Question

我正在尝试制作一个Python脚本，该脚本使用fbchat听Facebook聊天并搜索单词'cf'。如果在聊天中检测到该单词，我想发送一个预定义的消息Answer1。参见下面的代码和错误信息：

from fbchat import log, Client
from fbchat.models import *

wordImLookingFor = ['cf']

Answer1 =['Hello! how can i help you']

# Subclass fbchat.Client and override required methods
class EchoBot(Client):
    def onMessage(self, author_id, message_object, thread_id, thread_type, **kwargs):
        self.markAsDelivered(thread_id, message_object.uid)
        self.markAsRead(thread_id)

        log.info("{} from {} in {}".format(message_object, thread_id, thread_type.name))

        # If you're not the author, echo
        if author_id != self.uid:
            if word in message_object:
                print("The word is in the list!")
                self.send(Message(text=answer1), thread_id=thread_id, thread_type=ThreadType.USER)


             else:
                print("The word is not in the list!")



client = EchoBot('user', 'pass')


client.listen()

解析异常... 追溯（最近一次通话）： ... onMessage中的文件“ C：/Users/John/Downloads/fbd2.py”，第22行如果message_object中的单词： TypeError：“消息”类型的参数不可迭代

Answer 1

这是fbchat.models.Message的API

我相信您正在寻找文本字段

if word in message_object.text:
    print("The word is in the list!")

编辑：

简单解决方案

对于下一个错误，in关键字应使用单个字符串而不是字符串列表。由于您要在邮件中查找几个关键字，因此请对列表中的每个单词执行该case语句。

if author_id != self.uid:
    for word in wordImLookingFor:
        if word in message_object.text:
            print("The word is in the list!")
            self.send(Message(text=answer1), thread_id=thread_id, thread_type=ThreadType.USER)
        else:
            print("The word is not in the list!")

扩展解决方案

由于您最终将要搜索一组多个关键字，并且大概每个关键字都应引起一个不同的Answer，因此您可以通过创建一个关键字字典：answer来节省一些复杂性。

keywords = {'cf': 'Hello! how can i help you', 'key2': 'Answer2'}

class EchoBot(Client):
    def onMessage(self, author_id, message_object, thread_id, thread_type, **kwargs):
        self.markAsDelivered(thread_id, message_object.uid)
        self.markAsRead(thread_id)

        log.info("{} from {} in {}".format(message_object, thread_id, thread_type.name))

        if author_id != self.uid:
            for word in message_object.text.split():
                try:
                    self.send(Message(text=keywords[word]), thread_id=thread_id, thread_type=ThreadType.USER)
                    print(word + "is in the list!")
                except KeyError:
                    print(word + "is not in the list!")

Answer 2

您的例外情况是self.send（Message（text = answer1）

您似乎正在尝试以LIST而不是字符串的形式发送文本。

消息无法遍历列表，但需要一个简单的字符串。

Answer1 =['Hello! how can i help you']
Answer1 = 'Hello! How can I help you?'

Python 3：在列表中查找项目并根据该结果进行回答

2 个答案: