1

我对python相当陌生。我有一个需要理解的错误。

编码:

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"}]

twitterC.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

由于随机的性质,有时会出现错误:

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,随机数将从 8 个中选择一个...

有人可以告诉我这里发生了什么吗?

PS:对不起我的英语不好。

最好的祝福,

4

2 回答 2

5

使用range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,因为该列表包含少于 8 个项目(feeds_updates可能包含 8 个,但您没有迭代该列表)。

于 2011-04-01T09:56:46.963 回答
2

d.entries少于 8 个元素。直接迭代d.entries而不是一些断开的范围。

于 2011-04-01T09:53:19.400 回答