sql - 比较来自两个不同表的两列的逗号分隔值

Question

我想比较两个不同 Oracle 表的逗号分隔值的两个列(差异表)的值。我想找到与所有值匹配的行(NAME1 所有值 应与 NAME2 值匹配)。

注意:逗号分隔的值顺序不同。

例子:

T1:

ID_T1             NAME1
===================================


1      ASCORBIC ACID, PARACETAMOL, POTASSIUM HYDROGEN CARBONATE
2      SODIUM HYDROGEN CARBONATE, SODIUM CARBONATE ANHYDROUS, CITRIC ACID
3      CAFFEINE, PARACETAMOL PH. EUR.
4      PSEUDOEPHEDRINE HYDROCHLORIDE,DEXCHLORPHENIRAMINE MALEATE

T2:

ID_T2          NAME2
=================================

 4      POTASSIUM HYDROGEN CARBONATE, ASCORBIC ACID, PARACETAMOL
 5      SODIUM HYDROGEN CARBONATE, SODIUM CARBONATE ANHYDROUS
 6      PARACETAMOL PH. EUR.,CAFFEINE
 7      CODEINE PHOSPHATE, PARACETAMOL DC
 8      DEXCHLORPHENIRAMINE MALEATE, DEXTROMETHORPHAN HYDROBROMIDE 
10      DEXCHLORPHENIRAMINE MALEATE, PSEUDOEPHEDRINE HYDROCHLORIDE

我的结果应该只显示基于两个表中所有名称匹配的匹配行。

    ID_T1    ID_T2    MATCHING NAME
    ==================================
    1            4    POTASSIUM HYDROGEN CARBONATE, ASCORBIC ACID, PARACETAMOL
    3            6    PARACETAMOL PH. EUR.,CAFFEINE
    4           10    PSEUDOEPHEDRINE HYDROCHLORIDE,DEXCHLORPHENIRAMINE MALEATE

尝试使用 REGEXP_SUBST 但无法使其工作。

我使用下面的代码来解析这些值:

SELECT REGEXP_SUBSTR (NAME1, '[^,]+', 1, ROWNUM)
            FROM T1
           CONNECT BY ROWNUM <= LENGTH (NAME1) - 
           LENGTH (REPLACE (NAME, ',')) + 1

Answer 1

您可以将表格转换为第一范式，然后比较存储在每一行中的化合物。起点可以是:

{1} 标记每一行，并将标记写入新表。为每个标记赋予其原始 ID 加上 一个 3 字母前缀，指示该标记来自哪个表。 {2} 按 ID 对新(“规范化”)表的行进行分组，并执行 LISTAGG()。执行自连接，并找到匹配的“ token 组”。

{1} 标记化，创建表作为选择 (CTAS)

create table tokens
as 
select
  ltrim(        -- ltrim() and rtrim() remove leading/trailing spaces (blanks)
    rtrim( 
      substr( N.wrapped
      , instr( N.wrapped, ',', 1, T.pos ) + 1
      , ( instr( N.wrapped, ',', 1, T.pos + 1 ) - instr( N.wrapped, ',', 1, T.pos ) ) - 1 
      ) 
    )
  ) token
, N.id
from (        
  select ',' || name1 || ',' as wrapped, 'T1_' || to_char( id_t1 ) as id from t1 -- names wrapped in commas, (table)_id
  union all
  select ',' || name2 || ',' , 'T2_' || to_char( id_t2 ) from t2  
) N join (  
  select level as pos   -- (max) possible position of char in an existing token
  from dual 
  connect by level <= (
    select greatest(    -- find the longest string ie max position (query T1 and T2) 
      ( select max( length( name1 ) ) from t1 )
    , ( select max( length( name2 ) ) from t2 )
    ) as pos
    from dual
  )  
) T
  on T.pos <= ( length( N.wrapped ) - length( replace( N.wrapped, ',') ) ) - 1 
;

不使用 CONNECT BY 进行标记化的灵感来自 this SO answer .

TOKENS 表的内容如下所示:

SQL> select * from tokens ;
TOKEN                           ID       
ASCORBIC ACID                   T1_1     
SODIUM HYDROGEN CARBONATE       T1_2     
CAFFEINE                        T1_3     
PSEUDOEPHEDRINE HYDROCHLORIDE   T1_4     
PARACETAMOL                     T1_100   
sodium hydroxide                T1_110   
POTASSIUM HYDROGEN CARBONATE    T2_4     
SODIUM HYDROGEN CARBONATE       T2_5     
PARACETAMOL PH. EUR.            T2_6     
CODEINE PHOSPHATE               T2_7     
DEXCHLORPHENIRAMINE MALEATE     T2_8     
DEXCHLORPHENIRAMINE MALEATE     T2_10    
PARACETAMOL                     T2_200 
...

{2} GROUP BY、LISTAGG、自加入

select
  S1.id id1
, S2.id id2
, S1.tokengroup_T1
, S2.tokengroup_T2
from 
(
  select substr( id, 4, length( id ) - 3 ) id
  , listagg( token, ' + ' ) within group ( order by token ) tokengroup_T1
  from tokens
  group by id 
  having substr( id, 1, 3 ) = 'T1_'
) S1 
  join 
(
  select substr( id, 4, length( id ) - 3 ) id
  , listagg( token, ' + ' ) within group ( order by token ) tokengroup_T2
  from tokens
  group by id 
  having substr( id, 1, 3 ) = 'T2_'
) S2 
  on S1.tokengroup_T1 = S2.tokengroup_T2
;

-- result
ID1   ID2   TOKENGROUP_T1                                                 TOKENGROUP_T2                                                 
4     10    DEXCHLORPHENIRAMINE MALEATE + PSEUDOEPHEDRINE HYDROCHLORIDE   DEXCHLORPHENIRAMINE MALEATE + PSEUDOEPHEDRINE HYDROCHLORIDE   
110   210   potassium carbonate + sodium hydroxide                        potassium carbonate + sodium hydroxide                        
1     4     ASCORBIC ACID + PARACETAMOL + POTASSIUM HYDROGEN CARBONATE    ASCORBIC ACID + PARACETAMOL + POTASSIUM HYDROGEN CARBONATE    
3     6     CAFFEINE + PARACETAMOL PH. EUR.                               CAFFEINE + PARACETAMOL PH. EUR. 

当这样做时，您可以将物质按(字母)排序，您还可以在这里选择一个您喜欢的“分隔符”(我们使用了“+”)。

备选方案

如果所有这些对您都没有用，或者您认为这太复杂了，那么您可以尝试使用 TRANSLATE()。在这种情况下，我建议从您的数据集中去除所有空格/空白(在查询中 - 不改变原始数据!)像这样:

查询

select 
  id1, id2
, name1, name2
from (
  select 
    id_t1 id1
  , id_t2 id2
  , T1.name1 name1
  , T2.name2 name2
  from T1
    join T2 
      on  translate( replace( T1.name1, ' ', '' ), replace( T2.name2, ' ', '' ), '!' )
        = translate( replace( T2.name2, ' ', '' ), replace( T1.name1, ' ', '' ), '!' )
) ;

结果

  ID1   ID2 NAME1                                                                NAME2                                                        
    2     5 SODIUM HYDROGEN CARBONATE, SODIUM CARBONATE ANHYDROUS, CITRIC ACID   SODIUM HYDROGEN CARBONATE, SODIUM CARBONATE ANHYDROUS        
    3     6 CAFFEINE, PARACETAMOL PH. EUR.                                       PARACETAMOL PH. EUR.,CAFFEINE                                
  100    10 PARACETAMOL, DEXTROMETHORPHAN, PSEUDOEPHEDRINE, PYRILAMINE           DEXCHLORPHENIRAMINE MALEATE, PSEUDOEPHEDRINE HYDROCHLORIDE   
  110   210 sodium hydroxide, potassium carbonate                                sodium hydroxide, potassium carbonate

注意: 我已将以下行添加到您的示例数据中:

-- T1
110, 'sodium hydroxide, potassium carbonate'

-- T2
210, 'sodium hydroxide, potassium carbonate' 
211, 'potassium hydroxide, sodium carbonate'

我发现使用 TRANSLATE() 很容易产生“误报”，即 ID 为 110、210 和 211 的物质看起来“匹配”。 (换句话说:我认为这不是这项工作的正确工具。)

DBFIDDLE here

(点击链接查看示例表和查询)。