Question

我正在尝试从空格分隔的字符串创建一个数组，这个工作正常，直到我必须忽略双引号内的空格来分割字符串。

我试过：

inp='ROLE_NAME="Business Manager" ROLE_ID=67686'

arr=($(echo $inp | awk -F" " '{$1=$1; print}'))

这会将数组拆分为：

${arr[0]}: ROLE_NAME=Business
${arr[1]}: Manager
${arr[2]}: ROLE_ID=67686

实际上我想要它：

${arr[0]}: ROLE_NAME=Business Manager
${arr[1]}: ROLE_ID=67686

我对awk不是很好，所以无法弄清楚如何解决它。感谢

Answer 1

这是特定于bash的，可以使用ksh / zsh

inp='ROLE_NAME="Business Manager" ROLE_ID=67686'
set -- $inp
arr=()
while (( $# > 0 )); do
    word=$1
    shift
    # count the number of quotes
    tmp=${word//[^\"]/}
    if (( ${#tmp}%2 == 1 )); then
        # if the word has an odd number of quotes, join it with the next
        # word, re-set the positional parameters and keep looping
        word+=" $1"
        shift
        set -- "$word" "$@"
    else
        # this word has zero or an even number of quotes.
        # add it to the array and continue with the next word
        arr+=("$word")
    fi
done

for i in ${!arr[@]}; do printf "%d\t%s\n" $i "${arr[i]}"; done

0   ROLE_NAME="Business Manager"
1   ROLE_ID=67686

这特别会破坏任意空格上的单词，但会与单个空格连接，因此引号内的自定义空格将会丢失。

UNIX：从空格分隔的字符串创建数组，同时忽略引号中的空格

1 个答案: