What's the correct RegEx syntax for separating on a backward and forward slash?
用于分离后向和正向斜杠的正确RegEx语法是什么?
For Oracle I have tried:
对于Oracle,我尝试过:
string
word1\/word2\/word3
regexp_substr(string, '[^"\\/"]+', 1, 1) first
but it does not work correctly
但它无法正常工作
2 个解决方案
#1
You're using a character class [^"\\/"]; that selects all characters that aren't in the set {", /, \}. That's probably not what you want.
你正在使用一个字符类[^“\\ /”];选择不在集合{“,/,\}中的所有字符。这可能不是你想要的。
Instead, try
`.*?(\\/|$)`
This will match up to the \/ or the end of string, whichever comes first. Note that the \/ will be included in the match, because Oracle's regexp_substr doesn't appear to support lookahead assertions yet.
这将匹配字符串的\ /或结尾,以先到者为准。请注意,\ /将包含在匹配中,因为Oracle的regexp_substr似乎还不支持先行断言。
#2
If you want to get all the words you could just replace the \/ with space:
如果你想得到所有单词,你可以用空格替换\ /:
select REPLACE ('word1\/word2\/word3','\/',' ') "changes" FROM dual
Or if you want only the first word:
或者如果你只想要第一个词:
select REGEXP_SUBSTR('word1\/word2\/word3', '[^\\^/]+') "changes" FROM dual
#1
You're using a character class [^"\\/"]; that selects all characters that aren't in the set {", /, \}. That's probably not what you want.
你正在使用一个字符类[^“\\ /”];选择不在集合{“,/,\}中的所有字符。这可能不是你想要的。
Instead, try
`.*?(\\/|$)`
This will match up to the \/ or the end of string, whichever comes first. Note that the \/ will be included in the match, because Oracle's regexp_substr doesn't appear to support lookahead assertions yet.
这将匹配字符串的\ /或结尾,以先到者为准。请注意,\ /将包含在匹配中,因为Oracle的regexp_substr似乎还不支持先行断言。
#2
If you want to get all the words you could just replace the \/ with space:
如果你想得到所有单词,你可以用空格替换\ /:
select REPLACE ('word1\/word2\/word3','\/',' ') "changes" FROM dual
Or if you want only the first word:
或者如果你只想要第一个词:
select REGEXP_SUBSTR('word1\/word2\/word3', '[^\\^/]+') "changes" FROM dual