我在理解如何使用查询结果时遇到问题。我问了六个关于这个的问题,但我还是不明白。我从以前的代码复制,我让它以某种方式工作,但由于我不理解底层的概念,如果我做一个小的改变代码崩溃。如果你能告诉我你如何看待这里发生的事情并向我解释,我真的很感激。谢谢。
class ReceiveEmail(InboundMailHandler):
def receive(self, message):
logging.info("Received email from %s" % message.sender)
plaintext = message.bodies(content_type='text/plain')
for text in plaintext:
txtmsg = ""
txtmsg = text[1].decode()
logging.info("Body is %s" % txtmsg)
logging.info("CC email is %s" % ((message.cc).split(",")[1]))
query = User.all()
query.filter("userEmail =", ((message.cc).split(",")[1]))
results = query.fetch(1)
for result in results:
result.userScore += 1
um = results[0]
um.userScore = result.userScore
um.put()
在此代码中,据我所知,查询从cc列表中获取第二个电子邮件地址并获取结果。
然后我将userScore增加1.
接下来,我想更新数据存储区中的这个项目,所以我说
um = results[0]
um.userScore = result.userScore
um.put()
但这会导致索引超出范围错误:
um = results[0]
IndexError: list index out of range
为什么呢?我想象results[0]
是结果的零项。为什么它超出范围?我唯一能想到的是,列表可能是None
。但我不明白为什么。它必须有1个被提取的项目。
此外,如果我尝试通过将索引从[1]更改为[0]来测试第一个电子邮件地址
query.filter("userEmail =", ((message.cc).split(",")[0]))
然后我没有得到IndexError
。
我在这里做错了什么?
谢谢!
修改
见评论:
(message.cc).split(",")[0])
在电子邮件前留下了一个空格(从第二封电子邮件开始),因此查询与它们不匹配;
>>> cc.split(",")
['cc12@example.com', ' cc13@example.com', ' cc13@example.com']
用逗号修改问题后添加空格:
>>> listcc = cc.split(", ")
>>> listcc
['cc12@example.com', 'cc13@example.com', 'cc13@example.com']
>>>
答案 0 :(得分:1)
要理解代码,请将其分解并逐一查看:
class ReceiveEmail(InboundMailHandler):
def receive(self, message):
logging.info("Received email from %s" % message.sender)
# Get a list of CC addresses. This is basically a for loop.
cc_addresses = [address.strip() for address in message.cc.split(",")]
# The CC list goes with the message, not the bodies.
logging.info("CC email is %s" % (cc_addresses))
# Get and iterate over all of the *plain-text* bodies in the email.
plaintext = message.bodies(content_type='text/plain')
for text in plaintext:
txtmsg = ""
txtmsg = text[1].decode()
logging.info("Body is %s" % txtmsg)
# Setup a query object.
query = User.all()
# Filter the user objects to get only the emails in the CC list.
query.filter("userEmail IN", cc_addresses)
# But, only get at most 10 users.
users = query.fetch(10)
logging.info('Got %d user entities from the datastore.' % len(users))
# Iterate over each of the users increasing their score by one.
for user in users:
user.userScore += 1
# Now, write the users back to the datastore.
db.put(users)
logging.info('Wrote %d user entities.' % len(users))
我会调整您的模型结构。创建用户实体时,我会将key_name设置为电子邮件地址。您将能够提高查询效率。
一些参考文献: