Question

我有一张如下表：

-------------
ID   | NAME
-------------
1001 | A,B,C
1002 | D,E,F
1003 | C,E,G
-------------

我希望这些值显示为：

-------------
ID   | NAME
-------------
1001 | A
1001 | B
1001 | C
1002 | D
1002 | E
1002 | F
1003 | C
1003 | E
1003 | G
-------------

我尝试过：

select split('A,B,C,D,E,F', ',') from dual; -- WILL RETURN COLLECTION

select column_value
from table (select split('A,B,C,D,E,F', ',') from dual); -- RETURN COLUMN_VALUE

Answer 1

尝试使用以下查询：

 WITH T AS (SELECT 'A,B,C,D,E,F' STR  FROM DUAL)   SELECT    
 REGEXP_SUBSTR (STR, '[^,]+', 1, LEVEL) SPLIT_VALUES  FROM T 
 CONNECT BY LEVEL <= (SELECT LENGTH (REPLACE (STR, ',', NULL)) FROM T)

ID下方查询：

WITH TAB AS 
(SELECT '1001' ID, 'A,B,C,D,E,F' STR FROM DUAL
)
SELECT    ID, 
REGEXP_SUBSTR (STR, '[^,]+', 1, LEVEL) SPLIT_VALUES  FROM TAB 
CONNECT BY LEVEL <= (SELECT LENGTH (REPLACE (STR, ',', NULL)) FROM TAB);

修改尝试使用以下查询进行多个ID和多次分离：

WITH TAB AS (SELECT '1001' ID, 'A,B,C,D,E,F' STR FROM DUAL UNION SELECT '1002' ID, 'D,E,F' STR FROM DUAL UNION SELECT '1003' ID, 'C,E,G' STR FROM DUAL ) select id, substr(STR, instr(STR, ',', 1, lvl) + 1, instr(STR, ',', 1, lvl + 1) - instr(STR, ',', 1, lvl) - 1) name from ( select ',' || STR || ',' as STR, id from TAB ), ( select level as lvl from dual connect by level <= 100 ) where lvl <= length(STR) - length(replace(STR, ',')) - 1 order by ID, NAME

Answer 2

有多种选择。请参阅Split comma delimited strings in a table in Oracle。

使用 REGEXP_SUBSTR：

SQL> WITH sample_data AS(
  2  SELECT 10001 ID, 'A,B,C' str FROM dual UNION ALL
  3  SELECT 10002 ID, 'D,E,F' str FROM dual UNION ALL
  4  SELECT 10003 ID, 'C,E,G' str FROM dual
  5  )
  6  -- end of sample_data mimicking real table
  7  SELECT distinct id, trim(regexp_substr(str, '[^,]+', 1, LEVEL)) str
  8  FROM sample_data
  9  CONNECT BY LEVEL <= regexp_count(str, ',')+1
 10  ORDER BY ID
 11  /

        ID STR
---------- -----
     10001 A
     10001 B
     10001 C
     10002 D
     10002 E
     10002 F
     10003 C
     10003 E
     10003 G

9 rows selected.

SQL>

使用 XMLTABLE：

SQL> WITH sample_data AS(
  2  SELECT 10001 ID, 'A,B,C' str FROM dual UNION ALL
  3  SELECT 10002 ID, 'D,E,F' str FROM dual UNION ALL
  4  SELECT 10003 ID, 'C,E,G' str FROM dual
  5  )
  6  -- end of sample_data mimicking real table
  7  SELECT id,
  8        trim(COLUMN_VALUE) str
  9  FROM sample_data,
 10       xmltable(('"'
 11          || REPLACE(str, ',', '","')
 12          || '"'))
 13  /

        ID STR
---------- ---
     10001 A
     10001 B
     10001 C
     10002 D
     10002 E
     10002 F
     10003 C
     10003 E
     10003 G

9 rows selected.

Answer 3

您可以尝试这样的事情：

CREATE OR REPLACE TYPE "STR_TABLE"
as table of varchar2


create or replace function GetCollection( iStr varchar2, iSplit char default ',' ) return STR_TABLE as
pStr varchar2(4000) := trim(iStr);
rpart varchar(255);
pColl STR_TABLE := STR_TABLE();
begin
   while nvl(length(pStr),0) > 0 loop
         pos := inStr(pStr, iSplit );
         if pos > 0 then
            rpart := substr(pStr,1, pos-1);
            pStr  := substr(pStr,pos+1,length(pStr));
         else
            rpart := pStr;
            pStr := null;
         end if;
         if rpart is not null then
           pColl.Extend;
           pColl(pColl.Count) := rpart;
         end if;
   end loop;
   return pColl;
end;

Answer 4

不要使用CONNECT BY或REGEXP，这会在复杂查询中产生笛卡尔积。此外，上述解决方案希望您知道可能的结果（A，B，C，D，E，F）而不是组合列表

使用XMLTable：

SELECT c.fname, c.lname,
trim(COLUMN_VALUE) EMAIL_ADDRESS
 FROM 
  CONTACTS c, CONTACT_STATUS s,
  xmltable(('"'
  || REPLACE(EMAIL_ADDRESS, ';', '","')
  || '"'))
where  c.status = s.id

COLUMN_VALUE是属于xmltable的伪列。这是快速和正确的，允许您引用不知道其值的列。

这将获取列并创建一个值表＆＃34; item＆＃34;，＆＃34; item2＆＃34;，＆＃34; item3＆＃34;并自动加入其源表（CONTACTS）。这是在数千行上测试的

注意＆＃39;;＆＃39;在xmltable中是列字段中的分隔符。

Answer 5

我用这种方式解决了类似问题...

    select YT.ID,
           REPLACE(REGEXP_SUBSTR(','||YT.STR||',',',.*?,',1,lvl.lvl),',','') AS STR
    from YOURTABLE YT
    join (select level as lvl 
          from dual 
          connect by level <= (select max(regexp_count(STR,',')+1) from YOURTABLE)
         ) lvl on lvl.lvl <= regexp_count(YT.STR,',')+1

Answer 6

我尝试了Lalit Kumar B的解决方案，到目前为止一直有效。但是随着更多数据，我遇到了性能问题（> 60行，> 7级）。因此我使用了更静态的变体，我想作为替代方案分享。

WITH T AS (
      SELECT 1001 AS ID, 'A,B,C' AS NAME FROM DUAL
UNION SELECT 1002 AS ID, 'D,E,F' AS NAME FROM DUAL
UNION SELECT 1003 AS ID, 'C,E,G' AS NAME FROM DUAL
     )   --SELECT * FROM T
SELECT ID as ID,
       distinct_column AS NAME
  FROM ( SELECT t.ID,       
                trim(regexp_substr(t.NAME, '[^,]+', 1,1)) AS c1,
                trim(regexp_substr(t.NAME, '[^,]+', 1,2)) AS c2,
                trim(regexp_substr(t.NAME, '[^,]+', 1,3)) AS c3,
                trim(regexp_substr(t.NAME, '[^,]+', 1,4)) AS c4 -- etc.
           FROM T )
UNPIVOT ( distinct_column FOR cn IN ( c1, c2, c3, c4 ) )    


    ID NAME               
------ ------
  1001 A                    
  1001 B                    
  1001 C                    
  1002 D                    
  1002 E                    
  1002 F                    
  1003 C                    
  1003 E                    
  1003 G                    

9 Zeilen gewählt

Answer 7

此版本也适用于长度超过一个字符串的字符串：

select regexp_substr('A,B,C,Karl-Heinz,D','[^,]+', 1, level) from dual
  connect by regexp_substr('A,B,C,Karl-Heinz,D', '[^,]+', 1, level) is not null;

请参阅How to split comma separated string and pass to IN clause of select statement

通过Oracle SQL查询拆分行中列的逗号分隔值

7 个答案: