使用 regexp_substr 按顺序拆分 Oracle 中的字符串
我在 Oracle 数据库中有一个字符串，我的字符串是:'bbb;aaa;qqq;ccc'
I have a string in Oracle database, my string is: 'bbb;aaa;qqq;ccc'
我使用正则表达式来分割我的字符串:
I used regexp for split my string:
select distinct trim(regexp_substr('bbb;aaa;qqq;ccc','[^;]+', 1,level) ) as q
from dual
connect by regexp_substr('bbb;aaa;qqq;ccc', '[^;]+', 1, level) is not null ;
我想按顺序拆分它，我希望总是以下输出:
I want to split it in order, I expected the below output always:
bbb
aaa
qqq
ccc
因为 subString 的顺序对我来说非常重要.但是这个查询的结果不是有序的:
because order of the subString are very important for me. but the result of this query is not in order:
qqq
aaa
bbb
ccc
推荐答案
您不需要 DISTINCT 即可获得结果；此外，要以给定的顺序获得结果，您只需要一个 ORDER BY 子句:
You don't need a DISTINCT to get your result; besides, to get the result in a given order, all you need is an ORDER BY clause:
select trim(regexp_substr('bbb;aaa;qqq;ccc','[^;]+', 1,level) ) as q
from dual
connect by regexp_substr('bbb;aaa;qqq;ccc', '[^;]+', 1, level) is not null
order by level
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