Find and delete duplicate files
作用:查找指定目录(一个或多个)及子目录下的所有重复文件,分组列出,并可手动选择或自动随机删除多余重复文件,每组重复文件仅保留一份。(支持文件名有空格,例如:"file name" 等)
实现:find遍历指定目录查找所有文件,并对找到的所有文件进行MD5校验,通过比对MD5值分类处理重复文件。
不足: find 遍历文件耗时;
MD5校验大文件耗时;
对所有文件校验比对耗时(可考虑通过比对文件大小进行第一轮的重复性筛选,此方式针对存放大量大文件的目录效果明显,本脚本未采用);
演示:

注释:
脚本执行过程中显示MD5校验过程,完毕后,统计信息如下:
Files: 校验的文件总数
Groups: 重复文件组的数量
Size:此处统计的大小为,多余文件的总大小,即将要删除的多余的重复文件的大小,换句话说就是,删除重复文件后,磁盘空间会节省的空间。
可在“Show detailed information ?”提示后,按键“y”,进行重复文件组的查看,以便确认,也可直接跳过,进入删除文件方式的选择菜单:
删除文件方式有两种,一种是手动选择方式(默认的方式),每次列出一组重复文件,手动选择欲留下的文件,其他文件将会被删除,若没有选择 则默认保留列表的第一个文件,演示如下:

另一种方式是自动选择方式,默认保留每组文件的第一个文件,其他重复文件自动删除。(为防止删除重要文件,建议使用第一种方式),演示如下:

支持文件名空格的情况,演示如下:

代码专区:
#!/bin/bash
#Author: LingYi
#Date:
#Func: Delete duplicate files
#EG : $ [ DIR1 DIR2 ... DIRn ]
#Define the mnt file, confirming the write authority by yourself.
md5sum_result_log="/tmp/$(date +%Y%m%d%H%M%S)"
echo -e "\033[1;31mMd5suming ...\033[0m"
-I {} md5sum {} | tee -a $md5sum_result_log
files_sum=$(cat $md5sum_result_log | wc -l)
# Define array, using the value of md5 as index, filename as element.
# Firstly, you must do advance declaration to make sure the it's supported by bash.
declare -A md5sum_value_arry
while read md5sum_value md5sum_filename
do
#Space in a file name, in order to support this case ,using the ‘+’ as the segmentation charater.
#So, if '+' appears in a file name, there will be problems. The use should choose the manual mode to delete redundant files.
md5sum_value_arry[$md5sum_value]="${md5sum_value_arry[$md5sum_value]}+$md5sum_filename"
(( _${md5sum_value}+= ))
done <$md5sum_result_log
# counting the duplicate file groups and the size of redundant files in this loop.
groups_sum=
repfiles_size=
for md5sum_value_index in ${!md5sum_value_arry[@]}
do
]]; then
let groups_sum++
need_print_indexes="$need_print_indexes $md5sum_value_index"
eval repfile_sum=\$\(\( \$_$md5sum_value_index - \)\)
repfile_size=$( ls -lS "`echo ${md5sum_value_arry[$md5sum_value_index]}|awk -F'+' '{print $2}'`" | awk '{print $5}')
repfiles_size=$(( repfiles_size + repfile_sum*repfile_size ))
fi
done
#Outputing the statistical information.
echo -e "\033[1;31mFiles: $files_sum Groups: $groups_sum \
Size: ${repfiles_size}B $((repfiles_size/))K $((repfiles_size//))M\[0m"
[[ $groups_sum -eq ]] && exit
#The use chooses whether to check the file grouping or not.
read -n -s -t -p 'Show detailed information ?' user_ch
[[ $user_ch == 'n' ]] && echo || {
[[ $user_ch == 'q' ]] && exit
for print_value_index in $need_print_indexes
do
echo -ne "\n\033[1;35m$((++i)) \033[0m"
eval echo -ne "\\\033[1\;34m$print_value_index [ \$_${print_value_index} ]:\\\033[0m"
echo ${md5sum_value_arry[$print_value_index]} | tr '+' '\n'
done | more
}
#The user can choose the way of deleting file here.
echo -e "\n\nManual Selection by default !"
echo -e " 1 Manual selection\n 2 Random selection"
echo -ne "\033[1;31m"
read -t USER_CH
echo -ne "\033[0m"
[[ $USER_CH == 'q' ]] && exit
[[ $USER_CH -ne ]] && USER_CH= || {
echo -ne "\033[31mWARNING: you have choiced the Random Selection mode, files will be deleted at random !\nAre you sure ?\033[0m"
read -t yn
[[ $yn == 'q' ]] && exit
[[ $yn !=
}
#Handle files according to the user's selection
echo -e "\033[31m\nWarn: keep the first file by default.\033[0m"
for exec_value_index in $need_print_indexes
do
#This loop contains an array of files that are about to be deleted.
,j=;i<$(echo ${md5sum_value_arry[$exec_value_index]} | grep -o '+' | wc -l); i++,j++))
do
file_choices_arry[i]="$(echo ${md5sum_value_arry[$exec_value_index]}|awk -F'+' '{print $J}' J=$j)"
done
eval file_sum=\$_$exec_value_index
]]; then
#If the user selects a manual mode, handle the duplicate file group one by one in a loop.
echo -e "\033[1;34m$exec_value_index\033[0m"
; j<${#file_choices_arry[@]}; j++))
do
echo "[ $j ] ${file_choices_arry[j]}"
done
read -p "Number of the file you want to keep: " num_ch
[[ $num_ch == 'q' ]] && exit
$((${#file_choices_arry[@]}-)) |
else
num_ch=
fi
#If the user selects the automatic deletion mode, then delete the redundant files
; n<${#file_choices_arry[@]}; n++))
do
[[ $n -ne $num_ch ]] && {
echo -ne "\033[1mDeleting file \" ${file_choices_arry[n]} \" ... \033[0m"
rm -f "${file_choices_arry[n]}"
[[ $? -eq ]] && echo -e "\033[1;32mOK" || echo -e "\033[1;31mFAIL"
echo -ne "\033[0m"
}
done
done
Find and delete duplicate files的更多相关文章
- Compare, sort, and delete duplicate lines in Notepad ++
Compare, sort, and delete duplicate lines in Notepad ++ Organize Lines: Since version 6.5.2 the app ...
- Android Duplicate files copied in APK
今天调试 Android 应用遇到这么个问题: Duplicate files copied in APK META-INF/DEPENDENCIES File 1: httpmime-4.3.2.j ...
- com.android.build.api.transform.TransformException: com.android.builder.packaging.DuplicateFileException: Duplicate files copied in APK assets/com.xx.xx
完整的Error 信息(关键部分) Error:Execution failed for task ':fanwe_o2o_47_mgxz_dingzhi:transformResourcesWith ...
- AndroidStudio使用第三方jar包报错(Error: duplicate files during packaging of APK)
http://www.kwstu.com/ArticleView/android_201410252131196692 错误描述: Error: duplicate files during pack ...
- Android Studio 错误 Duplicate files copied in APK META-INF/LICENSE.txt
1 .Duplicate files copied in APK META-INF/LICENSE.txt android { packagingOptions { exclude 'META-I ...
- Duplicate files copied in APK META-INF/LICENSE.txt
Error:Execution failed for task ':app:packageDebug'. > Duplicate files copied in APK META-INF/LIC ...
- Android Studio 错误 Duplicate files copied in APK META-INF/LICENSE.txt解决方案
My logcat: log Execution failed for task ':Prog:packageDebug'. Duplicate files copied in APK META-IN ...
- List or delete hidden files from command prompt(CMD)
In Windows, files/folders have a special attribute called hidden attribute. By setting this attribut ...
- 解决DuplicateFileException: Duplicate files copied in APK META-INF/LICENSE(或META-INF/DEPENDENCIES)
导入eclipse项目时报 Error:Execution failed for task ':app:transformResourcesWithMergeJavaResForDebug'.> ...
随机推荐
- [原创]自己动手实现React-Native下拉框控件
因项目需要,自己动手实现了一个下拉框组件,最近得空将控件独立出来开源上传到了Github和npm. Github地址(求Star 求Star 求Star
- android Broadcast广播消息代码实现
我用的是Fragment , 发送写在一个类中,接收写在另外一个类的内部类中.代码动态实现注册. 代码: myReceiver = new zcd.netanything.MyCar.myReceiv ...
- Double Dispatch讲解与实例-面试题
引言 说实话,我看过GoF<Design Patterns>,也曾深深的被李建忠<设计模式>系列Webcast吸引.但是还没有见过“Double Dispatch模式”.的确G ...
- Linux环境搭建-在虚拟机中安装Centos7.0
最近在空闲时间学习Linux环境中各种服务的安装与配置,都属于入门级别的,这里把所有的学习过程记录下来,和大家一起分享. 我的电脑系统是win7,所以我需要在win7上安装一个虚拟机-VMware,然 ...
- BZOJ 1901: Zju2112 Dynamic Rankings[带修改的主席树]【学习笔记】
1901: Zju2112 Dynamic Rankings Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 7143 Solved: 2968[Su ...
- j2ee之Filter使用实例(页面跳转)
javax.servlet.Filter类中主要有三个方法. public void destroy(); //销毁对象 public void doFilter(ServletRequest req ...
- Referenced file contains errors (http://www.springframework.org/schema...错误
Referenced file contains errors (http://www.springframework.org/schema...错误 Referenced file contains ...
- [LeetCode] Sum Root to Leaf Numbers 求根到叶节点数字之和
Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number ...
- 做中学(Learning by Doing)之背单词-扇贝网推荐
做中学(Learning by Doing)之背单词-扇贝网推荐 看完杨贵福老师(博客,知乎专栏,豆瓣)的「继续背单词,8个月过去了」,我就有写这篇文章的冲动了,杨老师说: 有时候我会感觉非常后悔,如 ...
- 字节、字、bit、byte的关系
字 word 字节 byte 位 bit 字长是指字的长度 1字=2字节(1 word = 2 byte) 1字节=8位(1 byte = 8bit) 一个字的字长为16 一个字节的字长是8 bps ...