Question

过去24小时的每小时备份，
过去一个月的每日备份，
每月超过一个月的备份，直到卷空间不足。此时，Time Machine将删除最早的每周备份。

我已经拥有了每小时进行备份的bash脚本（rsync）。备份是名为“2015-01-01 08”的文件夹，其中“08”是小时。

在某些时候，需要删除超过24小时的文件夹。所以我在寻找这种魔力。我猜它会有点rm -R some_pattern。

模式应该如何？

Answer 1

Apple Time Machine的Backups.backupdb目录包含每个主机名的子文件夹，该文件夹包含名称为2017-12-21-190657的子文件夹。

对于这种情况，下面的shell脚本（用于修剪btrfs快照/子卷等效时间机器）将为您提供有关如何使用bash进行类似于Time Machine的备份细化/保留的一些灵感。

#!/bin/bash
# Copyright © 2017, Ceriel Jacobs <http://probackup.nl/>
# Licence: GPL v3
# This program is free software: you can redistribute it and/or modify
# it under the terms of the GNU General Public License as published by
# the Free Software Foundation, either version 3 of the License, or
# (at your option) any later version.
# 
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
# GNU General Public License for more details.
# 
# You should have received a copy of the GNU General Public License
# along with this program.  If not, see <http://www.gnu.org/licenses/>.

$DATE='/usr/bin/date'
$BTRFS='/usr/bin/btrfs'

pushd "$SNAPSHOT_DST/$HOST_SRC" >/dev/null
declare -a arrFiles
arrFiles=(2[0-9][0-9][0-9]-[01][0-9]-[0-3][0-9]-[0-2][0-9][0-5][0-9][0-5][0-9]) # BASH glob is auto sorted low to high
popd >/dev/null
# echo ${arrFiles[*]}

# Build epoch time array
declare -ai arrSinceEpoch
declare -i lastEpoch intTZ
# Calculate time zone correction in seconds
tmpTZ=$($DATE +%z) # f.e. +0100 -> +3600 seconds
(( intTZ = (${tmpTZ::3} * 3600) + (${tmpTZ:3:2} * 60) ))
unset tmpTZ

for c in ${!arrFiles[@]}; do
  # Replace dash between date and time by space
  # Replace dashes between time segments by colons
    # because:
    # date --date='2017-12-17-235559' +%s 
    # date: invalid date '2017-12-17-235559'
    # date --date="2017-12-17 23:55:59" +%s
    # 1513551359
  i=${arrFiles[$c]}
  strDateIn="${i::10} ${i:11:2}:${i:13:2}:${i:15:2}"
  #               -d --date   display time described by STRING, not 'now'
  #                                              %s   seconds since 1970-01-01 00:00:00 UTC
  strDateOut=$($DATE --date="$strDateIn" +'%F %T')
  (( arrSinceEpoch[$c]=$($DATE --date="$strDateIn" +%s) + intTZ ))
  # test sorting order
  [[ -n $lastEpoch && $lastEpoch -gt ${arrSinceEpoch[$c]} ]] && Error "Unexpected sorting order"
  lastEpoch=arrSinceEpoch[$c]
  # test date conversion
  [[ $strDateIn == $strDateOut ]] || Error "$strDateIn != $strDateOut"
done

if [[ $v='-v' ]]; then
  # First = oldest
  echo -n "Oldest backup: ${arrFiles[0]}"
  echo "  (${arrSinceEpoch[0]})"
  # Last = newest
  echo -n "Newest backup: ${arrFiles[-1]}"
  echo "  (${arrSinceEpoch[-1]})"
fi

declare -i intNewest intUpperLimit
declare -ai arrToBeRemoved
intNewest=${arrSinceEpoch[-1]}
intUpperLimitMinusSeven=$((intNewest-86400)) #24*3600 seconds
intUpperLimitMinusThirty=$((intNewest-2592000)) #30*86400 seconds

# Retention Rules
# - don't touch newest 24h versions
# - from old to new, keep only 1st version per day until newest-24 hours is reached
# - from old to new, keep oldest backup, and store versions that are at least 7 days newer as the previous
#   until newest-30 days is reached

# 1/2 Thinning to 1 version per day
lastday=''
for c in ${!arrFiles[@]}; do
  if [[ ${arrSinceEpoch[$c]} -gt $intUpperLimitMinusSeven ]]; then
    [[ $v='-v' ]] && echo "<24h: ${arrFiles[$c]}, [$c]"
    break # exit for loop because all (remaining) backups are made within 24 hours of the "latest" backup
  elif [[ -n $lastday && $lastday == ${arrFiles[$c]::10} ]]; then
     # duplicate day found -> add this snapshot to the to-be-removed array
     [[ $v='-v' ]] && echo "Remove snapshot: ${arrFiles[$c]}, [$c]"
     arrToBeRemoved[$c]=1
  fi
  lastday=${arrFiles[$c]::10}
done

# 2/3 Thinning to 1 version per week
# fulldate to day: 2017-12-21-164100 -> 1513541787 -> /86400 -> day# 17517
# save snap when snap day >= last week snap day + 7 days
# TODO: improve to allow some rounding: f.e. not delete #6 in 1,6,14,21
declare -i idxLastSavedWeekSnap intLastSavedWeekSnapDay intTemp
idxLastSavedWeekSnap=0
intLastSavedWeekSnapDay=$((${arrSinceEpoch[0]}/86400)) # epoch seconds to day number

for c in ${!arrFiles[@]}; do
  if [[ ${arrSinceEpoch[$c]} -gt $intUpperLimitMinusThirty ]]; then
    [[ $v='-v' ]] && echo "<30d: ${arrFiles[$c]}, [$c]"
    break # exit for loop because all (remaining) backups are made within 30 days of the "latest" backup
  elif [[ $c = 0 ]]; then
    continue # skip this snap, always keep 1st snapshot
  elif [[ ${arrToBeRemoved[$c]} = 1 ]]; then
    continue # skip this snap, already marked for deletion
  else
    intTemp=${arrSinceEpoch[$c]}/86400
    if [[ $intTemp -ge $intLastSavedWeekSnapDay ]]; then
      intLastSavedWeekSnapDay=$intTemp
    else
      arrToBeRemoved[$c]=1
    fi
  fi 
done

# print array count & indexes
if [[ $v='-v' && -n ${!arrToBeRemoved[@]} ]]; then
  echo -n "To be removed (${#arrToBeRemoved[@]}): "
  printf '#%s ' "${!arrToBeRemoved[@]}"
  echo
fi

# 3/3 If there are items to-be-removed
if [[ -n ${!arrToBeRemoved[@]} ]]; then
  # delete corresponding snapshots
  for c in ${!arrToBeRemoved[@]}; do
    [[ $v='-v' ]] && echo -n "btrfs subvolume delete $SNAPSHOT_DST/$HOST_SRC/${arrFiles[$c]};"
    #$BTRFS subvolume delete $SNAPSHOT_DST/$HOST_SRC/${arrFiles[$c]}
  done
  [[ $v='-v' ]] && echo
fi

Answer 2

这是使用OSX Terminal的最简单的解决方案。

获取TimeMachine中所有备份的列表。这还将显示您在步骤2中将需要的备份的完整目录路径...

$ tmutil listbackups

/Volumes/Time Machine Backups/Backups.backupdb/{your-macbook}/2018-10-02-213405
/Volumes/Time Machine Backups/Backups.backupdb/{your-macbook}/2018-10-09-192323
/Volumes/Time Machine Backups/Backups.backupdb/{your-macbook}/2018-10-19-212659

根据日期选择要删除的备份。请注意，使用通配符*并使用步骤1中的目录。例如，要删除所有2018年的备份，请使用以下命令：

$ sudo tmutil delete'/卷/时间机器 Backups / Backups.backupdb / {your-macbook} / 2018-'*

最后一步是缩小并恢复稀疏束中的空间。在备份驱动器中搜索.sparsebundle文件。

$ sudo hdiutil compact'/Volumes/{your-mac}.sparsebundle'

如何删除Apple的Time Machine？

2 个答案: