我按照本文中列出的说明将分隔的字符串拆分为表格的行:
Splitting string into multiple rows in Oracle
适用于该特定分隔字符串的答案在此小提琴中表示:
with temp as
(
select 108 Name, 'test' Project, 'Err1, Err2, Err3' Error from dual
union all
select 109, 'test2', 'Err1' from dual
)
select distinct
t.name, t.project,
trim(regexp_substr(t.error, '[^,]+', 1, levels.column_value)) as error
from
temp t,
table(cast(multiset(select level from dual connect by level <= length (
regexp_replace(t.error, '[^,]+')) + 1) as sys.OdciNumberList)) levels
order by name;
不幸的是我的字符串不是用逗号分隔的。它由子串':::'分隔。我尝试通过在下面的小提琴中编写SQL来改变答案以适应我的情况:
with temp as
(
select 108 Name, 'test' Project, 'Err1:::Err2:::Err3' Error from dual
union all
select 109, 'test2', 'Err1' from dual
)
select distinct
t.name, t.project,
trim(regexp_substr(t.error, '[^:::]+', 1, levels.column_value)) as error
from
temp t,
table(cast(multiset(select level from dual connect by level <= length (
regexp_replace(t.error, '[^:::]+')) + 1) as sys.OdciNumberList)) levels
order by name
正如您所看到的,我将测试字符串更改为由“:::”分隔并更改了正则表达式以匹配,但查询正在生成一个无关的行,其中返回的子字符串的值为Null。
任何人都可以帮助我理解为什么我所做的更改会产生具有Null值的无关行吗?
答案 0 :(得分:0)
只需使用REPLACE
和标准代码,
<强> SqlFiddleDemo 强>
with temp as
(
select 108 Name, 'test' Project, 'Err1:::Err2:::Err3' Error from dual
union all
select 109, 'test2', 'Err1' from dual
)
select distinct
t.name, t.project,
trim(regexp_substr(REPLACE(t.error, ':::', ', '), '[^,]+', 1, levels.column_value)) as error
from
temp t,
table(cast(multiset(select level from dual connect by level <= length (regexp_replace(REPLACE(t.error, ':::', ', '), '[^,]+')) + 1) as sys.OdciNumberList)) levels
order by name
或者您需要除以分隔符的长度:
<强> SqlFiddle 强>
with temp as
(
select 108 Name, 'test' Project, 'Err1:::Err2:::Err3' Error from dual
union all
select 109, 'test2', 'Err1:::Err2' from dual
)
select distinct
t.name, t.project,
trim(regexp_substr(t.error, '[^:::]+', 1, levels.column_value)) as error
from
temp t,
table(cast(multiset(select level from dual connect by level <= length (
regexp_replace(t.error, '[^:::]+'))/3 + 1) as sys.OdciNumberList)) levels
order by name
你可以看到执行的原因:
SELECT length (regexp_replace('Err1:::Err2:::Err3', '[^:::]+')) + 1 AS l
FROM dual
这将返回7和你的:
SELECT DISTINCT t.name, t.project,
trim(regexp_substr(t.error, '[^:::]+', 1, levels.column_value)) as error
将尝试获得regexp_substr
7次出现,其中4次为NULL
,最后4 NULL
将被NULL
压扁为DISTINCT
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