Question

我是webscraping和正则表达式的新手，在这里遇到问题。我的一个代码给了我一个HTML输出，但我需要从段落中提取某个部分而不是完整的段落。我需要帮助。以下是我的代码。

import mechanize
from bs4 import BeautifulSoup
import urllib2
br = mechanize.Browser()
response = br.open("http://www.consultadni.info/index.php")
br.select_form(name="form1")
br['APE_PAT']='PATRICIO'
br['APE_MAT']='GAMARRA'
br['NOMBRES']='MARCELINA'
req=br.submit().read()
soup = BeautifulSoup(req, "lxml")
for link in soup.findAll("a"):
     sub=link.get("href")
     soup1 = BeautifulSoup(sub, "lxml")
     print soup1.find_all('p')

屏幕输出：

[<p>/</p>]
[<p>datospersonales.php?nc=PATRICIO GAMARRA MARCELINA&amp;dni1=40772568&amp;dni2=12405868&amp;id1=12a40a58a68&amp;id2=30/06/1980&amp;dni3=40631880</p>]
[<p>datospersonales.php?nc=PATRICIO GAMARRA MARCELINA&amp;dni1=40772568&amp;dni2=12405868&amp;id1=12a40a58a68&amp;id2=30/06/1980&amp;dni3=40631880</p>]
[<p>http://www.infocorpperuconsultatusdeudas.blogspot.com/2015/05/infocorp-consulta-gratis-tu-reporte-de.html?ref=dnionline</p>]

我需要什么：30/06/1980＆amp; 40631880

Answer 1

清理解析URL的方法（Python 3）：

from urllib import parse

URL = "datospersonales.php?nc=PATRICIO GAMARRA MARCELINA&dni1=40772568&dni2=12405868&id1=12a40a58a68&id2=30/06/1980&dni3=40631880"

query_parts = parse.parse_qs(parse.urlparse(URL).query)

print(query_parts["id2"][0], query_parts["dni3"][0])

Answer 2

对于Python 2.7，请尝试这种方式：

from urlparse import parse_qs

result = set()

for link in soup.find_all("a"):
     sub = parse_qs(link.get("href"))

     if "id2" in sub:
         result.add((sub["id2"][0], sub["dni3"][0]))

print result

如何提取HTML段落的某些部分

2 个答案: