循环遍历每个子目录中的文件并应用条件

Question

我的目录包含许多子目录，每个子目录都有几个文件（仅对扩展名为 .ar 的文件感兴趣）。现在，我需要循环遍历每个子目录并检查，例如，如果文件数 = 4，则对这些文件执行某些操作，返回到第二个子目录检查文件，如果 = 3，则对它们执行另一个命令。请注意，我必须在 if 语句上应用非常复杂的条件。

与此类似的东西

dir=$1

for sub_dir in $dir; do
    if the number of files in $sub_dir = 4; then
        do something or command line 
    if the number of files in $sub_dir = 3; then
       do another command
    if the number of files in $sub_dir < 3; then
    escape them

    fi
done

我需要一个类似过程的模板。

Answer 1

假设子目录直接位于顶级目录下：

#!/bin/sh

topdir="$1"

for dir in "$topdir"/*/; do
    set -- "$dir"/*.ar

    if [ "$#" -eq 1 ] && [ ! -f "$1" ]; then
        # do things when no files were found
        # "$1" will be the pattern "$dir"/*.ar with * unexpanded
    elif [ "$#" -lt 3 ]; then
        # do things when less than 3 files were found
        # the filenames are in "$@"        
    elif [ "$#" -eq 3 ]; then
        # do things when 3 files were found
    elif [ "$#" -eq 4 ]; then
        # do things when 4 files were found
    else
        # do things when more than 4 files were found
    fi
done

或者，使用case：

#!/bin/sh

topdir="$1"

for dir in "$topdir"/*/; do
    set -- "$dir"/*.ar

    if [ "$#" -eq 1 ] && [ ! -f "$1" ]; then
        # no files found
    fi

    case "$#" in
        [12])
            # less than 3 files found
            ;;
        3)
            # 3 files found
            ;;
        4)
            # 4 files found
            ;;
        *)
            # more than 4 files found
    esac
done

需要文件名的代码分支用于"$@"引用子目录中的所有文件名，或者"$1"等等"$2"来引用各个文件。文件名将是包括$topdir开头目录的路径名。

Answer 2

你可以这样做：

dir=$1

subdirectories = $(find $dir -type d) # find only subdirectories in dir

for subdir in $subdirectories
do
   n_files=$(find $subdir -maxdepth 1 -type f | wc -l) # find ordinary files in subdir and get it quantity

   if [ $n_files -eq 4 ]
   then
      do_something_4
   fi

   if [ $n_files -eq 3 ]
   then
      do_something_3
   fi

   if [ $n_files -lt 3 ]
   then
      do_something_else
   fi
done