移动不使用的文件

Question

我如何移动不被任何进程使用的文件，使用bash脚本留下对通配符的支持?

基本概念是:

for file in $1..$n-1
  if ! fuser file
    mv file $n

其中$1..$n-1为源文件/目录，$n为目标路径。

EDIT: Working script

#!/bin/bash
# Move files which are not open by any process
dest=${@:$#}  # get last arg
for file in "${@:1:$#-1}"; do  # get all but last args
    fuser "$file" >/dev/null 2>&1 && continue
    mv "$file" "$dest"
done

谢谢大家的帮助!

EDIT2 这个脚本有一个bug，在某些目录下使用的文件不被fuser检查。

Answer 1

我将把目标目录作为第一个参数传递。那么你的伪代码就差不多完成了。

dest=$1
shift
for file; do    # shorthand for for file in "$@"; do
    fuser "$file" >/dev/null && continue
    mv "$file" "$dest"
done

Answer 2

lsof会告诉你，如果任何进程正在使用目录内的文件，如果不使用则将其移动到其他位置。

for file in $1..$n-1;do
   var=`lsof +D $file`
   if [[ -z "$var" ]]; then
        mv $file $n
   fi
done

Answer 3

for x in "$@"
do
  target="$x"
done
test -d "$target" ||exit       # last arg isn't a dir

for source in "$@"
do
  test "$source" = "$target" && continue
  if test -d "$source"
    then
      # source is a dir, check all files in it
      for f in "$source"/*
      do
        fuser "$f" || mv "$f" "$target"
      done
    else
      # source isn't a dir
      fuser "$source" ||mv "$source" "$target"
  fi
done