(
       'fn:string-join(tokenize($a,prev_product as "b"
     columns New_Products varchar(100) path '.'
    ) x

与我的相似，如果要重新透视，请在末尾添加一个listagg和group-by查询...

WITH
input(id,product) AS (
          SELECT 123,3'    FROM dual
UNION ALL SELECT 123,5'  FROM dual
UNION ALL SELECT 123,6'  FROM dual
UNION ALL SELECT 123,6'  FROM dual
),i(i) AS (
           SELECT 1 FROM dual
 UNION ALL SELECT 2 FROM dual
 UNION ALL SELECT 3 FROM dual
 UNION ALL SELECT 4 FROM dual
 UNION ALL SELECT 5 FROM dual
),unpivot AS (
  SELECT
    id,i,REGEXP_SUBSTR(product,'\d+',i) AS prd
  FROM input CROSS JOIN i
  WHERE REGEXP_SUBSTR(product,i) <> ''
)
SELECT 
  *,CASE 
    WHEN LAG(year_mth) OVER(PARTITION BY id,prd ORDER BY year_mth) IS NULL
    THEN 'new'
    ELSE 'old'
   END
FROM unpivot ORDER BY 3,4;
-- out  id  | i | year_mth | prd | case 
-- out -----+---+----------+-----+------
-- out  123 | 1 |   201901 | 1   | new
-- out  123 | 2 |   201901 | 2   | new
-- out  123 | 3 |   201901 | 3   | new
-- out  123 | 1 |   201902 | 1   | old
-- out  123 | 2 |   201902 | 2   | old
-- out  123 | 3 |   201902 | 4   | new
-- out  123 | 4 |   201902 | 5   | new
-- out  123 | 1 |   201903 | 2   | old
-- out  123 | 2 |   201903 | 3   | old
-- out  123 | 3 |   201903 | 4   | old
-- out  123 | 4 |   201903 | 6   | new
-- out  123 | 1 |   201904 | 1   | old
-- out  123 | 2 |   201904 | 4   | old
-- out  123 | 3 |   201904 | 5   | old
-- out  123 | 4 |   201904 | 6   | old

Oracle SQL将行中的一列与同一列中的前一行进行比较

如何解决Oracle SQL将行中的一列与同一列中的前一行进行比较

解决方法

相关推荐