AWS EC2：如何防止实例状态检查的无限循环？

Question

我有以下带有潜在无限while-loop python boto3代码。 通常，几分钟后， while-loop成功。 但是，如果AWS端发生故障，则该程序可能会无限期地挂起。

我确信这不是最合适的方法。

# credentials stored in ../.aws/credentials
# region stored in ../.aws/config

# builtins
from time import sleep
# plugins
import boto3

# Assign server instance IDs.
cye_production_web_server_2 = 'i-FAKE-ID'

# Setup EC2 client
ec2 = boto3.client('ec2')

# Start the second web server.
start_response = ec2.start_instances(
    InstanceIds=[cye_production_web_server_2, ],
    DryRun=False
)

print(
    'instance id:',
    start_response['StartingInstances'][0]['InstanceId'],
    'is',
    start_response['StartingInstances'][0]['CurrentState']['Name']
)

# Wait until status is 'ok'
status = None
while status != 'ok':
    status_response = ec2.describe_instance_status(
        DryRun=False,
        InstanceIds=[cye_production_web_server_2, ],
    )
    status = status_response['InstanceStatuses'][0]['SystemStatus']['Status']
    sleep(5)    # 5 second throttle

print(status_response)
print('status is', status.capitalize())

Answer 1

在循环中实现一个计数器，经过多次尝试后失败

status = None
counter = 5
while (status != 'ok' and counter > 0):
    status_response = ec2.describe_instance_status(
        DryRun=False,
        InstanceIds=[cye_production_web_server_2, ],
    )
    status = status_response['InstanceStatuses'][0]['SystemStatus']['Status']
    sleep(5)    # 5 second throttle
    counter=counter-1

print(status_response)
print('status is', status.capitalize())

Answer 2

您可以尝试在for循环中执行此操作，而不要使用固定的尝试次数。

例如：

MAX_RETRIES = 5

# Try until status is 'ok'
for x in range(MAX_RETRIES):
    status_response = ec2.describe_instance_status(
        DryRun=False,
        InstanceIds=[cye_production_web_server_2, ],
    )
    status = status_response['InstanceStatuses'][0]['SystemStatus']['Status']
    if status != 'ok':
        sleep(5)    # 5 second throttle
    else:
        break

Answer 3

使用超时可能是一个更好的主意

import time 

systemstatus = False
timeout = time.time() + 60*minute

while systemstatus is not True:
    status = ec2.describe_instance_status( \
                 DryRun = False,
                 InstanceIds = [instance_id]
             )

    if status['InstanceStatuses'][0]['SystemStatus']['Status'] == 'ok':
        systemstatus = True

    if time.time() > timeout:
        break
    else:
        time.sleep(10)