使用 python3 和 boto3 获取最新的 aws 卷快照 ID

Question

I am new to python and boto3 , I want to get the latest snapshot ID.

I am not sure if I wrote the sort with lambda correctly and how to access the last snapshot, or maybe I can do it with the first part of the script when I print the snapshot_id and snapshot_date ?

Thanks.

Here is my script

import boto3
ec2client = mysession.client('ec2', region_name=region)
ec2resource = mysession.resource('ec2', region_name=region)

def find_snapshots():
    for snapshot in ec2client_describe_snapshots['Snapshots']:
        snapshot_volume = snapshot['VolumeId']
        mnt_vol = "vol-xxxxx"
        if mnt_vol == snapshot_volume:
            snapshot_date = snapshot['StartTime']
            snapshot_id = snapshot['SnapshotId']
            print(snapshot_id)
            print(snapshot_date)

find_snapshots()

snapshots = ec2resource.snapshots.filter(Filters=[{'Name': 'volume-id', 'Values': [mnt_vol]}]).all()
print(snapshots)
snapshots = sorted(snapshots, key=lambda ss:ss.start_time)
print(snapshots)
snapshot_ids = map(lambda ss:ss.id, snapshots)
print(snapshot_ids)

last_snap_id = ?

output:

snap-05a8e27b15161d3d5
2016-12-25 05:00:17+00:00
snap-0b87285592e21f0
2016-12-25 03:00:17+00:00
snap-06fa39b86961ffa89
2016-12-24 03:00:17+00:00

ec2.snapshotsCollection(ec2.ServiceResource(), ec2.Snapshot)
[]
<map object at 0x7f8d91ea9cc0>

*update question to @roshan answer:

def find_snapshots():
    list_of_snaps = []
    for snapshot in ec2client_describe_snapshots['Snapshots']:
        snapshot_volume = snapshot['VolumeId']
        if mnt_vol == snapshot_volume:
            snapshot_date = snapshot['StartTime']
            snapshot_id = snapshot['SnapshotId']
            list_of_snaps.append({'date':snapshot['StartTime'], 'snap_id': snapshot['SnapshotId']})
    return(list_of_snaps)

find_snapshots()

print(find_snapshots())

#sort snapshots order by date
newlist = sorted(list_to_be_sorted, key=lambda k: k['name'])  
print(newlist)

output:

[{'date': datetime.datetime(2016, 12, 25, 14, 23, 37, tzinfo=tzutc()), 'snap_id': 'snap-0de26a40c1d1e53'}, {'date': datetime.datetime(2016, 12, 24, 22, 9, 34, tzinfo=tzutc()), 'snap_id': 'snap-0f0341c53f47a08'}]

Traceback (most recent call last):
  File "test.py", line 115, in <module>
    newlist = sorted(list_to_be_sorted, key=lambda k: k['name'])  
NameError: name 'list_to_be_sorted' is not defined

If I do:

list_to_be_sorted = (find_snapshots())
print(list_to_be_sorted)
#sort snapshots order by date
newlist = sorted(list_to_be_sorted, key=lambda k: k['snap_id'])
print(newlist[0])

the latest snapshot does not appear in the output:

{'date': datetime.datetime(2016, 12, 23, 3, 0, 18, tzinfo=tzutc()), 'snap_id': 'snap-0225cff1675c369'}

this one is the latest:

[{'date': datetime.datetime(2016, 12, 25, 5, 0, 17, tzinfo=tzutc()), 'snap_id': 'snap-05a8e27b15161d5'}

How do I get the latest snapshot ( snap_id ) ?

Thanks

Answer 1

You should it sort in reverse order.

sorted(list_to_be_sorted, key=lambda k: k['date'], reverse=True)[0]

sorts the list by the date (in ascending order - oldest to the latest) If you add reverse=True , then it sorts in descending order (latest to the oldest). [0] returns the first element in the list which is the latest snapshot.

If you want the snap_id of the latest snapshot, just access that key.

sorted(list_to_be_sorted, key=lambda k: k['date'], reverse=True)[0]['snap_id']

Answer 2

def find_snapshots():
    list_of_snaps = []
    for snapshot in ec2client_describe_snapshots['Snapshots']:
    snapshot_volume = snapshot['VolumeId']
    mnt_vol = "vol-xxxxx"
    if mnt_vol == snapshot_volume:
        snapshot_date = snapshot['StartTime']
        snapshot_id = snapshot['SnapshotId']
        list_of_snaps.append({'date':snapshot['StartTime'], 'snap_id': snapshot['SnapshotId']})
        print(snapshot_id)
        print(snapshot_date)

    #sort snapshots order by date
    newlist = sorted(list_of_snaps, key=lambda k: k['date'], reverse= True)
    latest_snap_id = newlist[0]['snap_id']
    #The latest_snap_id provides the exact output snapshot ID
    print(latest_snap_id)

Answer 3

最后一行需要是这样的：

newlist = sorted(list_of_snaps, key=lambda k: k['snap_id'])

Answer 4

You can append the snapshots to list, then sort the list based on date

def find_snapshots():
    list_of_snaps = []
    for snapshot in ec2client_describe_snapshots['Snapshots']:
    snapshot_volume = snapshot['VolumeId']
    mnt_vol = "vol-xxxxx"
    if mnt_vol == snapshot_volume:
        snapshot_date = snapshot['StartTime']
        snapshot_id = snapshot['SnapshotId']
        list_of_snaps.append({'date':snapshot['StartTime'], 'snap_id': snapshot['SnapshotId']})
        print(snapshot_id)
        print(snapshot_date)

    #sort snapshots order by date
    newlist = sorted(list_to_be_sorted, key=lambda k: k['name'])  

Hope it helps !!