Question

我有一个字符串'1,1,2,3,4,4,5,6,6,7'存储在一列中。我需要使用sql查询将不同的逗号分隔值作为输出。

e.g。对于给定的字符串输出应为'1,2,3,4,5,6,7'。输出中没有重复项。

Answer 1

没有正则表达式：

WITH t AS
( SELECT '1,2,3,3,3,4,5,6,7,7,7,7' AS num FROM dual
)
SELECT DISTINCT
       SUBSTR (
         num
       , instr(num, ',', 1, level) + 1
       , instr(num, ',', 1, level + 1) - instr(num, ',', 1, level) - 1)
        AS numbers
FROM (select ','||num||',' num from t)
CONNECT BY level <= length(num) - length(replace(num,',')) -1

使用正则表达式：

SELECT DISTINCT REGEXP_SUBSTR( '1,1,2,3,4,4,5,6,6,7' , '[^,]+', 1, lvl)
FROM DUAL, 
(SELECT LEVEL lvl
FROM DUAL
CONNECT BY LEVEL <= LENGTH( '1,1,2,3,4,4,5,6,6,7' ) - LENGTH(REPLACE( '1,1,2,3,4,4,5,6,6,7' , ','))+1)
WHERE lvl <= LENGTH( '1,1,2,3,4,4,5,6,6,7' ) - LENGTH(REPLACE( '1,1,2,3,4,4,5,6,6,7' , ',')) + 1

Answer 2

我们可以使用regex_substr和connect by来做到这一点。请试试这个。

 select distinct num from
(SELECT REGEXP_SUBSTR('1,1,2,3,4,4,5,6,6,7','[^,]+',1,level) as num
FROM DUAL 
CONNECT BY LEVEL<= LENGTH(REGEXP_REPLACE('1,1,2,3,4,4,5,6,6,7','[^,]','')));

没有正则表达式：经过一些问题的澄清

 with t as (SELECT distinct substr(replace('1,1,2,3,4,4,5,6,6,7',','),level,1)||',' as num
 FROM DUAL
 CONNECT BY LEVEL<=  LENGTH( '1,1,2,3,4,4,5,6,6,7' ) - LENGTH(REPLACE( '1,1,2,3,4,4,5,6,6,7' , ','))+1)
 select listagg(num) within group (order by num) from t;

Answer 3

尝试

select
  regexp_replace('1,1,2,3,4,4,5,6,6,7', '([^,]+),\1', '\1')
from 
   dual;

但是，如果您的输入字符串包含的图形超过两次，则此方法不起作用。如果这困扰你，你可能想尝试

select
  regexp_replace('1,1,2,3,4,4,4,5,6,6,6,6,6,6,7', '([^,]+)(,\1)+', '\1')
from dual;

Answer 4

根据我的观点：

select wm_concat(distinct substr(replace('1,1,2,3,4,4,5,6,6,7',',',''),level,1)) as out     
from dual connect by level <= length('1,1,2,3,4,4,5,6,6,7');

Answer 5

SELECT 
  listagg(ra,',')  WITHIN GROUP (ORDER BY ra)
FROM
(
  SELECT
  DISTINCT (REGEXP_SUBSTR('02,02,02,02,02,03,04,03', '[^,]+', 1, LEVEL) )ra
  FROM DUAL
  CONNECT BY REGEXP_SUBSTR('02,02,02,02,02,03,04,03', '[^,]+', 1, LEVEL) IS 
  NOT NULL
);

oracle中的逗号分隔值不同

5 个答案: