arrays - Bashスクリプト - 配列を埋める方法は?

Question

このディレクトリ構造があるとしましょう：

DIRECTORY:

.........a

.........b

.........c

.........d

私がやりたいことは: ディレクトリの要素を配列に格納したい

何かのようなもの ： array = ls /home/user/DIRECTORY

array[0]最初のファイルの名前（つまり「a」）が含まれています

array[1] == 'b' 等

手伝ってくれてありがとう

Answer 1

You can't simply do array = ls /home/user/DIRECTORY, because - even with proper syntax - it wouldn't give you an array, but a string that you would have to parse, and Parsing ls is punishable by law. You can, however, use built-in Bash constructs to achieve what you want :

#!/usr/bin/env bash

readonly YOUR_DIR="/home/daniel"

if [[ ! -d $YOUR_DIR ]]; then
    echo >&2 "$YOUR_DIR does not exist or is not a directory"
    exit 1
fi

OLD_PWD=$PWD
cd "$YOUR_DIR"

i=0
for file in *
do
    if [[ -f $file ]]; then
        array[$i]=$file
        i=$(($i+1))
    fi
done

cd "$OLD_PWD"
exit 0

This small script saves the names of all the regular files (which means no directories, links, sockets, and such) that can be found in $YOUR_DIR to the array called array.

Hope this helps.

Answer 2

オプション 1、手動ループ:

dirtolist=/home/user/DIRECTORY
shopt -s nullglob    # In case there aren't any files
contentsarray=()
for filepath in "$dirtolist"/*; do
    contentsarray+=("$(basename "$filepath")")
done
shopt -u nullglob    # Optional, restore default behavior for unmatched file globs

オプション 2、bash 配列の策略を使用:

dirtolist=/home/user/DIRECTORY
shopt -s nullglob
contentspaths=("$dirtolist"/*)   # This makes an array of paths to the files
contentsarray=("${contentpaths[@]##*/}")  # This strips off the path portions, leaving just the filenames
shopt -u nullglob    # Optional, restore default behavior for unmatched file globs