bash：将 grep 正则表达式结果分配给数组
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bash: assign grep regex results to array
提问by Ryan
I am trying to assign a regular expression result to an array inside of a bash script but I am unsure whether that's possible, or if I'm doing it entirely wrong. The below is what I want to happen, however I know my syntax is incorrect:
我正在尝试将正则表达式结果分配给 bash 脚本内的数组，但我不确定这是否可能，或者我是否完全错误。下面是我想要发生的事情，但是我知道我的语法不正确：
indexes[4]=$(echo b5f1e7bfc2439c621353d1ce0629fb8b | grep -o '[a-f0-9]\{8\}')
such that:
使得：
index[1]=b5f1e7bf
index[2]=c2439c62
index[3]=1353d1ce
index[4]=0629fb8b
Any links, or advice, would be wonderful :)
任何链接或建议都会很棒:)
回答by theosp
here
这里
array=( $(echo b5f1e7bfc2439c621353d1ce0629fb8b | grep -o '[a-f0-9]\{8\}') )
$ echo ${array[@]}
b5f1e7bf c2439c62 1353d1ce 0629fb8b
回答by Paused until further notice.
#!/bin/bash
# Bash >= 3.2
hexstring="b5f1e7bfc2439c621353d1ce0629fb8b"
# build a regex to get four groups of eight hex digits
for i in {1..4}
do
regex+='([[:xdigit:]]{8})'
done
[[ $hexstring =~ $regex ]] # match the regex
array=(${BASH_REMATCH[@]}) # copy the match array which is readonly
unset array[0] # so we can eliminate the full match and only use the parenthesized captured matches
for i in "${array[@]}"
do
echo "$i"
done
回答by ghostdog74
here's a pure bash way, no external commands needed
这是一种纯粹的 bash 方式，不需要外部命令
#!/bin/bash
declare -a array
s="b5f1e7bfc2439c621353d1ce0629fb8b"
for((i=0;i<=${#s};i+=8))
do
array=(${array[@]} ${s:$i:8})
done
echo ${array[@]}
output
输出
$ ./shell.sh
b5f1e7bf c2439c62 1353d1ce 0629fb8b
