linux 查找名称中带有可选字符的文件

Question

suppose I have two files: ac and abc . I want to find a regex to match both files. Normally I would expect the following regex to work, but it never does:

find ./ -name ab?c

I have tried escaping or not the questionmark, this never seems to work. Normally in the regex documentations I have found; ? means: previous character repeated 0 or 1 times , but find doesn't seem to understand this.

I have tried this on two different find versions: GNU find version 4.2.31 and find (GNU findutils) 4.6.0

PS: this works with * , but I specifically would like to match just one optional character .

find ./ -name a*c

gives

./ac
./abc

Answer 1

The expression passed to -name is not a regex, it is a glob expression. A (single) glob expression can't be used for your use case but you can use regular expressions using -regex :

find -regex '.*/ab?c'

Btw, the default regular expression language is Emacs Style as explained here : https://www.emacswiki.org/emacs/RegularExpression . You can change the regex language using -regextype .

Answer 2

To match the expression with only one optional character try using or option:

touch abc ac abbc
find . -name "a?c" -or -name "ac"

Gives you only: abc and ac names.

Generally you can build pretty complex find queries using or and and options =)

Answer 3

The find -name option uses a glob pattern, which is not the same as a regex. For globs, ? means any single character. If you want a character to be optional, you need to use two patterns:

find ./ -name abc -o -name ac

Answer 4

Other answers are good enough to have a solution but knowing find 's -regex option matches on whole entry is essential. So you can't just do a partial match:

-regex 'ab?c'

You have to use one or two dot-stars:

-regex '.*ab?c.*'

Also without wildcards this would be possible using grep :

ls . | grep 'ab\?c'