在oracle sql中用','从字符串分隔符中获取数字

时间:2013-11-11 08:39:15

标签: sql regex oracle

我有一个varchar collumn,它可能包含这样的格式:

  

123124125126

现在我想获取所有数字,并将它放在select list

中的这一列中
  

123
  124个
  125个
  126

有什么想法吗?

3 个答案:

答案 0 :(得分:1)

试试这个,

with test as 
(
SELECT '123,124,125,126' str FROM dual  
)  
SELECT regexp_substr (str, '[^,]+', 1, ROWNUM) SPLIT  
FROM   TEST  
CONNECT BY LEVEL <= LENGTH (regexp_replace (str, '[^,]+'))  + 1;

如果最后还有一个额外的逗号,请尝试此操作

with test as 
(
SELECT '123,124,125,126,' str FROM dual  
)
SELECT regexp_substr(str,'[^,]+', 1, LEVEL) FROM test
connect by regexp_substr(str, '[^,]+', 1, level) is not null;

答案 1 :(得分:0)

无数次回答......

WITH CTE
    AS (SELECT
             '123,124,125,126' AS COL1
        FROM
             DUAL)
SELECT
      REGEXP_SUBSTR ( COL1,
                   '[^,]+',
                   1,
                   RN )
          COL1
FROM
          CTE
      CROSS JOIN
          (SELECT
                ROWNUM RN
           FROM
                (SELECT
                       MAX ( LENGTH ( REGEXP_REPLACE ( COL1,
                                                '[^,]+' ) ) )
                       + 1
                           MAX_L
                 FROM
                       CTE)
           CONNECT BY
                LEVEL <= MAX_L)
WHERE
      REGEXP_SUBSTR ( COL1,
                   '[^,]+',
                   1,
                   RN )
          IS NOT NULL
ORDER BY
      COL1;

答案 2 :(得分:0)

或者; substrinstrlagregexp_count在一起:

select substr(str,second,first-second) as "Result String"       
  from
  (
    with t(str) as
    (   
     select '123,124,125,126' from dual
    )
     select replace(instr(str,',',1,level),0,length(str)+1) first,
            nvl(lag(instr(str,',',1,level)) over (order by level),0)+1 second,
            str              
       from dual
       cross join ( select str from t )
     connect by level <= regexp_count(str,',')+1
  );

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