删除最旧的AWS EC2快照

时间:2016-05-22 19:22:18

标签: amazon-web-services amazon-ec2

我尝试使用此脚本删除除最后6个以外的所有AWS EC2快照:

#!/bin/bash
PATH=/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games:/usr/local/games

# Backup script

Volume="{VOL-DATA}"
Owner="{OWNER}"
Description="{DESCRIPTION}"
Local_numbackups=6
Local_region="us-west-1"

# Remove old snapshots associated to a description, keep the last $Local_numbackups
aws ec2 describe-snapshots --filters Name=description,Values=$Description | grep "SnapshotId" | head -n -$Local_numbackups | awk '{print $2}' | sed -e 's/,//g' | xargs -n 1 -t aws ec2 delete-snapshot --snapshot-id

然而它并不起作用。它删除实例,但不删除最旧的实例。为什么呢?

3 个答案:

答案 0 :(得分:1)

你正试图做一些太复杂的事情,无法在一行中处理(优雅),所以我们需要将其分解一下。首先,让我们按年龄排序快照,从最旧到最新:

aws ec2 describe-snapshots --filters Name=description,Values=$Description --query 'Snapshots[*].[StartTime,SnapshotId]' --output text | sort -n

然后我们可以删除StartTime字段以单独获取快照ID:

aws ec2 describe-snapshots --filters Name=description,Values=$Description --query 'Snapshots[*].[StartTime,SnapshotId]' --output text | sort -n | sed -e 's/^.*\t//'

head(或tail)不适合丢弃我们想要保留的固定数量的快照。我们需要以另一种方式过滤掉那些。所以,完全放在一起:

# Get array of snapshot IDs sorted by age (oldest to newest)
snapshots=($(aws ec2 describe-snapshots --filters Name=description,Values=$Description --query 'Snapshots[*].[StartTime,SnapshotId]' --output text | sort -n | sed -e 's/^.*\t//'))
# Get number of snapshots
count=${#snapshots[@]}

if [ "$count" -lt "$Local_numbackups" ]; then
  echo "We already have less than $Local_numbackups snapshots"
  exit 0
else
  # Drop the last (newest) $Local_numbackups IDs from the array
  snapshots=(${snapshots[@]:0:$((count - Local_numbackups))})
  # Loop through the remaining snapshots and delete
  for snapshot in ${snapshots[@]}; do
    aws ec2 delete-snapshot --snapshot-id $snapshot
  done
fi

(虽然显然可以使用AWS CLI在bash中执行此操作,但它非常复杂,我个人更喜欢使用更强大的语言和AWS SDK。)

答案 1 :(得分:1)

以下是一个示例。

  days2keep="30"
  region="us-west-2"
  name="jdoe"

  #date - -v is for Osx

  cutoffdate=`date -j -v-${days2keep}d '+%Y-%m-%d'`

  echo "Finding list of snapshots before $cutoffdate "
  oldsnapids=$(aws ec2 describe-snapshots --region $region --filters Name=tag:Name,Values=$name --query Snapshots[?StartTime\<=\`$cutoffdate\`].SnapshotId --output text)

  for snapid in $oldsnapids
  do
    echo Deleting snapshot $snapid
    aws ec2 delete-snapshot --snapshot-id $snapid --region $region
  done

答案 2 :(得分:0)

我们可以使用以下步骤删除所有旧快照: -

  • 列出所有快照ID以及它们是旧的并放在一个文件中,如: - /opt/snapshot.txt

  • 然后使用&#34; aws configure&#34;从命令行访问AWS账户的命令,此时我们需要提供凭证: -

如:

AWS Access Key ID [None]: XXXXXXXXXXXXXXXXXX    
AWS Secret Access Key [None]: XXXXXXXXXXXXXXXXXXXXX    
Default region name [None]: XXXXXXXXXXXXXXXX
  • 之后我们可以使用下面的shell脚本,我们需要提供快照ID的文件名

代码:

#!/bin/bash 

list=$(cat /opt/snapshot.txt) 

for i in $list 

do 

   aws ec2 delete-snapshot --snapshot-id $i 

   if [ $? -eq 0 ]; then 

     echo Going Good 

   else 

     echo FAIL 

   fi 

done

由于