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在Oracle中拆分逗号分隔的值

在Oracle中拆分逗号分隔的值

非常适合我-

sql> WITH dummy_table AS(
  2  SELECT '3862,3654,3828' dummy FROM dual
  3  )
  4  SELECT trim(regexp_substr(dummy,'[^,]+',1,Level)) AS dummycol
  5  FROM dummy_table
  6    CONNECT BY level <= LENGTH(REGEXP_REPLACE(dummy,'[^,]+'))+1
  7  /

DUMMYCOL
--------------
3862
3654
3828

sql>

还有许多其他方法可以实现它。阅读将单个逗号分隔的字符串拆分成行

。只见PRIOR子句中使用DBMS_RANDOM的摆脱循环回路的在这里

尝试以下方法

sql> WITH dummy_table AS
  2    ( SELECT 1 rn, '3862,3654,3828' dummy FROM dual
  3    UNION ALL
  4    SELECT 2, '1234,5678' dummy FROM dual
  5    )
  6  SELECT trim(regexp_substr(dummy,'[^,]+',1,Level)) AS dummycol
  7  FROM dummy_table
  8    CONNECT BY LEVEL          <= LENGTH(REGEXP_REPLACE(dummy,'[^,]+'))+1
  9  AND prior rn                 = rn
 10  AND PRIOR DBMS_RANDOM.VALUE IS NOT NULL
 11  /

DUMMYCOL
--------------
3862
3654
3828
1234
5678

sql>

其他方式,

sql> WITH dummy_table AS
  2    ( SELECT 1 rn, '3862,3654,3828' dummy FROM dual
  3    UNION ALL
  4    SELECT 2, '1234,5678,xyz' dummy FROM dual
  5    )
  6  SELECT trim(regexp_substr(t.dummy, '[^,]+', 1, levels.column_value)) AS dummycol
  7  FROM dummy_table t,
  8    TABLE(CAST(MULTISET
  9    (SELECT LEVEL
 10    FROM dual
 11      CONNECT BY LEVEL <= LENGTH (regexp_replace(t.dummy, '[^,]+')) + 1
 12    ) AS sys.OdciNumberList)) LEVELS
 13    /

DUMMYCOL
--------------
3862
3654
3828
1234
5678
xyz

6 rows selected.

sql>
Oracle 2022/1/1 18:39:48 有347人围观

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