移動未使用的文件

Question

任何想法如何移動未使用任何進程使用bash腳本支持通配符的文件？

基本概念是：

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

其中$1..$n-1是源文件/目錄和$n是目標路徑。

編輯：工作腳本

#!/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 

感謝您的幫助傢伙！

EDIT2 該腳本存在一個錯誤，在某個目錄下使用的文件沒有被fuser檢查。稍後會檢查。

Answer 1

我會通過目的地目錄作爲第一個參數。那麼你的僞代碼幾乎就在那裏。 ，

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

Answer 2

lsof會給你，如果任何進程正在使用的目錄裏面的文件，如果不使用它，然後將其移動到其他位置。

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