如何使用 Bash 提取字符串的一部分

Question

I have been trying to extract part of string in bash. I'm using it on Windows 10. Basically I want to remove "artifacts/" sfscripts_artifact _ and ".zip"

Original String

artifacts/online-order-api_sfscripts_artifact_1.5.6-6.zip

I've tried

input="artifacts/online-order-api_sfscripts_artifact_1.5.6-6.zip"
echo "${input//[^0-9.-]/}"

Output

--1.5.6-6.

Expected Output

online-order-api 1.5.6-6

Answer 1

You may use this awk solution:

s='artifacts/online-order-api_sfscripts_artifact_1.5.6-6.zip'
awk -F_ '{gsub(/^[^\/]*\/|\.[^.]*$/, ""); print $1, $NF}' <<< "$s"

online-order-api 1.5.6-6

Or else this sed solution:

sed -E 's~^[^/]*/|\.[^.]+$~~g; s~(_[^_]+){2}_~ ~;' <<< "$s"

online-order-api 1.5.6-6

Answer 2

As a general solution using only variable expansion, consider:

input='artifacts/online-order-api_sfscripts_artifact_1.5.6-6.zip'

part0=${input%%_*}
part0=${part0##*/}
part1=${input##*_}
part1=${part1%.*}

echo "${part0} ${part1}"

Output:

online-order-api 1.5.6-6

Answer 3

Similar to the answer of adebayo10k, but in the order indicated by the user:

# Remove .zip from the end
tmp0="${input%.zip}"
# Remove path
tmp1="${tmp0##*/}"
# Extract version (remove everything before last underscore)
version="${tmp1##*_}"
# Extract name (remove everything after first underscore)
name="${tmp1%%_*}"
# print stuff
echo "${name}" "${version}"

Answer 4

A solution in pure bash using the =~ operator.

[[ $input =~ .*/([^_]*).*_(.*)\.[^.]*$ ]] &&
    echo "${BASH_REMATCH[1]} ${BASH_REMATCH[2]}"

prints out

online-order-api 1.5.6-6

with the given input.

Answer 5

Given your input,

echo ${input#artifacts/}

seems to me the simplest approach. This uses your assumption that you know already that the preceding path name is artifacts and leaves input unchagned if it has a different structure. If you want to remove any starting directory name, you can do a

echo ${input#*/}

Answer 6

mawk 'sub("_.+_"," ",$.(NF=NF))' OFS= FS='^.+/|[.][^.]+$'

online-order-api 1.5.6-6