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Last active Jan 13, 2018

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aws lambdaではてなのホッテントリのRSSを取得するpython
#coding: UTF-8
import json
import sys
import urllib.request
import xmltodict
def hotentry(category):
url = ""
if category=="all" or category=="":
url += "?mode=rss"
elif category=="social":
url += "/social.rss"
elif category=="it":
url += "/it.rss"
elif category=="life":
url += "/life.rss"
elif category=="knowledge":
url += "/knowledge.rss"
elif category=="fun":
url += "/fun.rss"
elif category=="entertainment":
url += "/entertainment.rss"
elif category=="game":
url += "/game.rss"
elif category=="economics":
url += "/economics.rss"
return url
def getCategory(event):
category = ""
if (isinstance(event, dict)):
if ("queryStringParameters" in event):
if ("category" in event["queryStringParameters"]):
category = event["queryStringParameters"]["category"]
return category
def lambda_handler(event, context):
category = getCategory(event)
url = hotentry(category)
data = ""
with urllib.request.urlopen(url) as res:
xml ="utf-8")
dict = xmltodict.parse(xml)
data = json.dumps(dict, indent=4)
return {
"statusCode": "200",
'headers': {
'Content-Type': 'application/json',
'Access-Control-Allow-Origin' : '*'
"body": data
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