Question

我目前正在尝试应用正则表达式，以便从包含链接的列表中过滤掉某些链接。

我现在已经尝试了几种方式，但我总是遇到这个错误：

Traceback (most recent call last):
  File "/Users/User/Documents/pyp/pushbullet_updater/DoDa/test.py", line 20, in <module>
    print(get_chapter_links(links))
  File "/Users/User/Documents/pyp/pushbullet_updater/DoDa/test.py", line 15, in get_chapter_links
    match = re.findall(r"https://bluesilvertranslations\.wordpress\.com/\d{4}/\d{2}/\d{2}/douluo-dalu-\d{1,3}-\s*/", link)
  File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/re.py", line 210, in findall
    return _compile(pattern, flags).findall(string)
TypeError: expected string or buffer

我做错了什么？

以下是代码：

import requests
from bs4 import BeautifulSoup
import re

#Gets chapter links
def get_chapter_links(index_url):
    r = requests.get(index_url)
    soup = BeautifulSoup(r.content, 'lxml')
    links = soup.find_all('a')
    url_list = []
    for url in links:
        url_list.append((url.get('href')))

    for link in url_list: # Iterates through every line and looks for a match:
        match = re.findall(r"https://bluesilvertranslations\.wordpress\.com/\d{4}/\d{2}/\d{2}/douluo-dalu-\d{1,3}-\s*/", link)
    return match

links = 'https://bluesilvertranslations.wordpress.com/chapter-list/'

print(get_chapter_links(links))

Answer 1

来自re文档

re.findall(pattern, string, flags=0)
Return all non-overlapping matches of pattern in string, as a list of strings. The string is scanned left-to-right, and matches are returned in the order found. If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group. Empty matches are included in the result unless they touch the beginning of another match.

New in version 1.5.2.

Changed in version 2.4: Added the optional flags argument.

注意：

第一个参数应该是一个模式，第二个参数应该是一个字符串

修改后的代码：

import requests from bs4 import BeautifulSoup import re #Gets chapter links def get_chapter_links(index_url): r = requests.get(index_url) soup = BeautifulSoup(r.content, 'lxml') links = soup.find_all('a') url_list = [] for url in links: url_list.append((url.get('href'))) match = [] # Create a list and append to it the matched links for link in url_list: # Iterates through every line and looks for a match: if link: # I have added this to see in there are value in link. match += re.findall(r"https://bluesilvertranslations\.wordpress\.com/\d{4}/\d{2}/\d{2}/douluo-dalu-\d{1,3}-.*/", link) # I have changed the regex a bit since your did not match return match links = 'https://bluesilvertranslations.wordpress.com/chapter-list/' print(get_chapter_links(links))

＆＃34; TypeError：期望的字符串或缓冲区＆＃34;使用正则表达式过滤列表时

1 个答案: