从 python 请求中的 txt 文件中的链接获取所有分页 URL 的列表

Question

大家好，定义一个函数，从 python.

的 txt 文件中的 links 中获取底部的所有分页 URL 列表

这是我需要完成的示例。

输入link

http://www.apartmentguide.com/apartments/Alabama/Hartselle/

期望的输出

www.apartmentguide.com/apartments/Alabama/Hartselle/?page=2
www.apartmentguide.com/apartments/Alabama/Hartselle/?page=3
www.apartmentguide.com/apartments/Alabama/Hartselle/?page=4
www.apartmentguide.com/apartments/Alabama/Hartselle/?page=5
www.apartmentguide.com/apartments/Alabama/Hartselle/?page=6
www.apartmentguide.com/apartments/Alabama/Hartselle/?page=7
www.apartmentguide.com/apartments/Alabama/Hartselle/?page=8
www.apartmentguide.com/apartments/Alabama/Hartselle/?page=9

以此类推每个输入 Url 具有的任何限制。

这是我到目前为止编写的函数，但它不起作用我也不擅长 Python。

import requests
#from bs4 import BeautifulSoup
from scrapy import Selector as Se
import urllib2


lists = open("C:\Users\Administrator\Desktop\3.txt","r")
read_list = lists.read()
line = read_list.split("\n")


def get_links(line):
    for each in line:
        r = requests.get(each)
        sel = Se(text=r.text, type="html")
        next_ = sel.xpath('//a[@class="next sprite"]//@href').extract()
        for next_1 in next_:
            next_2 = "http://www.apartmentguide.com"+next_1
            print next_2
        get_links(next_1)

get_links(line)

Answer 1

以下是执行此操作的两种方法。

import mechanize

import requests
from bs4 import BeautifulSoup, SoupStrainer
import urlparse

import pprint

#-- Mechanize --
br = mechanize.Browser()

def get_links_mechanize(root):
    links = []
    br.open(root)

    for link in br.links():
        try:
            if dict(link.attrs)['class'] == 'page':
                links.append(link.absolute_url)
        except:
            pass
    return links


#-- Requests / BeautifulSoup / urlparse --
def get_links_bs(root):
    links = []
    r = requests.get(root)

    for link in BeautifulSoup(r.text, parse_only=SoupStrainer('a')):
        if link.has_attr('href') and link.has_attr('class') and 'page' in link.get('class'):
            links.append(urlparse.urljoin(root, link.get('href')))

    return links


#with open("C:\Users\Administrator\Desktop\3.txt","r") as f:
#    for root in f:
#        links = get_links(root) 
#        # <Do something with links>
root = 'http://www.apartmentguide.com/apartments/Alabama/Hartselle/'

print "Mech:"
pprint.pprint( get_links_mechanize(root) )
print "Requests/BS4/urlparse:"
pprint.pprint( get_links_bs(root) )

一个人使用 mechanize -- 它使用 URL 更智能一些，但速度要慢得多，并且可能会过大，具体取决于您在做什么。

另一个使用 requests 获取页面（urllib2 就足够了），BeautifulSoup 解析标记，urlparse 从页面中的相对 URL 形成绝对 URL已列出。

请注意，这两个函数 return 以下列表：

['http://www.apartmentguide.com/apartments/Alabama/Hartselle/?page=2',
 'http://www.apartmentguide.com/apartments/Alabama/Hartselle/?page=3',
 'http://www.apartmentguide.com/apartments/Alabama/Hartselle/?page=4',
 'http://www.apartmentguide.com/apartments/Alabama/Hartselle/?page=5',
 'http://www.apartmentguide.com/apartments/Alabama/Hartselle/?page=2',
 'http://www.apartmentguide.com/apartments/Alabama/Hartselle/?page=3',
 'http://www.apartmentguide.com/apartments/Alabama/Hartselle/?page=4',
 'http://www.apartmentguide.com/apartments/Alabama/Hartselle/?page=5']

其中有重复项。您可以通过更改

来删除重复项

return links

到

return list(set(links))

无论您选择哪种方法。

编辑：

我注意到上述功能仅 return 链接到第 2-5 页，您必须浏览这些页面才能看到实际上有 10 页。

一种完全不同的方法是抓取 "root" 页面的结果数量，然后预测会产生多少页面，然后从中构建链接。

由于每页有 20 个结果，计算出多少页很简单，请考虑：

import requests, re, math, pprint

def scrape_results(root):
    links = []
    r = requests.get(root)

    mat = re.search(r'We have (\d+) apartments for rent', r.text)
    num_results = int(mat.group(1))                     # 182 at the moment
    num_pages = int(math.ceil(num_results/20.0))        # ceil(182/20) => 10

    # Construct links for pages 1-10
    for i in range(num_pages):
        links.append("%s?page=%d" % (root, (i+1)))

    return links

pprint.pprint(scrape_results(root))

这将是 3 种方法中最快的方法，但可能更容易出错。

编辑 2:

可能是这样的：

import re, math, pprint
import requests, urlparse
from bs4 import BeautifulSoup, SoupStrainer

def get_pages(root):
    links = []
    r = requests.get(root)

    mat = re.search(r'We have (\d+) apartments for rent', r.text)
    num_results = int(mat.group(1))                     # 182 at the moment
    num_pages = int(math.ceil(num_results/20.0))        # ceil(182/20) => 10

    # Construct links for pages 1-10
    for i in range(num_pages):
        links.append("%s?page=%d" % (root, (i+1)))

    return links

def get_listings(page):
    links = []
    r = requests.get(page)

    for link in BeautifulSoup(r.text, parse_only=SoupStrainer('a')):
        if link.has_attr('href') and link.has_attr('data-listingid') and 'name' in link.get('class'):
            links.append(urlparse.urljoin(root, link.get('href')))

    return links

root='http://www.apartmentguide.com/apartments/Alabama/Hartselle/'
listings = []
for page in get_pages(root):
    listings += get_listings(page)

pprint.pprint(listings)
print(len(listings))

Answer 2

Re 我不确定，所以尝试了 xpath。

links = open("C:\Users\ssamant\Desktop\Anida\Phase_II\Apartmentfinder\2.txt","r")
read_list = links.read()
line = read_list.split("\n")

for each in line:
    lines = []
    r = requests.get(each)
    sel = Selector(text=r.text,type="html")
    mat = sel.xpath('//h1//strong/text()').extract()
    mat = str(mat)
    mat1 = mat.replace(" apartments for rent']","")
    mat2 = mat1.replace("[u'","")
    mat3 = int(mat2)
    num_pages = int(math.ceil(mat3/20.0))
    for i in range(num_pages):
        lines.append("%s/Page%d" % (each, (i+1)))
    with open('C:\Users\ssamant\Desktop\Anida\Phase_II\Apartmentfinder\test.csv', 'ab') as f:
        writer = csv.writer(f)
        for val in lines:
            writer.writerow([val])

从 python 请求中的 txt 文件中的链接获取所有分页 URL 的列表

Get list of all paginated URL's from links in txt file in python requests

python

function

list

python-requests