在Python 3.6 AWS Lambda函数中输出事件数据
我试图在AWS Lambda Python函数中输出一个名为sheetID的事件,但似乎无法获得正确的语法 以下是一段代码片段:在Python 3.6 AWS Lambda函数中输出事件数据,python,amazon-web-services,beautifulsoup,aws-lambda,python-3.6,Python,Amazon Web Services,Beautifulsoup,Aws Lambda,Python 3.6,我试图在AWS Lambda Python函数中输出一个名为sheetID的事件,但似乎无法获得正确的语法 以下是一段代码片段: def lambda_handler(event, context): scraper = Scraper() scraper.run() return { "Message": "Scrape function ran correctly!" } class Scraper(): def __init__(self):
def lambda_handler(event, context):
scraper = Scraper()
scraper.run()
return { "Message": "Scrape function ran correctly!" }
class Scraper():
def __init__(self):
log.debug("Starting scraper...")
scope = ['https://spreadsheets.google.com/feeds', 'https://www.googleapis.com/auth/drive']
credentials = sac.from_json_keyfile_name('src/h9d7246486f2.json', scope)
log.debug("Authorizing gspread...")
self.gc = gspread.authorize(credentials)
self.spreadsheet = self.gc.open_by_key(event['sheetID'])
...
这就是问题所在:
self.spreadsheet = self.gc.open_by_key(event['sheetID'])
它以前有一个静态的谷歌表单名称
self.spreadsheet = self.gc.open("CorStats")
testlambda函数可以很好地处理事件数据,但不确定是否可以或应该在上述函数中使用格式部分
def my_handler(event, context):
message = 'Hello {} {}!'.format(event['key1'],
event['key3'])
return {
'message' : message
}
以下是错误日志:
{
"errorMessage": "name 'event' is not defined",
"errorType": "NameError",
"stackTrace": [
[
"/var/task/src/scrape_lambda.py",
18,
"lambda_handler",
"scraper = Scraper()"
],
[
"/var/task/src/scrape_lambda.py",
30,
"__init__",
"self.spreadsheet = self.gc.open_by_key(event['sheetID'])"
]
]
}
您需要在_init__方法中包含事件,例如def _init__self,event:,然后使用scraper=Scraperevent实例化它
执行def\uu init\uu self、sheet\u id,然后执行scraper=Scraperevent['sheetID'和self.spreadsheet=self.gc.open\u by\u keysheet\u id可能会更干净。这样,您就可以将scraper类与lambda事件解耦,并可以单独测试它。您能分享收到的错误消息吗?@maafk错误日志已添加。谢谢