如何将逗号分隔值转换为 oracle 中的行？

Question

这是 DDL——

create table tbl1 (
   id number,
   value varchar2(50)
);

insert into tbl1 values (1, 'AA, UT, BT, SK, SX');
insert into tbl1 values (2, 'AA, UT, SX');
insert into tbl1 values (3, 'UT, SK, SX, ZF');

注意，这里的值是逗号分隔的字符串。

但是，我们需要如下结果 -

ID VALUE
-------------
1  AA
1  UT
1  BT
1  SK
1  SX
2  AA
2  UT
2  SX
3  UT
3  SK
3  SX
3  ZF

我们怎么写SQL呢？

Answer 1

我同意这是一个非常糟糕的设计。 如果您无法更改该设计，请尝试以下操作：

select distinct id, trim(regexp_substr(value,'[^,]+', 1, level) ) value, level
  from tbl1
   connect by regexp_substr(value, '[^,]+', 1, level) is not null
   order by id, level;

输出

id value level
1   AA  1
1   UT  2
1   BT  3
1   SK  4
1   SX  5
2   AA  1
2   UT  2
2   SX  3
3   UT  1
3   SK  2
3   SX  3
3   ZF  4

归功于此

以更优雅和有效的方式删除重复项（感谢@mathguy）

select id, trim(regexp_substr(value,'[^,]+', 1, level) ) value, level
  from tbl1
   connect by regexp_substr(value, '[^,]+', 1, level) is not null
      and PRIOR id =  id 
      and PRIOR SYS_GUID() is not null  
   order by id, level;

如果您想要“ANSIer”方法，请使用 CTE：

with t (id,res,val,lev) as (
           select id, trim(regexp_substr(value,'[^,]+', 1, 1 )) res, value as val, 1 as lev
             from tbl1
            where regexp_substr(value, '[^,]+', 1, 1) is not null
            union all           
            select id, trim(regexp_substr(val,'[^,]+', 1, lev+1) ) res, val, lev+1 as lev
              from t
              where regexp_substr(val, '[^,]+', 1, lev+1) is not null
              )
select id, res,lev
  from t
order by id, lev;

输出

id  val lev
1   AA  1
1   UT  2
1   BT  3
1   SK  4
1   SX  5
2   AA  1
2   UT  2
2   SX  3
3   UT  1
3   SK  2
3   SX  3
3   ZF  4

MT0 的另一种递归方法，但没有正则表达式：

WITH t ( id, value, start_pos, end_pos ) AS
  ( SELECT id, value, 1, INSTR( value, ',' ) FROM tbl1
  UNION ALL
  SELECT id,
    value,
    end_pos                    + 1,
    INSTR( value, ',', end_pos + 1 )
  FROM t
  WHERE end_pos > 0
  )
SELECT id,
  SUBSTR( value, start_pos, DECODE( end_pos, 0, LENGTH( value ) + 1, end_pos ) - start_pos ) AS value
FROM t
ORDER BY id,
  start_pos;

我尝试了 3 种方法，使用 30000 行数据集和 118104 行返回，并得到以下平均结果：

• 我的递归方法：5 秒
• MT0 进场：4 秒
• Mathguy 方法：16 秒
• MT0 递归方法无正则表达式：3.45 秒

@Mathguy 还使用更大的数据集进行了测试：

在所有情况下，递归查询（我只测试了带有常规 substr 和 instr 的查询）效果更好，提高了 2 到 5 倍。以下是每个字符串的字符串数/标记数和分层与递归的 CTAS 执行时间的组合，层次第一。 所有时间以秒为单位

• 30,000 x 4：5 / 1。
• 30,000 x 10：15 / 3。
• 30,000 x 25：56 / 37。
• 5,000 x 50：33 / 14。
• 5,000 x 100：160 / 81。
• 10,000 x 200：1,924 / 772

Answer 2

这将获得值，而无需您删除重复项或必须使用在CONNECT BY中包含SYS_GUID()或DBMS_RANDOM.VALUE() ：

SELECT t.id,
       v.COLUMN_VALUE AS value
FROM   TBL1 t,
       TABLE(
         CAST(
           MULTISET(
             SELECT TRIM( REGEXP_SUBSTR( t.value, '[^,]+', 1, LEVEL ) )
             FROM   DUAL
             CONNECT BY LEVEL <= REGEXP_COUNT( t.value, '[^,]+' )
           )
           AS SYS.ODCIVARCHAR2LIST
         )
       ) v

更新：

返回列表中元素的索引：

选项 1 - 返回 UDT：

CREATE TYPE string_pair IS OBJECT( lvl INT, value VARCHAR2(4000) );
/

CREATE TYPE string_pair_table IS TABLE OF string_pair;
/

SELECT t.id,
       v.*
FROM   TBL1 t,
       TABLE(
         CAST(
           MULTISET(
             SELECT string_pair( level, TRIM( REGEXP_SUBSTR( t.value, '[^,]+', 1, LEVEL ) ) )
             FROM   DUAL
             CONNECT BY LEVEL <= REGEXP_COUNT( t.value, '[^,]+' )
           )
           AS string_pair_table
         )
       ) v;

选项 2 - 使用ROW_NUMBER() ：

SELECT t.id,
       v.COLUMN_VALUE AS value,
       ROW_NUMBER() OVER ( PARTITION BY id ORDER BY ROWNUM ) AS lvl
FROM   TBL1 t,
       TABLE(
         CAST(
           MULTISET(
             SELECT TRIM( REGEXP_SUBSTR( t.value, '[^,]+', 1, LEVEL ) )
             FROM   DUAL
             CONNECT BY LEVEL <= REGEXP_COUNT( t.value, '[^,]+' )
           )
           AS SYS.ODCIVARCHAR2LIST
         )
       ) v;

Answer 3

Vercelli 发布了正确答案。 但是，要拆分的字符串不止一个， connect by将生成呈指数增长的行数，其中包含许多重复项。 （只需尝试没有distinct的查询。）这将破坏非平凡大小的数据的性能。

克服这个问题的一种常见方法是使用prior条件和附加检查以避免层次结构中的循环。 像这样：

select id, trim(regexp_substr(value,'[^,]+', 1, level) ) value, level
  from tbl1
   connect by regexp_substr(value, '[^,]+', 1, level) is not null
          and prior id = id
          and prior sys_guid() is not null
   order by id, level;

例如，参见关于 OTN 的讨论： https : //community.oracle.com/thread/2526535

Answer 4

另一种方法是定义一个简单的 PL/SQL 函数：

CREATE OR REPLACE FUNCTION split_String(
  i_str    IN  VARCHAR2,
  i_delim  IN  VARCHAR2 DEFAULT ','
) RETURN SYS.ODCIVARCHAR2LIST DETERMINISTIC
AS
  p_result       SYS.ODCIVARCHAR2LIST := SYS.ODCIVARCHAR2LIST();
  p_start        NUMBER(5) := 1;
  p_end          NUMBER(5);
  c_len CONSTANT NUMBER(5) := LENGTH( i_str );
  c_ld  CONSTANT NUMBER(5) := LENGTH( i_delim );
BEGIN
  IF c_len > 0 THEN
    p_end := INSTR( i_str, i_delim, p_start );
    WHILE p_end > 0 LOOP
      p_result.EXTEND;
      p_result( p_result.COUNT ) := SUBSTR( i_str, p_start, p_end - p_start );
      p_start := p_end + c_ld;
      p_end := INSTR( i_str, i_delim, p_start );
    END LOOP;
    IF p_start <= c_len + 1 THEN
      p_result.EXTEND;
      p_result( p_result.COUNT ) := SUBSTR( i_str, p_start, c_len - p_start + 1 );
    END IF;
  END IF;
  RETURN p_result;
END;
/

那么SQL就变得很简单了：

SELECT t.id,
       v.column_value AS value
FROM   TBL1 t,
       TABLE( split_String( t.value ) ) v

Answer 5

--converting row of data into comma sepaerated string
SELECT
    department_id,
    LISTAGG(first_name, ',') WITHIN GROUP(
        ORDER BY
            first_name
    ) comma_separted_data
FROM
    hr.employees
GROUP BY
    department_id;

--comma-separated string into row of data

CREATE TABLE t (
    deptno          NUMBER,
    employee_name   VARCHAR2(255)
);

INSERT INTO t VALUES (
    10,
    'mohan,sam,john'
);

INSERT INTO t VALUES (
    20,
    'manideeep,ashok,uma'
);

INSERT INTO t VALUES (
    30,
    'gopal,gopi,manoj'
);

SELECT
    deptno,
    employee_name,
    regexp_count(employee_name, ',') + 1,
    regexp_substr(employee_name, '\w+', 1, 1)
FROM
    t,
    LATERAL (
        SELECT
            level l
        FROM
            dual
        CONNECT BY
            level < regexp_count(employee_name, ',') + 1
    );

DROP TABLE t;

Answer 6

SELECT  COL1,   COL2
FROM    (   SELECT INDX, MY_STR1, MY_STR2, COL1_ELEMENTS, COL1, COL2_ELEMENTS, COL2
            FROM    (   SELECT 0 "INDX", COL1 "MY_STR1", COL1_ELEMENTS, COL1, '' "MY_STR2", COL2_ELEMENTS, COL2
                        FROM(
                                SELECT
                                    REPLACE(COL1, ', ', ',') "COL1",    -- In case there is a space after comma
                                    Trim(Length(Replace(COL1, ' ', ''))) - Trim(Length(Translate(REPLACE(COL1, ', ', ','), 'A,', 'A'))) + 1 "COL1_ELEMENTS",    -- Number of elements
                                    Replace(COL2, ', ', ',') "COL2",    -- In case there is a space after comma
                                    Trim(Length(Replace(COL2, ' ', ''))) - Trim(Length(Translate(REPLACE(COL2, ', ', ','), 'A,', 'A'))) + 1 "COL2_ELEMENTS"     -- Number of elements
                                FROM
                                    (SELECT 'aaa,bbb,ccc' "COL1", 'qq, ww, ee' "COL2" FROM DUAL)        -- Your example data
                            )
                    )
                MODEL       -- Modeling --> INDX = 0    COL1='aaa,bbb,ccc'      COL2='qq,ww,ee'
                    DIMENSION BY(0 as INDX)
                    MEASURES(COL1, COL1_ELEMENTS, COL2, CAST('a' as VarChar2(4000)) as MY_STR1, CAST('a' as VarChar2(4000)) as MY_STR2)
                    RULES ITERATE (10)      --UNTIL (ITERATION_NUMBER <= COL1_ELEMENTS[ITERATION_NUMBER + 1]) -- If you don't know the number of elements this should be bigger then you aproximation. Othewrwise it will split given number of elements
                    (
                        COL1_ELEMENTS[ITERATION_NUMBER + 1] = COL1_ELEMENTS[0],
                        MY_STR1[0] = COL1[CV()],
                        MY_STR1[ITERATION_NUMBER + 1] = SubStr(MY_STR1[ITERATION_NUMBER], InStr(MY_STR1[ITERATION_NUMBER], ',', 1) + 1),
                        COL1[ITERATION_NUMBER + 1] = SubStr(MY_STR1[ITERATION_NUMBER], 1, CASE WHEN InStr(MY_STR1[ITERATION_NUMBER], ',') <> 0 THEN InStr(MY_STR1[ITERATION_NUMBER], ',')-1 ELSE Length(MY_STR1[ITERATION_NUMBER]) END),
                        MY_STR2[0] = COL2[CV()],
                        MY_STR2[ITERATION_NUMBER + 1] = SubStr(MY_STR2[ITERATION_NUMBER], InStr(MY_STR2[ITERATION_NUMBER], ',', 1) + 1),
                        COL2[ITERATION_NUMBER + 1] = SubStr(MY_STR2[ITERATION_NUMBER], 1, CASE WHEN InStr(MY_STR2[ITERATION_NUMBER], ',') <> 0 THEN InStr(MY_STR2[ITERATION_NUMBER], ',')-1 ELSE Length(MY_STR2[ITERATION_NUMBER]) END)
                    )
        )
WHERE INDX > 0 And INDX <= COL1_ELEMENTS    -- INDX 0 contains starting strings
--
--  COL1  COL2
--  ----  ----
--  aaa   qq
--  bbb   ww
--  ccc   ee