如何在没有find的情况下在linux shell脚本中根据日期查找和删除文件？

PLEASE NOTE THAT I CANNOT USE 'find' IN THE TARGET ENVIRONMENT

请注意,我不能在目标环境中使用“找到”

I need to delete all files more than 7 days old in a linux shell script. SOmething like:

我需要在linux shell脚本中删除超过7天的所有文件。就像是:

FILES=./path/to/dir
for f in $FILES
do
  echo "Processing $f file..."
  # take action on each file. $f store current file name
  # perhaps stat each file to get the last modified date and then delete files with date older than today -7 days.

done

Can I use 'stat' to do this? I was trying to use

我可以用'stat'来做这个吗?我试图用

find *.gz -mtime +7 -delete

but discovered that I cannot use find on the target system (there is no permission for the cron user and this can't be changed). Target system is Redhat Enterprise.

但发现我不能在目标系统上使用find(没有cron用户的权限,这是无法更改的)。目标系统是Redhat Enterprise。

The file names are formatted like this:

文件名的格式如下:

gzip > /mnt/target03/rest-of-path/web/backups/DATABASENAME_date "+%Y-%m-%d".gz

gzip> / mnt / target03 / rest-of-path / web / backups / DATABASENAME_date“+%Y-%m-%d”.gz

4 个解决方案

#1

Since you have time in the filename then use that to time the deletion heres some code that does that :

由于你在文件名中有时间,然后使用它来删除时间,这是一些代码:

This script gets the current time in seconds since epoch and then calculates the timestamp 7 days ago. Then for each file parses the filename and converts the date embeded in each filename to a timestamp then compares timestamps to determine which files to delete. Using timestamps gets rid of all hassles with working with dates directly (leap year, different days in months, etc )

此脚本获取自纪元以来的当前时间(以秒为单位),然后计算7天前的时间戳。然后,对于每个文件,解析文件名并将每个文件名中嵌入的日期转换为时间戳,然后比较时间戳以确定要删除的文件。使用时间戳消除了直接处理日期的所有麻烦(闰年,几个月不同的日子等)

The actual remove is commented out so you can test the code.

实际删除已注释掉,因此您可以测试代码。

#funciton to get timestamp X days prior to input timestamp
# arg1 = number of days past input timestamp
# arg2 = timestamp ( e.g. 1324505111 ) seconds past epoch
getTimestampDaysInPast () {
    daysinpast=$1
    seconds=$2
    while [ $daysinpast -gt 0 ] ; do
    daysinpast=`expr $daysinpast - 1`
    seconds=`expr $seconds - 86400`
    done
# make midnight
    mod=`expr $seconds % 86400`
    seconds=`expr $seconds - $mod`
    echo $seconds
} 
# get current time in seconds since epoch
getCurrentTime() {
    echo `date +"%s"`
}

# parse format and convert time to timestamp
# e.g. 2011-12-23 -> 1324505111
# arg1 = filename with date string in format %Y-%m-%d
getFileTimestamp () {
    filename=$1
    date=`echo $filename |  sed "s/[^0-9\-]*\([0-9\-]*\).*/\1/g"`
    ts=`date -d $date | date +"%s"`
    echo $ts
}

########################### MAIN ############################
# Expect directory where files are to be deleted to be first 
# arg on commandline. If not provided then use current working
# directory

FILEDIR=`pwd`
if [ $# -gt 0 ] ; then 
    FILEDIR=$1
fi
cd $FILEDIR

now=`getCurrentTime`
mustBeBefore=`getTimestampDaysInPast 7 $now`
SAVEIFS=$IFS
# need this to loop around spaces with filenames
IFS=$(echo -en "\n\b")
# for safety change this glob to something more restrictive
for f in * ; do 
    filetime=`getFileTimestamp $f`
    echo "$filetime lt $mustBeBefore"
    if [ $filetime -lt $mustBeBefore ] ; then
    # uncomment this when you have tested this on your system
    echo "rm -f $f"
    fi
done
# only need this if you are going to be doing something else
IFS=$SAVEIFS

#2

This should work:

这应该工作:

#!/bin/sh

DIR="/path/to/your/files"
now=$(date +%s)
DAYS=30

for file in "$DIR/"*
do
    if [ $(((`stat $file -c '%Y'`) + (86400 * $DAYS))) -lt $now ]
    then
    # process / rm / whatever the file...
    fi
done

A bit of explanation: stat <file> -c '%Z' gives the modification time of the file as seconds since the UNIX epoch for a file, and $(date +%s) gives the current UNIX timestamp. Then there's just a simple check to see whether the file's timestamp, plus seven days' worth of seconds, is greater than the current timestamp.

一点解释:stat -c'%Z'给出文件的修改时间,自文件的UNIX纪元以来为秒,$(date +%s)给出当前的UNIX时间戳。然后只需要一个简单的检查,看看文件的时间戳加上七天的秒数是否大于当前的时间戳。

#3

If you prefer to rely on the date in the filenames, you can use this routine, that checks if a date is older than another:

如果您更喜欢依赖文件名中的日期,则可以使用此例程检查日期是否早于另一个:

is_older(){
    local dtcmp=`date -d "$1" +%Y%m%d`; shift
    local today=`date -d "$*" +%Y%m%d`
    return `test $((today - dtcmp)) -gt 0`
}

and then you can loop through filenames, passing '-7 days' as the second date:

然后你可以遍历文件名,传递'-7天'作为第二个日期:

for filename in *;
do
    dt_file=`echo $filename | grep -o -E '[12][0-9]{3}(-[0-9]{2}){2}'`
    if is_older "$dt_file" -7 days; then
        # rm $filename or whatever
    fi
done

In is_older routine, date -d "-7 days" +%Y%m%d will return the date of 7 days before, in numeric format ready for the comparison.

在is_older例程中,日期-d“-7天”+%Y%m%d将返回7天前的日期,以数字格式准备进行比较。

#4

DIR=''

now=$(date +%s)

for file in "$DIR/"*
do
echo $(($(stat "$file" -c '%Z') + $((86400 * 7))))
echo "----------"
echo $now

done

#1