比较两个不同的oracle表中的2个逗号分隔的列

Question

我想比较表1的列n和表2的列b。 两者都有逗号分隔的值。
就我而言，如果列B中的所有值都存在于列A中，则我必须返回TRUE
否则为假。 请指教。 例如：

1.列A具有a，b，c，def列B具有a，c，b，其应返回TRUE
2.第1列具有vvv，ccc，rr第2列具有ccc，rr，125，应返回FALSE
3.第1列为空，第2列为空，应返回TRUE

请帮忙

Answer 1

可以使用以下SQL完成：

    with t1 as( select '1111,1985,433435,5765' as colA from dual union all
                select '232323,7777,8888,9999' as colA from dual union all
                select '111,2000,433435,5765' as colA from dual),
      t2 as ( select '1111,466576,24323' as colB from dual union all
              select '11111,1985,435657' as colb from dual union all
              select '2222,232323,24989323' as colb from dual )

    select colA from t1 where exists(
select 1 from t2 where REGEXP_COUNT(','||t1.colA||',' , ',('||REPLACE(t2.colB, ',' , '|') || '),' ) > 0)

Answer 2

希望此代码段对您有所帮助。

SELECT DECODE(NULLIF(
  (SELECT LISTAGG(x.col_val,',') WITHIN GROUP(
  ORDER BY x.col_val)intrsctnVal
  FROM
    (SELECT SUBSTR(REPLACE(COL2,',',''),LEVEL,1) col_val
    FROM
      (SELECT 'a,b,c' AS COL2 FROM DUAL
      )B
      CONNECT BY level <= regexp_count(col2,',')+1
    INTERSECT
    SELECT SUBSTR(REPLACE(COL1,',',''),LEVEL,1) col_val
    FROM
      (SELECT 'a,b,c,d' AS COL1 FROM DUAL
      )A
      CONNECT BY LEVEL <= REGEXP_COUNT(COL1,',')+1
    )X
  ),
  (SELECT 'a,b,c' AS COL2 FROM DUAL
  )),NULL,'TRUE','FALSE') MATCH
FROM DUAL;

Answer 3

最终，我得到了它）对我来说，仅使用SQL来解决它是一个有趣的任务：

         B          N SORTED                                                                           BOOL
---------- ---------- -------------------------------------------------------------------------------- -----
         1          1 [a],[b],[c]                                                                      true
         1          2 [ccc],[rr],[vvv]                                                                 false
         1          3 []                                                                               true

第1步：合并两个表：

SQL> with
  2  t1 as (
  3    select 1 n, 'c,b,a' c from dual
  4    union all
  5    select 2 n, 'vvv,ccc,rr' c from dual
  6    union all
  7    select 3 n, null c from dual),
  8  t2 as (
  9    select 1 n, 'a,b,c' c from dual
 10    union all
 11    select 2 n, 'ccc,rr,125' c from dual
 12    union all
 13    select 3 n, null c from dual),
 14  t12 as (select 1 b, n, c from t1 union all select 2 b, n, c from t2)
 15  select * from t12
 16  /

         B          N C
---------- ---------- ----------
         1          1 c,b,a
         1          2 vvv,ccc,rr
         1          3 
         2          1 a,b,c
         2          2 ccc,rr,125
         2          3 

6 rows selected

第2步：计数分隔符：

SQL> with
  2  t1 as (
  3    select 1 n, 'c,b,a' c from dual
  4    union all
  5    select 2 n, 'vvv,ccc,rr' c from dual
  6    union all
  7    select 3 n, null c from dual),
  8  t2 as (
  9    select 1 n, 'a,b,c' c from dual
 10    union all
 11    select 2 n, 'ccc,rr,125' c from dual
 12    union all
 13    select 3 n, null c from dual),
 14  t12 as (select 1 b, n, c from t1 union all select 2 b, n, c from t2),
 15  t as (select b, n, c, nvl(length(regexp_replace(c, '[^,]+')) + 1, 1) cnt from t12)
 16  select * from t
 17  /

         B          N C                 CNT
---------- ---------- ---------- ----------
         1          1 c,b,a               3
         1          2 vvv,ccc,rr          3
         1          3                     1
         2          1 a,b,c               3
         2          2 ccc,rr,125          3
         2          3                     1

6 rows selected

第3步：计算边界值

SQL> with
  2  t1 as (
  3    select 1 n, 'c,b,a' c from dual
  4    union all
  5    select 2 n, 'vvv,ccc,rr' c from dual
  6    union all
  7    select 3 n, null c from dual),
  8  t2 as (
  9    select 1 n, 'a,b,c' c from dual
 10    union all
 11    select 2 n, 'ccc,rr,125' c from dual
 12    union all
 13    select 3 n, null c from dual),
 14  t12 as (select 1 b, n, c from t1 union all select 2 b, n, c from t2),
 15  t as (select b, n, c, nvl(length(regexp_replace(c, '[^,]+')) + 1, 1) cnt from t12),
 16  tbrd as (select b, n, c, cnt, sum(cnt) over (order by b,n) brd from t)
 17  select * from tbrd
 18  /

         B          N C                 CNT        BRD
---------- ---------- ---------- ---------- ----------
         1          1 c,b,a               3          3
         1          2 vvv,ccc,rr          3          6
         1          3                     1          7
         2          1 a,b,c               3         10
         2          2 ccc,rr,125          3         13
         2          3                     1         14

6 rows selected

第4步：为行生成新查询

SQL> with
  2  t1 as (
  3    select 1 n, 'c,b,a' c from dual
  4    union all
  5    select 2 n, 'vvv,ccc,rr' c from dual
  6    union all
  7    select 3 n, null c from dual),
  8  t2 as (
  9    select 1 n, 'a,b,c' c from dual
 10    union all
 11    select 2 n, 'ccc,rr,125' c from dual
 12    union all
 13    select 3 n, null c from dual),
 14  t12 as (select 1 b, n, c from t1 union all select 2 b, n, c from t2),
 15  t as (select b, n, c, nvl(length(regexp_replace(c, '[^,]+')) + 1, 1) cnt from t12),
 16  tbrd as (select b, n, c, cnt, sum(cnt) over (order by b,n) brd from t),
 17  allrec as (select level lv from dual connect by level <= (select sum(cnt) from t))
 18  select * from allrec
 19  /

        LV
----------
         1
         2
         3
         4
         5
         6
         7
         8
         9
        10
        11
        12
        13
        14

14 rows selected

第5步：主要查询

SQL> with
  2  t1 as (
  3    select 1 n, 'c,b,a' c from dual
  4    union all
  5    select 2 n, 'vvv,ccc,rr' c from dual
  6    union all
  7    select 3 n, null c from dual),
  8  t2 as (
  9    select 1 n, 'a,b,c' c from dual
 10    union all
 11    select 2 n, 'ccc,rr,125' c from dual
 12    union all
 13    select 3 n, null c from dual),
 14  t12 as (select 1 b, n, c from t1 union all select 2 b, n, c from t2),
 15  t as (select b, n, c, nvl(length(regexp_replace(c, '[^,]+')) + 1, 1) cnt from t12),
 16  tbrd as (select b, n, c, cnt, sum(cnt) over (order by b,n) brd from t),
 17  allrec as (select level lv from dual connect by level <= (select sum(cnt) from t)),
 18  step5 as (select r.b, r.n, r.c, regexp_substr(c, '[^,]+', 1, r.brd - a.lv + 1) s from tbrd r, allrec a
 19  where r.brd - a.lv between 0 and r.cnt - 1)
 20  select * from step5
 21  /

         B          N C          S
---------- ---------- ---------- ----------
         1          1 c,b,a      a
         1          1 c,b,a      b
         1          1 c,b,a      c
         1          2 vvv,ccc,rr rr
         1          2 vvv,ccc,rr ccc
         1          2 vvv,ccc,rr vvv
         1          3            
         2          1 a,b,c      c
         2          1 a,b,c      b
         2          1 a,b,c      a
         2          2 ccc,rr,125 125
         2          2 ccc,rr,125 rr
         2          2 ccc,rr,125 ccc
         2          3            

14 rows selected

第6步：对值进行排序

SQL> with
  2  t1 as (
  3    select 1 n, 'c,b,a' c from dual
  4    union all
  5    select 2 n, 'vvv,ccc,rr' c from dual
  6    union all
  7    select 3 n, null c from dual),
  8  t2 as (
  9    select 1 n, 'a,b,c' c from dual
 10    union all
 11    select 2 n, 'ccc,rr,125' c from dual
 12    union all
 13    select 3 n, null c from dual),
 14  t12 as (select 1 b, n, c from t1 union all select 2 b, n, c from t2),
 15  t as (select b, n, c, nvl(length(regexp_replace(c, '[^,]+')) + 1, 1) cnt from t12),
 16  tbrd as (select b, n, c, cnt, sum(cnt) over (order by b,n) brd from t),
 17  allrec as (select level lv from dual connect by level <= (select sum(cnt) from t)),
 18  step5 as (select r.b, r.n, r.c, regexp_substr(c, '[^,]+', 1, r.brd - a.lv + 1) s from tbrd r, allrec a
 19  where r.brd - a.lv between 0 and r.cnt - 1),
 20  step6 as (
 21  select b, n, listagg('[' || s || ']', ',') within group (order by s) sorted
 22  from step5
 23  group by b, n)
 24  select * from step6
 25  /

         B          N SORTED
---------- ---------- --------------------------------------------------------------------------------
         1          1 [a],[b],[c]
         1          2 [ccc],[rr],[vvv]
         1          3 []
         2          1 [a],[b],[c]
         2          2 [125],[ccc],[rr]
         2          3 []

6 rows selected

第7步：结果

SQL> with
  2  t1 as (
  3    select 1 n, 'c,b,a' c from dual
  4    union all
  5    select 2 n, 'vvv,ccc,rr' c from dual
  6    union all
  7    select 3 n, null c from dual),
  8  t2 as (
  9    select 1 n, 'a,b,c' c from dual
 10    union all
 11    select 2 n, 'ccc,rr,125' c from dual
 12    union all
 13    select 3 n, null c from dual),
 14  t12 as (select 1 b, n, c from t1 union all select 2 b, n, c from t2),
 15  t as (select b, n, c, nvl(length(regexp_replace(c, '[^,]+')) + 1, 1) cnt from t12),
 16  tbrd as (select b, n, c, cnt, sum(cnt) over (order by b,n) brd from t),
 17  allrec as (select level lv from dual connect by level <= (select sum(cnt) from t)),
 18  step5 as (select r.b, r.n, r.c, regexp_substr(c, '[^,]+', 1, r.brd - a.lv + 1) s from tbrd r, allrec a
 19  where r.brd - a.lv between 0 and r.cnt - 1),
 20  step6 as (
 21  select b, n, listagg('[' || s || ']', ',') within group (order by s) sorted
 22  from step5
 23  group by b, n)
 24  select t.*, decode(
 25    (select 1 from step6 z
 26     where z.b = 2 and z.sorted = t.sorted and rownum < 2), 1, 'true', 'false') bool
 27  from step6 t where b = 1
 28  /

         B          N SORTED                                                                           BOOL
---------- ---------- -------------------------------------------------------------------------------- -----
         1          1 [a],[b],[c]                                                                      true
         1          2 [ccc],[rr],[vvv]                                                                 false
         1          3 []                                                                               true

当然，使用函数要好得多，但是，我只是想找到一种仅使用SQL的方法。

（2017-06-05：增加功能）

功能可以是：

create or replace function comparesepval(vArg1 in varchar2, vArg2 in varchar2) return number result_cache deterministic
as
  type tpStore is table of number index by varchar2(32);

  tStore1 tpStore;
  tStore2 tpStore;
  nCnt number;
  vVal varchar2(32);
begin
  if (vArg1 is null and vArg2 is null) or (vArg1 = vArg2) then
    return 1;
  end if;

  if vArg1 is null then
    return 0;
  end if;

  nCnt := nvl(length(regexp_replace(vArg1, '[^,]+')), 0) + 1;

  if nCnt = 1 or nCnt <> nvl(length(regexp_replace(vArg2, '[^,]+')), 0) + 1 then -- vArg1 <> vArg2
    return 0;
  end if;

  -- parse string
  for i in 1..nCnt
  loop
    declare

      procedure prcAgr(vArg in varchar2, tStore in out tpStore)
      as
        vVal varchar2(32) := nvl(regexp_substr(vArg, '(.*?)(,|$)', 1, i, null, 1), ','); -- ',' - for null
      begin
        tStore(vVal) := case when tStore.exists(vVal) then tStore(vVal) + 1 else 1 end;
      end;

    begin
      prcAgr(vArg1, tStore1);
      prcAgr(vArg2, tStore2);
    end;
  end loop;

  if tStore1.count <> tStore2.count then
    return 0;
  end if;

  vVal := tStore1.first;
  loop
    exit when vVal is null;
    if not tStore2.exists(vVal) then
      return 0;
    end if;
    if tStore2(vVal) <> tStore1(vVal) then
      return 0;
    end if;
    vVal := tStore1.next(vVal);
  end loop;

  return 1;
end;
/

测试：

SQL> with tbl as (
  2    select null c1, null c2 from dual union all
  3    select 'a' c1, 'a' c2 from dual union all
  4    select 'a' c1, 'b' c2 from dual union all
  5    select 'a,a' c1, 'a,b' c2 from dual union all
  6    select 'a,a' c1, 'a,b,a' c2 from dual union all
  7    select 'a,a,b' c1, 'a,b,a' c2 from dual union all
  8    select 'a,,b' c1, ',b,a' c2 from dual union all
  9    select 'a,' c1, 'd,' c2 from dual
 10    )
 11  select comparesepval(c1, c2) c, c1, c2 from tbl;

         C C1    C2
---------- ----- -----
         1       
         1 a     a
         0 a     b
         0 a,a   a,b
         0 a,a   a,b,a
         1 a,a,b a,b,a
         1 a,,b  ,b,a
         0 a,    d,

8 rows selected