linux - Ubuntu/Linux 庆典 : traverse directory and subdirectories to work with files

Question

让我从我需要的开始；该程序被赋予一个目录，然后它将检查目录中的所有文件(工作)并对文件执行操作(等待它可以找到这部分的所有文件)。然后它将查找子目录并为每个子目录重新运行它自己。

我正在测试的目录如下所示:

desktop/test_files/ (starting directory)
desktop/test_files/folder 1/
desktop/test_files/folder 1>folder 2/
desktop/test_files/folder 1>folder 2/<files, 20 or so>
desktop/test_files/folder 3/
desktop/test_files/folder 3/<more files, 20 or so>

文件夹和文件的名称中确实包含空格

输出是:

$ ./x007_shorter.sh Desktop/test_files/

Desktop/test_files/"folder 1"/
Desktop/test_files/folder 1/"folder 2"/
ls: cannot access */: No such file or directory
Desktop/test_files/folder 1/folder 2/"folder 3"/
./x007_shorter.sh: line 4: cd: ./folder 3/: No such file or directory
ls: cannot access */: No such file or directory

程序如下:

#!/bin/bash
function findir {
    newDir=$1
    eval cd $newDir
    ARRAY=( $(ls -d */) )
    declare -a diry
    count=0
    a=0
    while [ $a -lt ${#ARRAY[@]} ]; do
        diry[$count]="${ARRAY[$a]}"
        noSpace=true
        while [ true ]; do
            if [[ ${diry[$count]} == */* ]] ; then
                if [ $noSpace = false ]; then
                diry[$count]="${diry[$count]:0:((${#diry[$count]}-1))}\"/"
                fi
                break
                noSpace=true
            fi
            let "a=$a+1"
            if [ $noSpace = false ]; then
                diry[$count]="${diry[$count]} ${ARRAY[$a]}"
            else
                diry[$count]="\"${diry[$count]} ${ARRAY[$a]}"
            fi
            noSpace=false
        done
        let "count=$count+1"
        let "a=$a+1"
    done
    for a in `seq 1 ${#diry[@]}`; do
        eval cd .$newDir
#        list "${diry[($a-1)]}"
        where=`pwd`
#        eval cd $newDir
        #findir "${diry[($a-1)]}"
        #findir "$where${diry[($a-1)]:1}"
        #Right option won,  echo "${diry[($a-1)]} Vs $where/${diry[($a-1)]}"
        echo "$where/${diry[($a-1)]}"
        findir "./${diry[($a-1)]}"
    done
}
function list {
    input_file_directory=$1
    eval cd $input_file_directory
    ARRAY=( $(find . -maxdepth 1 -type f -print) )
    declare -a files
    count=0
    a=0
    while [ $a -lt ${#ARRAY[@]} ]; do
        files[$count]="${ARRAY[$a]}"
        while [ true ]; do
            if [[ ${ARRAY[(($a+1))]} == ./* ]] ; then
                break
            fi
            if [[ "${ARRAY[(($a+1))]}" == "" ]] ; then
                break
            fi
            let "a=$a+1"
            files[$count]="${files[$count]} ${ARRAY[$a]}"
        done
        let "count=$count+1"
        let "a=$a+1"
    done
    where=`pwd`
    for a in `seq 1 ${#files[@]}`; do
        echo "$where${files[($a-1)]:1}"
        #going to work on each file, just echoing file till lists all files
    done
}

clear
dar=""
if [[ $1 = "" ]]; then
    read -p "Please enter a directory for me to scan" newdir
    dar=$newdir
    list $newdir
    findir $newdir
else
    dar=$1
    list $1
    findir $1
fi

Answer 1

您有什么理由不能为此使用查找？将您想要的每个文件操作粘贴到它自己的脚本中(我在下面将其称为 dostufftomyfile.sh)，然后执行:

find $dir -type f -print0 | xargs -0 dostufftomyfile.sh

将 $dir 替换为您将从中搜索的顶级目录...

编辑添加...当您编写 shell 脚本时，请确保将 $@ 放在双引号中...例如，您希望您的 dostufftomyfile.sh 脚本具有以下结构:

#!/bin/sh
for f in "$@"
do
    echo "Processing file: $f"
    # Do something to file $f
done

如果你不引用 $@ 那么文件名中的空格将被忽略(我怀疑你不会想要的):-)