使用 python 在单个 Cloudwatch Synthetics Canary 中检查多个 URL

如何解决使用 python 在单个 Cloudwatch Synthetics Canary 中检查多个 URL

我想在单个 CloudWatch Synthetics Canary 中检查多个 (100+) URL。我从这篇博文开始我的脚本： https://aws.amazon.com/blogs/mt/create-canaries-in-python-and-selenium-using-amazon-cloudwatch-synthetics/

但是经过一些小的修改，我得到了以下错误：

TypeError: 'nonetype' object is not callable
ERROR: Canary execution exception.Traceback (most recent call last):
File "/var/task/index.py",line 74,in handle_canary response = await customer_canary.handler(event,context) 
File "/opt/python/bm_check.py",line 36,in handler return await main() File "/opt/python/bm_check.py",line 28,in main await webdriver.execute_step("Navigate to home",navigate_to_home(url)) 
File "/opt/python/aws_synthetics/core/base_synthetics.py",line 231,in execute_step return_value = function_to_execute()

我的金丝雀脚本：

from selenium.webdriver.support.ui import webdriverwait
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.common.by import By
import selenium.common.exceptions
from aws_synthetics.selenium import synthetics_webdriver as webdriver
from aws_synthetics.common import synthetics_logger as logger
from aws_synthetics.common import synthetics_configuration

TIMEOUT = 5


async def main():
    browser = webdriver.Chrome()

    synthetics_configuration.set_config(
        {
            "screenshot_on_step_start": False,"screenshot_on_step_success": True,"screenshot_on_step_failure": True
        }
    )
    
    def navigate_to_home(url):
        browser.implicitly_wait(TIMEOUT)
        browser.get(url)
    
    url = "https://d2h3ljlsmzojxz.cloudfront.net/"
    await webdriver.execute_step("Navigate to home",navigate_to_home(url))

    url = "youtube.com"
    await webdriver.execute_step("Navigate to home",navigate_to_home(url))
    
    logger.info("---------Finished the execution---------")
    
async def handler(event,context):
    return await main()

我的最终目标是使用 for 循环检查这些 URL，并将每个 URL 视为执行步骤，截取成功和失败的屏幕截图。 我做错了什么？

提前致谢！

编辑： 我咨询了 aws 合作伙伴并帮助我找到了解决方案。 在 url 中需要使用 http(s) 协议。 我的工作脚本：

from selenium.webdriver.support.ui import webdriverwait
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.common.by import By
import selenium.common.exceptions
from aws_synthetics.selenium import synthetics_webdriver as webdriver
from aws_synthetics.common import synthetics_logger as logger
from aws_synthetics.common import synthetics_configuration

TIMEOUT = 5


async def main():
    browser = webdriver.Chrome()

    synthetics_configuration.set_config(
        {
        "screenshot_on_step_start": False,"screenshot_on_step_failure": True
        }
    )

    def navigate_to_home():
        browser.implicitly_wait(TIMEOUT)
        browser.get(url)
    
    url_list = [
        "https://d2h3ljlsmzojxz.cloudfront.net/","https://youtube.com"
    ]

    for index,url in enumerate(url_list):
        await webdriver.execute_step("URL check {}".format(index),navigate_to_home)

    logger.info("---------Finished the execution---------")

    
async def handler(event,context):
    return await main()

解决方法

您当前的代码正在尝试使用 navigate_to_home(url) 返回的值作为回调函数。此处该值为 None。

请尝试以下操作以确保传入正确的回调函数：

await webdriver.execute_step("Navigate to home",lambda: navigate_to_home(url))