在 Oracle DB 中拆分字符串

Question

例如，我有 3 行的表格：

|-Id-||--props--|
1        aa = black
2        aa = change
         d = eexplore
3        xxx = yield
         d = int
         mmmm = no

我想编写带有结果的 SQL 语句：

aa    
d
xxx
mmm

即输出对 x=y 的所有第一个元素。给定表的每一行中的对可以用newline分隔。

我尝试了以下 SQL 但结果不正确

    SELECT REGEXP_SUBSTR ('a = b, b = c',
                          '[^=]',
                          1,
                          LEVEL) as name
      FROM DUAL
CONNECT BY LEVEL <= (SELECT REGEXP_COUNT ('a = b, b = c', '=') FROM DUAL)

Answer 1

请试试下面的 sql fiddle here

with names as ( SELECT REGEXP_SUBSTR (regexp_replace('aa = b, bb = c , eee = d','[[:space:]]*',''), '[^=]+', 1, level ) as name FROM DUAL CONNECT BY LEVEL <= (SELECT REGEXP_COUNT ('a = b, b = c , e = d ', '=') FROM DUAL)) select nvl(regexp_substr(name,'[^,]+',1,2),regexp_substr(name,'[^,]+',1,1) ) from names;

Update1：​​-可以用逗号替换新行

在 Unix 上，LINEFEED (chr(10)) 是“行尾”标记。 在 windows 回车/LINEFEED (chr(13)||chr(10)) 应该是行尾标记

with names as
(
 SELECT REGEXP_SUBSTR (regexp_replace(replace('aa = black' ||chr(13)||chr(10)||'bb = change'||chr(13)||chr(10)||'mmmm=no',chr(13)||chr(10),','),'[[:space:]]*',''),
                          '[^=]+',
                          1,
                         level ) as name
      FROM DUAL
CONNECT BY LEVEL <= (SELECT REGEXP_COUNT 
('aa = black' ||chr(13)||chr(10)||'bb = change'||chr(13)||chr(10)||'mmmm=no', '=')
 FROM DUAL)

)
select nvl(regexp_substr(name,'[^,]+',1,2),
regexp_substr(name,'[^,]+',1) ) from names;

Answer 2

使用substr 返回第一个字符：

select substr(LEVEL, 1, 1) from tablename group by substr(LEVEL, 1, 1)