从方括号中提取字符串

Question

I need to get the value myfile.ext from this example string: 'My message [file:myfile.ext]'

I saw a similar question , but couldn't manage to make the string returns empty string if the brackets are not in the string.

I need to check if the string contains [file:***] and return to the variable the *** otherwise must be empty value.

Answer 1

You can use grep for this:

$ grep -Po '(?<=\[file:)[^]]*(?=])' file
myfile.ext

Which is the same as (depending if you are reading from a file or piping):

$ echo "'My message [file:myfile.ext]'" | grep -Po '(?<=\[file:)[^]]*(?=])'
myfile.ext

Explanation

It looks for the string coming after [file: and up to next ] . To store it into a variable, use the var=$(command) expression:

result=$(grep -Po '(?<=\[file:)[^]]*(?=])' file)

it will be empty as default, and myfile.ext otherwise.

Sample

$ cat a
My message [file:myfile.ext
My message [file:myfile.ext]]a]

$ grep -Po '(?<=\[file:)[^]]*(?=])' a
myfile.ext

Answer 2

您可以使用sed的-n标志来禁止打印所有行，然后在正则表达式中专门使用p标志来打印匹配项：

sed -n 's/\[file:\([^\]*\)]/\1/p' file

Answer 3

You can use this BASH regex:

[[ "$s" == *[* ]] && [[ "$s" =~ \[[^:]*:([^\]]+)\] ]] && echo "${BASH_REMATCH[1]}"
myfile.ext