linux 查找正则表达式
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linux find regex
提问by Dijkstra
I'm having trouble using the regex of the findcommand. Probably something I don't understand about escaping on the command line.
我在使用find命令的正则表达式时遇到问题。可能我不明白在命令行上转义。
Why are these not the same?
为什么这些不一样？
find -regex '.*[1234567890]'
find -regex '.*[[:digit:]]'
Bash, Ubuntu
巴什，Ubuntu
采纳答案by dogbane
Regular expressions with character classes (e.g. [[:digit:]]) are not supported in the default regular expression syntax used by find. You need to specify a different regex type such as posix-extendedin order to use them.
使用[[:digit:]]的默认正则表达式语法不支持具有字符类（例如）的正则表达式find。您需要指定不同的正则表达式类型posix-extended才能使用它们。
Take a look at GNU Find's Regular Expression documentationwhich shows you all the regex types and what they support.
查看 GNU Find 的正则表达式文档，它向您展示了所有正则表达式类型及其支持的内容。
回答by bmk
You should have a look on the -regextypeargument of find, see manpage:
您应该查看 的-regextype参数find，请参阅联机帮助页：
-regextype type
Changes the regular expression syntax understood by -regex and -iregex
tests which occur later on the command line. Currently-implemented
types are emacs (this is the default), posix-awk, posix-basic,
posix-egrep and posix-extended.
I guess the emacstype doesn't support the [[:digit:]]construct. I tried it with posix-extendedand it worked as expected:
我猜该emacs类型不支持该[[:digit:]]构造。我试过了posix-extended，它按预期工作：
find -regextype posix-extended -regex '.*[1234567890]'
find -regextype posix-extended -regex '.*[[:digit:]]'
回答by StKiller
Well, you may try this '.*[0-9]'
嗯，你可以试试这个 '.*[0-9]'
回答by kurumi
Note that -regexdepends on whole path.
请注意，这-regex取决于整个路径。
-regex pattern
File name matches regular expression pattern.
This is a match on the whole path, not a search.
You don't actually have to use -regexfor what you are doing.
您实际上不必使用-regex您正在做的事情。
find . -iname "*[0-9]"
