bash - 使用递归 - 计算目录中的可执行文件

Question

我不是在寻找一些高级别的答案，只有一行，我正在尝试制作一个可以工作的脚本，并使其与递归一起工作并理解它:)

我正在尝试学习 bash 脚本，但遇到了一些问题：

我正在尝试计算目录及其子目录中可执行文件的数量

这就是我的想法

文件名countFiles：

#!/bin/bash
counter = 0
for FILE in ($1/*)  /// going over all the files in the input path
do
 if (-d $FILE); then        /// checking if dir
  ./countFiles $FILE  /// calling the script again with ned path
  counter= $($counter + $0) /// addint to the counter the number of exe files from FILE
 fi
 if (-f $FILE); then /// checking if file
  if (-x $FILE); then /// checking id exe file
   counter = $counter + 1 // adding counter by 1
  fi
 fi
done
exit($counter) // return the number for exe files

Answer 1

如果你真的想使用递归（这在 Bash 中是个坏主意）：首先，不要递归调用你的脚本。相反，递归调用函数。这将更有效率（没有分叉开销）。尝试修正你的语法：

#!/bin/bash

shopt -s nullglob

count_x_files() {
   # Counts number of executable (by user, regular and non-hidden) files
   # in directory given in first argument
   # Return value in variable count_x_files_ret
   # This is a recursive function and will fail miserably if there are
   # too deeply nested directories
   count_x_files_ret=0
   local file counter=0
   for file in "$1"/*; do
      if [[ -d $file ]]; then
         count_x_files "$file"
         ((counter+=count_x_files_ret))
      elif [[ -f $file ]] && [[ -x $file ]]; then
         ((++counter))
      fi
   done
   count_x_files_ret=$counter
}

count_x_files "$1"
echo "$count_x_files_ret"