我有以下查询:
SELECT DISTINCT A.REZ FROM
(
SELECT REGEXP_SUBSTR(P_EQUATION, '([A-Z|a-z|0-9]+)\{([0-9|\+|\-| |\*|\/\)\(]+)\}#([A-Z|a-z|0-9|_]+)#',1, LEVEL) AS REZ FROM DUAL
CONNECT BY REGEXP_SUBSTR(P_EQUATION, '([A-Z|a-z|0-9]+)\{([0-9|\+|\-| |\*|\/\)\(]+)\}#([A-Z|a-z|0-9|_]+)#',1, LEVEL) IS NOT NULL
) A;
如果我提供了以下输入:
P_EQUATION := 'A123{(01+02)*2}#ACCOUNT_BALANCE# + B123{(20+10)/20}#ACCOUNT_BALANCE#';
它给了我以下内容:
REZ
-------------------------------------
A123{(01+02)*2}#ACCOUNT_BALANCE#
B123{(20+10)/20}#ACCOUNT_BALANCE#
但是,虽然减号包含在模式中,但如果我将其添加到花括号内,它将不再识别文本作为匹配!
例如:
P_EQUATION := 'A123{(01-02)*2}#ACCOUNT_BALANCE#';
我无法找到解决方案,这让我感到非常震惊,尤其是当我尝试单独使用减号时,如果我尝试单独匹配数字,它也能正常工作:(
答案 0 :(得分:1)
Oracle似乎正在使用POSIX样式的正则表达式:https://docs.oracle.com/cd/B12037_01/server.101/b10759/ap_posix001.htm#i690819
反斜杠不是POSIX括号表达式中的元字符。所以在POSIX中,正则表达式
[\d]
匹配\
或d
反斜杠可能搞乱了,而且他们没有必要。你也没有意识到|
是一个char类中的文字(评论也指出了这一点)。我已经修复了这些问题,并将-
移动到char类的开头,这允许它被解释为文字。
你走了:
([A-Za-z0-9]+)\{([-0-9+ */)(]+)\}#([A-Za-z0-9_]+)#
答案 1 :(得分:1)
无法弄清楚代码的问题,但这是一种方法:
with temp as
(
select 'A123{(01+02)*2}#ACCOUNT_BALANCE# + B123{(20+10)/20}#ACCOUNT_BALANCE#' P_EQUATION from dual union all
select 'A123{(01-02)*2}#ACCOUNT_BALANCE#' P_EQUATION from dual
)
SELECT DISTINCT A.REZ FROM
(
SELECT REGEXP_SUBSTR(P_EQUATION, '[[:alpha:]]+[[:digit:]]+{\([[:digit:]]+\S[[:digit:]]+\)\S[[:digit:]]+}#[[:alpha:]]+_[[:alpha:]]+#',1, LEVEL) AS REZ FROM temp
CONNECT BY REGEXP_SUBSTR(P_EQUATION, '[[:alpha:]]+[[:digit:]]+{\([[:digit:]]+\S[[:digit:]]+\)\S[[:digit:]]+}#[[:alpha:]]+_[[:alpha:]]+#',1, LEVEL) IS NOT NULL
) A;
输出:
REZ
---------------------------------------
B123{(20+10)/20}#ACCOUNT_BALANCE#
A123{(01-02)*2}#ACCOUNT_BALANCE#
A123{(01+02)*2}#ACCOUNT_BALANCE#