我是 python 的新手。我有一个错误需要了解。
代码:
配置文件:
# Vou definir os feeds
feeds_updates = [{"feedurl": "http://aaa1.com/rss/punch.rss", "linktoourpage": "http://www.ha.com/fun.htm"},
{"feedurl": "http://aaa2.com/rss.xml", "linktoourpage": "http://www.ha.com/fun.htm"},
{"feedurl": "http://aaa3.com/Heaven", "linktoourpage": "http://www.ha.com/fun.htm"},
{"feedurl": "http://aaa4.com/feed.php", "linktoourpage": "http://www.ha.com/fun.htm"},
{"feedurl": "http://aaa5.com/index.php?format=feed&type=rss", "linktoourpage": "http://www.ha.com/fun.htm"},
{"feedurl": "http://aaa6.com/rss.xml", "linktoourpage": "http://www.ha.com/fun.htm"},
{"feedurl": "http://aaa7.com/?format=xml", "linktoourpage": "http://www.ha.com/fun.htm"},
{"feedurl": "http://aaa8/site/component/rsssyndicator/?feed_id=1", "linktoourpage": "http://www.ha.com/fun.htm"}]
推特C.py
# -*- coding: utf-8 -*-
import config # Ficheiro de configuracao
import twitter
import random
import sqlite3
import time
import bitly_api #https://github.com/bitly/bitly-api-python
import feedparser
...
# Vou escolher um feed ao acaso
feed_a_enviar = random.choice(config.feeds_updates)
# Vou apanhar o conteudo do feed
d = feedparser.parse(feed_a_enviar["feedurl"])
# Vou definir quantos feeds quero ter no i
i = range(8)
print i
# Vou meter para "updates" 10 entradas do feed
updates = []
for i in range(8):
updates.append([{"url": feed_a_enviar["linktoourpage"], "msg": d.entries[i].title + ", "}])
# Vou escolher ums entrada ao acaso
print updates # p debug so
update_to_send = random.choice(updates)
print update_to_send # Para efeitos de debug
并且由于随机的性质有时会出现的错误:
Traceback (most recent call last):
File "C:\Users\anlopes\workspace\redes_sociais\src\twitterC.py", line 77, in <module>
updates.append([{"url": feed_a_enviar["linktoourpage"], "msg": d.entries[i].title + ", "}])
IndexError: list index out of range
我没有遇到错误，列表“feeds_updates”是一个包含 8 个元素的列表，我认为声明得很好，RANDOM 将从 8 个元素中选择一个...
谁能告诉我这里发生了什么？
PS:抱歉我的英语不好。
最好的问候，
请您参考如下方法:
使用 range 进行迭代几乎总是不是最好的方法。在 Python 中，您可以直接遍历列表、字典、集合等:
for item in d.entries:
updates.append([{"url": feed_a_enviar["linktoourpage"], "msg": item.title + ", "}])
显然 d.entries[i] 会触发错误，因为该列表包含的项目少于 8 个(feeds_updates 可能包含 8 个，但您没有迭代该列表) .
