Lambda函数报错“Object of type function is not JSON serializable”求助

问题描述 #

将Tweepy机器人部署在AWS Lambda中，其他功能均正常运行，仅非问候场景的回复分支触发报错：Object of type function is not JSON serializable。以下是机器人完整代码及错误栈信息：

完整代码 #

import tweepy
import random
import os
import json
import boto3

read = tweepy.Client(os.getenv('btok'))
write = tweepy.Client(
    consumer_key=os.getenv('ckey'), 
    consumer_secret=os.getenv('csec'), 
    access_token=os.getenv('atok'), 
    access_token_secret=os.getenv('atos')
)

# 声明变量
user_id = 1568306756798775297
greet = random.choice(list(open('greetings.txt'))) # 从列表中随机选择问候语
reply = random.choice(list(open('replies.txt'))) # 从列表中随机选择回复语
strings = {'hi', 'hello', 'hey', 'hai'} # 识别的问候关键词
dynamodb = boto3.resource('dynamodb')
table = dynamodb.Table('lastid')

def main(event, context):
    # TODO implement
    reply()
    return {
        'statusCode': 200,
        'body': json.dumps('Hello from Lambda!')
    }
def reply():
    
    value = table.get_item(Key={'lastidkey': '1'})
    last = value['Item']['lidval']
    response = read.get_users_mentions(user_id, since_id=last) # 获取新提及
    for tweet in response.data:
        text = tweet.text
        id = tweet.id
        if any(x in text.lower() for x in strings):
            write.create_tweet(in_reply_to_tweet_id=id, text=greet) # 问候场景回复问候语
        else:
            write.create_tweet(in_reply_to_tweet_id=id, text=reply) # 非问候场景回复随机内容
        table.put_item(Item={'lastidkey': '1', 'lidval': str(tweet.id)}) # 将最新ID写入DynamoDB

错误栈信息 #

{
  "errorMessage": "Object of type function is not JSON serializable",
  "errorType": "TypeError",
  "requestId": "cb6423eb-e27e-4229-b322-c6df9fad3b5c",
  "stackTrace": [
    "  File \"/var/task/reply.py\", line 25, in main\n    reply()\n",
    "  File \"/var/task/reply.py\", line 41, in reply\n    write.create_tweet(in_reply_to_tweet_id=id, text=reply) # otherwise, it's a random reply\n",
    "  File \"/opt/python/tweepy/client.py\", line 824, in create_tweet\n    return self._make_request(\n",
    "  File \"/opt/python/tweepy/client.py\", line 126, in _make_request\n    response = self.request(method, route, params=request_params,\n",
    "  File \"/opt/python/tweepy/client.py\", line 83, in request\n    with self.session.request(\n",
    "  File \"/opt/python/requests/sessions.py\", line 573, in request\n    prep = self.prepare_request(req)\n",
    "  File \"/opt/python/requests/sessions.py\", line 484, in prepare_request\n    p.prepare(\n",
    "  File \"/opt/python/requests/models.py\", line 371, in prepare\n    self.prepare_body(data, files, json)\n",
    "  File \"/opt/python/requests/models.py\", line 511, in prepare_body\n    body = complexjson.dumps(json, allow_nan=False)\n",
    "  File \"/var/runtime/simplejson/__init__.py\", line 398, in dumps\n    return cls(\n",
    "  File \"/var/runtime/simplejson/encoder.py\", line 296, in encode\n    chunks = self.iterencode(o, _one_shot=True)\n",
    "  File \"/var/runtime/simplejson/encoder.py\", line 378, in iterencode\n    return _iterencode(o, 0)\n",
    "  File \"/var/runtime/simplejson/encoder.py\", line 272, in default\n    raise TypeError('Object of type %s is not JSON serializable' %\n"
  ]
}

问题原因与修复方案 #

核心问题 #

变量名与函数名冲突：你先定义了存储随机回复文本的变量reply，随后又定义了同名的reply()函数。当代码执行到非问候分支的write.create_tweet(..., text=reply)时，reply已经被后面的函数覆盖，实际传入的是函数对象而非预期的文本字符串，导致Tweepy尝试将函数序列化为JSON时失败。

修复步骤 #

1. 重命名回复文本变量：将存储随机回复的变量名改为reply_text（或其他不与函数冲突的名称）。
2. 更新代码中引用该变量的位置：在非问候分支中使用新的变量名。

修改后的关键代码片段：

# 变量声明部分
greet = random.choice([line.strip() for line in open('greetings.txt')]) # 读取时去除换行符
reply_text = random.choice([line.strip() for line in open('replies.txt')]) # 重命名变量并去除换行符

# 非问候回复分支
else:
    write.create_tweet(in_reply_to_tweet_id=id, text=reply_text) # 使用新变量名

额外优化建议 #

• 读取文件时添加strip()方法，去除每行末尾的换行符，避免回复内容带有多余的空白行。
• 避免在Python中使用相同名称命名变量和函数，这类命名冲突容易引发难以排查的逻辑错误。

内容的提问来源于stack exchange，提问作者Jack