在 BASH 中对另一个数组进行迭代期间使用变量动态命名数组

Question

我有一个数组

FIRST_ARRAY=(NEWYORK CALIFORNIA TEXAS)

接受一个州的脚本可以 return 该州的城市

例如，以下将 return：

user@localhost:~$ search NEWYORK cities
newyorkcity
buffalo
albany

user@localhost:~$ search CALIFORNIA cities
sanfrancisco 
paloalto 
losangeles

user@localhost:~$ search TEXAS cities
houston 
dallas 
austin

我想遍历 FIRST_ARRAY

for state in ${FIRST_ARRAY[@]}
   do
     cities=`search ${FIRST_ARRAY[state]} cities`
     ARRAY_$state=($cities}
done

最后我希望创建以下数组，它们将包含以下值

ARRAY_NEWYORK=(newyorkcity buffalo albany)
ARRAY_CALIFORNIA=(sanfrancisco paloalto losangeles)
ARRAY_TEXAS=(houston dallas austin)

如果这可行，那么例如，我希望能够通过以下方式调用动态创建的数组来访问 austin

echo ${ARRAY_TEXAS[2]}

谢谢！

Answer 1

在 bash 4.0 或更高版本上，readarray / mapfile 可用，以下是简洁正确的实现：

for state in "${FIRST_ARRAY[@]}"; do
    readarray -t "ARRAY_$state" < <(search "$state" cities)
done

在 bash 4.3 中，可以使用安全的直译代码：

for state in "${FIRST_ARRAY[@]}"; do
    readarray -t cities < <(search "$state" cities)

    # make "dest" a namevar -- an alias -- for your intended destination
    # skip to next state if current state name is invalid
    # otherwise we could be assigning to an utterly different "dest"
    declare -n dest="ARRAY_$state" || continue

    # assign to that namevar
    dest=( "$cities" )

    # and discard it
    unset -n dest
done

在 bash 3.x 中，安全地执行此操作需要一些 printf %q 魔法来准备要由 eval:

解析的内容

for state in "${FIRST_ARRAY[@]}"; do

    # why this, and not array=( $cities )? Try a city with spaces in its name.
    # or look at what happens when you have a "city name" that's a wildcard.
    cities=( )
    while read -r city; do
        cities+=( "$city" )
    done < <(search "$state" cities)

    # generate eval-safe replacement for the cities array
    printf -v cities_str '%q ' "${cities[@]}"

    # extra paranoia: make sure we fail with a parse error rather than doing something
    # unexpected if the state name is not a valid shell variable
    printf -v eval_str 'ARRAY_%q=( %s )' "$state" "$cities_str"

    eval "$eval_str" # and evaluate that
done

原始问题没有提供 search 的实现来使答案可测试。对于这个答案，我使用以下内容：

search() {
  case  in
    NEWYORK) printf '%s\n' newyorkcity buffalo albany ;;
    CALIFORNIA) printf '%s\n' sanfrancisco paloalto losangeles ;;
    TEXAS) printf '%s\n' houston dallas austin ;;
  esac
}

根据上面的定义，可以验证结果如下：

$ declare -p ARRAY_NEWYORK ARRAY_CALIFORNIA ARRAY_TEXAS
declare -a ARRAY_NEWYORK='([0]="newyorkcity" [1]="buffalo" [2]="albany")'
declare -a ARRAY_CALIFORNIA='([0]="sanfrancisco" [1]="paloalto" [2]="losangeles")'
declare -a ARRAY_TEXAS='([0]="houston" [1]="dallas" [2]="austin")'

在 BASH 中对另一个数组进行迭代期间使用变量动态命名数组

Dynamically name array with variable during iteration over another array in BASH

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associative-array