问题:使用python和BeautifulSoup从网页检索链接

如何使用Python检索网页的链接并复制链接的URL地址?

How can I retrieve the links of a webpage and copy the url address of the links using Python?


回答 0

这是在BeautifulSoup中使用SoupStrainer类的一小段代码:

import httplib2
from bs4 import BeautifulSoup, SoupStrainer

http = httplib2.Http()
status, response = http.request('http://www.nytimes.com')

for link in BeautifulSoup(response, parse_only=SoupStrainer('a')):
    if link.has_attr('href'):
        print(link['href'])

BeautifulSoup文档实际上非常好,涵盖了许多典型方案:

https://www.crummy.com/software/BeautifulSoup/bs4/doc/

编辑:请注意,我使用了SoupStrainer类,因为如果您事先知道要解析什么,它会更有效率(在内存和速度方面)。

Here’s a short snippet using the SoupStrainer class in BeautifulSoup:

import httplib2
from bs4 import BeautifulSoup, SoupStrainer

http = httplib2.Http()
status, response = http.request('http://www.nytimes.com')

for link in BeautifulSoup(response, parse_only=SoupStrainer('a')):
    if link.has_attr('href'):
        print(link['href'])

The BeautifulSoup documentation is actually quite good, and covers a number of typical scenarios:

https://www.crummy.com/software/BeautifulSoup/bs4/doc/

Edit: Note that I used the SoupStrainer class because it’s a bit more efficient (memory and speed wise), if you know what you’re parsing in advance.


回答 1

为了完整起见,BeautifulSoup 4版本还使用了服务器提供的编码:

from bs4 import BeautifulSoup
import urllib.request

parser = 'html.parser'  # or 'lxml' (preferred) or 'html5lib', if installed
resp = urllib.request.urlopen("http://www.gpsbasecamp.com/national-parks")
soup = BeautifulSoup(resp, parser, from_encoding=resp.info().get_param('charset'))

for link in soup.find_all('a', href=True):
    print(link['href'])

或Python 2版本:

from bs4 import BeautifulSoup
import urllib2

parser = 'html.parser'  # or 'lxml' (preferred) or 'html5lib', if installed
resp = urllib2.urlopen("http://www.gpsbasecamp.com/national-parks")
soup = BeautifulSoup(resp, parser, from_encoding=resp.info().getparam('charset'))

for link in soup.find_all('a', href=True):
    print link['href']

以及使用requestslibrary的版本,该版本在Python 2和3中均可使用:

from bs4 import BeautifulSoup
from bs4.dammit import EncodingDetector
import requests

parser = 'html.parser'  # or 'lxml' (preferred) or 'html5lib', if installed
resp = requests.get("http://www.gpsbasecamp.com/national-parks")
http_encoding = resp.encoding if 'charset' in resp.headers.get('content-type', '').lower() else None
html_encoding = EncodingDetector.find_declared_encoding(resp.content, is_html=True)
encoding = html_encoding or http_encoding
soup = BeautifulSoup(resp.content, parser, from_encoding=encoding)

for link in soup.find_all('a', href=True):
    print(link['href'])

soup.find_all('a', href=True)调用将查找<a>具有href属性的所有元素。没有属性的元素将被跳过。

BeautifulSoup 3于2012年3月停止开发;新项目确实应该始终使用BeautifulSoup 4。

请注意,您应该将HTML从字节解码为BeautifulSoup。您可以将HTTP响应标头中找到的字符集告知BeautifulSoup以帮助解码,但这可能是错误的,并且与<meta>HTML本身中的标头信息相冲突,因此,以上内容使用BeautifulSoup内部类方法EncodingDetector.find_declared_encoding()来确保这样的嵌入式编码提示可以胜过配置错误的服务器。

使用requests,即使response.encoding响应text/*未返回任何字符集,如果响应具有mimetype ,该属性也会默认为Latin-1 。这与HTTP RFC一致,但是在与HTML解析一起使用时会很麻烦,因此当charsetContent-Type标头中设置为no时,应忽略该属性。

For completeness sake, the BeautifulSoup 4 version, making use of the encoding supplied by the server as well:

from bs4 import BeautifulSoup
import urllib.request

parser = 'html.parser'  # or 'lxml' (preferred) or 'html5lib', if installed
resp = urllib.request.urlopen("http://www.gpsbasecamp.com/national-parks")
soup = BeautifulSoup(resp, parser, from_encoding=resp.info().get_param('charset'))

for link in soup.find_all('a', href=True):
    print(link['href'])

or the Python 2 version:

from bs4 import BeautifulSoup
import urllib2

parser = 'html.parser'  # or 'lxml' (preferred) or 'html5lib', if installed
resp = urllib2.urlopen("http://www.gpsbasecamp.com/national-parks")
soup = BeautifulSoup(resp, parser, from_encoding=resp.info().getparam('charset'))

for link in soup.find_all('a', href=True):
    print link['href']

and a version using the requests library, which as written will work in both Python 2 and 3:

from bs4 import BeautifulSoup
from bs4.dammit import EncodingDetector
import requests

parser = 'html.parser'  # or 'lxml' (preferred) or 'html5lib', if installed
resp = requests.get("http://www.gpsbasecamp.com/national-parks")
http_encoding = resp.encoding if 'charset' in resp.headers.get('content-type', '').lower() else None
html_encoding = EncodingDetector.find_declared_encoding(resp.content, is_html=True)
encoding = html_encoding or http_encoding
soup = BeautifulSoup(resp.content, parser, from_encoding=encoding)

for link in soup.find_all('a', href=True):
    print(link['href'])

The soup.find_all('a', href=True) call finds all <a> elements that have an href attribute; elements without the attribute are skipped.

BeautifulSoup 3 stopped development in March 2012; new projects really should use BeautifulSoup 4, always.

Note that you should leave decoding the HTML from bytes to BeautifulSoup. You can inform BeautifulSoup of the characterset found in the HTTP response headers to assist in decoding, but this can be wrong and conflicting with a <meta> header info found in the HTML itself, which is why the above uses the BeautifulSoup internal class method EncodingDetector.find_declared_encoding() to make sure that such embedded encoding hints win over a misconfigured server.

With requests, the response.encoding attribute defaults to Latin-1 if the response has a text/* mimetype, even if no characterset was returned. This is consistent with the HTTP RFCs but painful when used with HTML parsing, so you should ignore that attribute when no charset is set in the Content-Type header.


回答 2

其他人推荐了BeautifulSoup,但是使用lxml更好。尽管它的名字,它也用于解析和抓取HTML。它比BeautifulSoup快得多,而且甚至比BeautifulSoup(他们声名fa起)更好地处理“断”的HTML。如果您不想学习lxml API,它也具有BeautifulSoup的兼容性API。

伊恩·布里金(Ian Blicking)同意

除非您使用的是Google App Engine或不允许使用任何非Python的工具,否则没有理由再使用BeautifulSoup。

lxml.html还支持CSS3选择器,因此这种事情很简单。

具有lxml和xpath的示例如下所示:

import urllib
import lxml.html
connection = urllib.urlopen('http://www.nytimes.com')

dom =  lxml.html.fromstring(connection.read())

for link in dom.xpath('//a/@href'): # select the url in href for all a tags(links)
    print link

Others have recommended BeautifulSoup, but it’s much better to use lxml. Despite its name, it is also for parsing and scraping HTML. It’s much, much faster than BeautifulSoup, and it even handles “broken” HTML better than BeautifulSoup (their claim to fame). It has a compatibility API for BeautifulSoup too if you don’t want to learn the lxml API.

Ian Blicking agrees.

There’s no reason to use BeautifulSoup anymore, unless you’re on Google App Engine or something where anything not purely Python isn’t allowed.

lxml.html also supports CSS3 selectors so this sort of thing is trivial.

An example with lxml and xpath would look like this:

import urllib
import lxml.html
connection = urllib.urlopen('http://www.nytimes.com')

dom =  lxml.html.fromstring(connection.read())

for link in dom.xpath('//a/@href'): # select the url in href for all a tags(links)
    print link

回答 3

import urllib2
import BeautifulSoup

request = urllib2.Request("http://www.gpsbasecamp.com/national-parks")
response = urllib2.urlopen(request)
soup = BeautifulSoup.BeautifulSoup(response)
for a in soup.findAll('a'):
  if 'national-park' in a['href']:
    print 'found a url with national-park in the link'
import urllib2
import BeautifulSoup

request = urllib2.Request("http://www.gpsbasecamp.com/national-parks")
response = urllib2.urlopen(request)
soup = BeautifulSoup.BeautifulSoup(response)
for a in soup.findAll('a'):
  if 'national-park' in a['href']:
    print 'found a url with national-park in the link'

回答 4

以下代码使用urllib2和检索网页中所有可用的链接BeautifulSoup4

import urllib2
from bs4 import BeautifulSoup

url = urllib2.urlopen("http://www.espncricinfo.com/").read()
soup = BeautifulSoup(url)

for line in soup.find_all('a'):
    print(line.get('href'))

The following code is to retrieve all the links available in a webpage using urllib2 and BeautifulSoup4:

import urllib2
from bs4 import BeautifulSoup

url = urllib2.urlopen("http://www.espncricinfo.com/").read()
soup = BeautifulSoup(url)

for line in soup.find_all('a'):
    print(line.get('href'))

回答 5

现在,BeautifulSoup现在使用lxml。请求,lxml和列表理解能力是一个杀手comb。

import requests
import lxml.html

dom = lxml.html.fromstring(requests.get('http://www.nytimes.com').content)

[x for x in dom.xpath('//a/@href') if '//' in x and 'nytimes.com' not in x]

在list comp中,“ if’//’和’url.com’不是x”是一种简单的方法,可以清理网站“内部”导航网址等的网址列表。

Under the hood BeautifulSoup now uses lxml. Requests, lxml & list comprehensions makes a killer combo.

import requests
import lxml.html

dom = lxml.html.fromstring(requests.get('http://www.nytimes.com').content)

[x for x in dom.xpath('//a/@href') if '//' in x and 'nytimes.com' not in x]

In the list comp, the “if ‘//’ and ‘url.com’ not in x” is a simple method to scrub the url list of the sites ‘internal’ navigation urls, etc.


回答 6

仅用于获取链接,而无需B.soup和regex:

import urllib2
url="http://www.somewhere.com"
page=urllib2.urlopen(url)
data=page.read().split("</a>")
tag="<a href=\""
endtag="\">"
for item in data:
    if "<a href" in item:
        try:
            ind = item.index(tag)
            item=item[ind+len(tag):]
            end=item.index(endtag)
        except: pass
        else:
            print item[:end]

对于更复杂的操作,当然BSoup仍然是首选。

just for getting the links, without B.soup and regex:

import urllib2
url="http://www.somewhere.com"
page=urllib2.urlopen(url)
data=page.read().split("</a>")
tag="<a href=\""
endtag="\">"
for item in data:
    if "<a href" in item:
        try:
            ind = item.index(tag)
            item=item[ind+len(tag):]
            end=item.index(endtag)
        except: pass
        else:
            print item[:end]

for more complex operations, of course BSoup is still preferred.


回答 7

该脚本可以满足您的需求,而且可以将相对链接解析为绝对链接。

import urllib
import lxml.html
import urlparse

def get_dom(url):
    connection = urllib.urlopen(url)
    return lxml.html.fromstring(connection.read())

def get_links(url):
    return resolve_links((link for link in get_dom(url).xpath('//a/@href')))

def guess_root(links):
    for link in links:
        if link.startswith('http'):
            parsed_link = urlparse.urlparse(link)
            scheme = parsed_link.scheme + '://'
            netloc = parsed_link.netloc
            return scheme + netloc

def resolve_links(links):
    root = guess_root(links)
    for link in links:
        if not link.startswith('http'):
            link = urlparse.urljoin(root, link)
        yield link  

for link in get_links('http://www.google.com'):
    print link

This script does what your looking for, But also resolves the relative links to absolute links.

import urllib
import lxml.html
import urlparse

def get_dom(url):
    connection = urllib.urlopen(url)
    return lxml.html.fromstring(connection.read())

def get_links(url):
    return resolve_links((link for link in get_dom(url).xpath('//a/@href')))

def guess_root(links):
    for link in links:
        if link.startswith('http'):
            parsed_link = urlparse.urlparse(link)
            scheme = parsed_link.scheme + '://'
            netloc = parsed_link.netloc
            return scheme + netloc

def resolve_links(links):
    root = guess_root(links)
    for link in links:
        if not link.startswith('http'):
            link = urlparse.urljoin(root, link)
        yield link  

for link in get_links('http://www.google.com'):
    print link

回答 8

要查找所有链接,在本示例中,我们将urllib2模块与re.module一起使用 * re模块中最强大的功能之一是“ re.findall()”。使用re.search()查找模式的第一个匹配项时,re.findall()查找所有 匹配项并将它们作为字符串列表返回,每个字符串代表一个匹配项*

import urllib2

import re
#connect to a URL
website = urllib2.urlopen(url)

#read html code
html = website.read()

#use re.findall to get all the links
links = re.findall('"((http|ftp)s?://.*?)"', html)

print links

To find all the links, we will in this example use the urllib2 module together with the re.module *One of the most powerful function in the re module is “re.findall()”. While re.search() is used to find the first match for a pattern, re.findall() finds all the matches and returns them as a list of strings, with each string representing one match*

import urllib2

import re
#connect to a URL
website = urllib2.urlopen(url)

#read html code
html = website.read()

#use re.findall to get all the links
links = re.findall('"((http|ftp)s?://.*?)"', html)

print links

回答 9

为什么不使用正则表达式:

import urllib2
import re
url = "http://www.somewhere.com"
page = urllib2.urlopen(url)
page = page.read()
links = re.findall(r"<a.*?\s*href=\"(.*?)\".*?>(.*?)</a>", page)
for link in links:
    print('href: %s, HTML text: %s' % (link[0], link[1]))

Why not use regular expressions:

import urllib2
import re
url = "http://www.somewhere.com"
page = urllib2.urlopen(url)
page = page.read()
links = re.findall(r"<a.*?\s*href=\"(.*?)\".*?>(.*?)</a>", page)
for link in links:
    print('href: %s, HTML text: %s' % (link[0], link[1]))

回答 10

链接可以位于各种属性中,因此您可以传递这些属性的列表以进行选择

例如,使用src和href属性(在这里,我使用^开头的运算符来指定这些属性值之一以http开头。您可以根据需要定制此属性

from bs4 import BeautifulSoup as bs
import requests
r = requests.get('https://stackoverflow.com/')
soup = bs(r.content, 'lxml')
links = [item['href'] if item.get('href') is not None else item['src'] for item in soup.select('[href^="http"], [src^="http"]') ]
print(links)

属性=值选择器

[attr ^ = value]

表示属性名称为attr的元素,其值以值为前缀(前缀)。

Links can be within a variety of attributes so you could pass a list of those attributes to select

for example, with src and href attribute (here I am using the starts with ^ operator to specify that either of these attributes values starts with http. You can tailor this as required

from bs4 import BeautifulSoup as bs
import requests
r = requests.get('https://stackoverflow.com/')
soup = bs(r.content, 'lxml')
links = [item['href'] if item.get('href') is not None else item['src'] for item in soup.select('[href^="http"], [src^="http"]') ]
print(links)

Attribute = value selectors

[attr^=value]

Represents elements with an attribute name of attr whose value is prefixed (preceded) by value.


回答 11

下面是使用@ars接受的答案和一个例子BeautifulSoup4requestswget模块来处理下载。

import requests
import wget
import os

from bs4 import BeautifulSoup, SoupStrainer

url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/eeg-mld/eeg_full/'
file_type = '.tar.gz'

response = requests.get(url)

for link in BeautifulSoup(response.content, 'html.parser', parse_only=SoupStrainer('a')):
    if link.has_attr('href'):
        if file_type in link['href']:
            full_path = url + link['href']
            wget.download(full_path)

Here’s an example using @ars accepted answer and the BeautifulSoup4, requests, and wget modules to handle the downloads.

import requests
import wget
import os

from bs4 import BeautifulSoup, SoupStrainer

url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/eeg-mld/eeg_full/'
file_type = '.tar.gz'

response = requests.get(url)

for link in BeautifulSoup(response.content, 'html.parser', parse_only=SoupStrainer('a')):
    if link.has_attr('href'):
        if file_type in link['href']:
            full_path = url + link['href']
            wget.download(full_path)

回答 12

经过以下更正(发现无法正常运行的情况),我发现了@ Blairg23工作的答案:

for link in BeautifulSoup(response.content, 'html.parser', parse_only=SoupStrainer('a')):
    if link.has_attr('href'):
        if file_type in link['href']:
            full_path =urlparse.urljoin(url , link['href']) #module urlparse need to be imported
            wget.download(full_path)

对于Python 3:

urllib.parse.urljoin 必须使用以获得完整的URL。

I found the answer by @Blairg23 working , after the following correction (covering the scenario where it failed to work correctly):

for link in BeautifulSoup(response.content, 'html.parser', parse_only=SoupStrainer('a')):
    if link.has_attr('href'):
        if file_type in link['href']:
            full_path =urlparse.urljoin(url , link['href']) #module urlparse need to be imported
            wget.download(full_path)

For Python 3:

urllib.parse.urljoin has to be used in order to obtain the full URL instead.


回答 13

BeatifulSoup自己的解析器可能很慢。使用能够直接从URL进行解析的lxml(可能会在下面提到一些限制)可能更为可行。

import lxml.html

doc = lxml.html.parse(url)

links = doc.xpath('//a[@href]')

for link in links:
    print link.attrib['href']

上面的代码将按原样返回链接,并且在大多数情况下,它们将是相对链接或从站点根目录开始的绝对链接。由于我的用例仅提取某种类型的链接,因此下面是一个将链接转换为完整URL的版本,该版本可以选择接受glob模式(如)*.mp3。它不会处理相对路径中的单点和双点,但是到目前为止,我并不需要它。如果您需要解析包含../或的URL片段,./urlparse.urljoin可能会派上用场。

注意:直接的lxml url解析不处理来自加载,https也不进行重定向,因此,出于这个原因,以下版本使用urllib2+ lxml

#!/usr/bin/env python
import sys
import urllib2
import urlparse
import lxml.html
import fnmatch

try:
    import urltools as urltools
except ImportError:
    sys.stderr.write('To normalize URLs run: `pip install urltools --user`')
    urltools = None


def get_host(url):
    p = urlparse.urlparse(url)
    return "{}://{}".format(p.scheme, p.netloc)


if __name__ == '__main__':
    url = sys.argv[1]
    host = get_host(url)
    glob_patt = len(sys.argv) > 2 and sys.argv[2] or '*'

    doc = lxml.html.parse(urllib2.urlopen(url))
    links = doc.xpath('//a[@href]')

    for link in links:
        href = link.attrib['href']

        if fnmatch.fnmatch(href, glob_patt):

            if not href.startswith(('http://', 'https://' 'ftp://')):

                if href.startswith('/'):
                    href = host + href
                else:
                    parent_url = url.rsplit('/', 1)[0]
                    href = urlparse.urljoin(parent_url, href)

                    if urltools:
                        href = urltools.normalize(href)

            print href

用法如下:

getlinks.py http://stackoverflow.com/a/37758066/191246
getlinks.py http://stackoverflow.com/a/37758066/191246 "*users*"
getlinks.py http://fakedomain.mu/somepage.html "*.mp3"

BeatifulSoup’s own parser can be slow. It might be more feasible to use lxml which is capable of parsing directly from a URL (with some limitations mentioned below).

import lxml.html

doc = lxml.html.parse(url)

links = doc.xpath('//a[@href]')

for link in links:
    print link.attrib['href']

The code above will return the links as is, and in most cases they would be relative links or absolute from the site root. Since my use case was to only extract a certain type of links, below is a version that converts the links to full URLs and which optionally accepts a glob pattern like *.mp3. It won’t handle single and double dots in the relative paths though, but so far I didn’t have the need for it. If you need to parse URL fragments containing ../ or ./ then urlparse.urljoin might come in handy.

NOTE: Direct lxml url parsing doesn’t handle loading from https and doesn’t do redirects, so for this reason the version below is using urllib2 + lxml.

#!/usr/bin/env python
import sys
import urllib2
import urlparse
import lxml.html
import fnmatch

try:
    import urltools as urltools
except ImportError:
    sys.stderr.write('To normalize URLs run: `pip install urltools --user`')
    urltools = None


def get_host(url):
    p = urlparse.urlparse(url)
    return "{}://{}".format(p.scheme, p.netloc)


if __name__ == '__main__':
    url = sys.argv[1]
    host = get_host(url)
    glob_patt = len(sys.argv) > 2 and sys.argv[2] or '*'

    doc = lxml.html.parse(urllib2.urlopen(url))
    links = doc.xpath('//a[@href]')

    for link in links:
        href = link.attrib['href']

        if fnmatch.fnmatch(href, glob_patt):

            if not href.startswith(('http://', 'https://' 'ftp://')):

                if href.startswith('/'):
                    href = host + href
                else:
                    parent_url = url.rsplit('/', 1)[0]
                    href = urlparse.urljoin(parent_url, href)

                    if urltools:
                        href = urltools.normalize(href)

            print href

The usage is as follows:

getlinks.py http://stackoverflow.com/a/37758066/191246
getlinks.py http://stackoverflow.com/a/37758066/191246 "*users*"
getlinks.py http://fakedomain.mu/somepage.html "*.mp3"

回答 14

import urllib2
from bs4 import BeautifulSoup
a=urllib2.urlopen('http://dir.yahoo.com')
code=a.read()
soup=BeautifulSoup(code)
links=soup.findAll("a")
#To get href part alone
print links[0].attrs['href']
import urllib2
from bs4 import BeautifulSoup
a=urllib2.urlopen('http://dir.yahoo.com')
code=a.read()
soup=BeautifulSoup(code)
links=soup.findAll("a")
#To get href part alone
print links[0].attrs['href']

回答 15

外部链接和内部链接都可以有很多重复的链接。要区分两者并仅使用集合获得唯一链接:

# Python 3.
import urllib    
from bs4 import BeautifulSoup

url = "http://www.espncricinfo.com/"
resp = urllib.request.urlopen(url)
# Get server encoding per recommendation of Martijn Pieters.
soup = BeautifulSoup(resp, from_encoding=resp.info().get_param('charset'))  
external_links = set()
internal_links = set()
for line in soup.find_all('a'):
    link = line.get('href')
    if not link:
        continue
    if link.startswith('http'):
        external_links.add(link)
    else:
        internal_links.add(link)

# Depending on usage, full internal links may be preferred.
full_internal_links = {
    urllib.parse.urljoin(url, internal_link) 
    for internal_link in internal_links
}

# Print all unique external and full internal links.
for link in external_links.union(full_internal_links):
    print(link)

There can be many duplicate links together with both external and internal links. To differentiate between the two and just get unique links using sets:

# Python 3.
import urllib    
from bs4 import BeautifulSoup

url = "http://www.espncricinfo.com/"
resp = urllib.request.urlopen(url)
# Get server encoding per recommendation of Martijn Pieters.
soup = BeautifulSoup(resp, from_encoding=resp.info().get_param('charset'))  
external_links = set()
internal_links = set()
for line in soup.find_all('a'):
    link = line.get('href')
    if not link:
        continue
    if link.startswith('http'):
        external_links.add(link)
    else:
        internal_links.add(link)

# Depending on usage, full internal links may be preferred.
full_internal_links = {
    urllib.parse.urljoin(url, internal_link) 
    for internal_link in internal_links
}

# Print all unique external and full internal links.
for link in external_links.union(full_internal_links):
    print(link)

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