Question

我有一个python类的作业，我必须从特定位置的特定链接开始，然后按照该链接特定次数。据说第一个链接的位置为1。这是链接：http://python-data.dr-chuck.net/known_by_Fikret.html

traceback error picture 我找不到链接，错误＆＃34;索引超出范围＆＃34;出来。任何人都可以帮助找出如何找到链接/位置？这是我的代码：

import urllib
from BeautifulSoup import *

url = raw_input('Enter - ')
html = urllib.urlopen(url).read()
soup = BeautifulSoup(html)
count = int(raw_input('Enter count: '))+1
position = int(raw_input('Enter position: '))


tags = soup('a')
tags_lst = list()
for tag in tags:
    needed_tag = tag.get('href', None)
    tags_lst.append(needed_tag)
    for i in range(0,count):
        print 'retrieving: ',tags_lst[position]

好的，我写了这段代码，它有点工作：

import urllib
from BeautifulSoup import *

url = raw_input('Enter - ')
html = urllib.urlopen(url).read()
soup = BeautifulSoup(html)
count = int(raw_input('Enter count: '))+1
position = int(raw_input('Enter position: '))


tags = soup('a')
tags_lst = list()
for tag in tags:
    needed_tag = tag.get('href', None)
    tags_lst.append(needed_tag)
for i in range(0,count):    
    print 'retrieving: ',tags_lst[position]
    position = position + 1

我仍然获得除示例中的其他链接，但是当我打印整个链接列表时，位置匹配，所以我不知道。非常奇怪。

Answer 1

[编辑：剪切+粘贴评论中的这一行]嗨！我不得不在类似的练习中工作，因为我有些疑惑，我找到了你的问题。这是我的代码，我认为它的工作原理。我希望它对你有所帮助

import urllib
from bs4 import BeautifulSoup

url = 'http://py4e-data.dr-chuck.net/known_by_Fikret.html'
html = urllib.urlopen(url).read()
soup = BeautifulSoup(html, 'html.parser')
count = 8
position = 18
tags_lst = []

for x in xrange(count-1):
    tags = soup('a')
    my_tags = tags[position-1]
    needed_tag = my_tags.get('href', None)
    tags_lst.append(needed_tag)
    url = str(needed_tag)
    html = urllib.urlopen(url).read()
    soup = BeautifulSoup(html, 'html.parser')

Answer 2

您的BeautifulSoup导入错误。我不认为它适用于您展示的代码。你的下部循环也令人困惑。您可以通过切片完整检索的列表来获取所需的网址列表。

我已经在我的代码中对您的网址进行了硬编码，因为它比在每次运行中输入更容易。

试试这个：

import urllib
from bs4 import BeautifulSoup

#url = raw_input('Enter - ')
url = 'http://python-data.dr-chuck.net/known_by_Fikret.html'
html = urllib.urlopen(url).read()
soup = BeautifulSoup(html)
# print soup
count = int(raw_input('Enter count: '))+1
position = int(raw_input('Enter position: '))


tags = soup('a')
# next line gets count tags starting from position
my_tags = tags[position: position+count]
tags_lst = []
for tag in my_tags:
    needed_tag = tag.get('href', None)
    tags_lst.append(needed_tag)
print tags_lst

Answer 3

此任务的几乎所有解决方案都有两个部分来加载网址。相反，我定义了一个函数，用于打印任何给定URL的相关链接。

最初，该函数将使用Fikret.html url作为输入。后续输入依赖于出现在所需位置的刷新URL。重要的代码是这一行：url = allerretour(url)[position-1]这会获得新的url，它会为循环提供另一轮。

import urllib
from bs4 import BeautifulSoup
url = 'http://py4e-data.dr-chuck.net/known_by_Fikret.html' # raw_input('Enter URL : ')

position = 3 # int(raw_input('Enter position : '))
count = 4 #int(raw_input('Enter count : '))

def allerretour(url):
    print('Retrieving: ' + url)
    soup = BeautifulSoup(urllib.urlopen(url).read())
    link = list()
    for tag in soup('a'):
        link.append(tag.get('href', None))
    return(link)


for x in range(1, count + 2):
    url = allerretour(url)[position-1]

Answer 4

这是我的解决方案：

import urllib.request, urllib.parse, urllib.error
from bs4 import BeautifulSoup
import ssl

ctx = ssl.create_default_context()
ctx.check_hostname = False
ctx.verify_mode = ssl.CERT_NONE

url = input('Enter: ')
link_line = int(input("Enter position: ")) - 1 
relative to first link
count = int(input("Enter count: "))

html = urllib.request.urlopen(url, context=ctx).read()
soup = BeautifulSoup(html, 'html.parser')

while count >= 0:
   html = urllib.request.urlopen(url, context=ctx).read()
   soup = BeautifulSoup(html, 'html.parser')
   tags = soup('a')
   print(url)
   url = tags[link_line].get("href", None)
   count = count - 1

Answer 5

这是我在Python 2.7中的解决方案：

import urllib
from BeautifulSoup import *

URL = raw_input("Enter the URL:") #Enter main URL
link_line = int(raw_input("Enter position:")) - 1 #The position of link relative to first link
count = int(raw_input("Enter count:")) #The number of times to be repeated

while count >= 0:
    html = urllib.urlopen(URL).read()
    soup = BeautifulSoup(html)
    tags = soup('a')
    print URL
    URL = tags[link_line].get("href", None)
    count = count - 1

Answer 6

这是提供所需输出的工作代码

import urllib.request, urllib.parse, urllib.error
from bs4 import BeautifulSoup
import ssl

# Ignore SSL certificate errors
ctx = ssl.create_default_context()
ctx.check_hostname = False
ctx.verify_mode = ssl.CERT_NONE
n=1
url = input('Enter - ')
count= int(input('Enter count'))+1
pos=int(input('Enter position'))
new=url
while n<count:
    if new == url:
        html = urllib.request.urlopen(url, context=ctx).read()
        print('Retrieving', url)
    html = urllib.request.urlopen(new, context=ctx).read()
    soup = BeautifulSoup(html, 'html.parser')
    tags = soup('a')
    my_tags=tags[pos-1]
    new=my_tags.get('href', None)
    print('Retrieving' , new)
    n=n+1

Answer 7

我将解决方案放在下面，经过测试，直到今天为止工作良好。

导入需求模块

import urllib.request, urllib.parse, urllib.error
from bs4 import BeautifulSoup
import re

访问网站

url = "http://py4e-data.dr-chuck.net/known_by_Vairi.html"
html = urllib.request.urlopen(url).read()
soup = BeautifulSoup(html, 'html.parser')
all_num_list = list()
link_position = 18
Process_repeat = 7

检索所有锚标签

tags = soup('a')

while Process_repeat - 1  >= 0 :
    print("Process round", Process_repeat)
    target = tags[link_position - 1]
    print("target:", target)
    url = target.get('href', 2)
    print("Current url", url)
    html = urllib.request.urlopen(url).read()
    soup = BeautifulSoup(html, 'html.parser')
    tags = soup('a')
    Process_repeat = Process_repeat - 1

Answer 8

import urllib.error, urllib.request
from bs4 import BeautifulSoup

#url = 'http://py4e-data.dr-chuck.net/known_by_Fikret.html'
url = input('Enter link - ')
count = int(input('Enter count - '))
position = int(input('position - ') )
html = urllib.request.urlopen(url).read()
soup = BeautifulSoup(html, 'html.parser')

tags = soup('a')
my_tags = tags[position-1]
needed_tag = my_tags.get('href', None)
print("------ : ", tags[position-1].contents[0])

for x in range(count-1):

    url = str(needed_tag)
    html = urllib.request.urlopen(url).read()
    soup = BeautifulSoup(html, 'html.parser')

    tags = soup('a')
    my_tags = tags[position-1]
    needed_tag = my_tags.get('href', None)
    print("------ : ", tags[position-1].contents[0])

Answer 9

尝试一下。您可以保留输入URL的权限。有您以前链接的示例。祝你好运！

var a = ['barclays', 'google', 'vod'];
var b = [];

var dataLength = 4;

for (let i = 0; i < dataLength; i++) {
   for (let o = 0; o < a.length; o++){
       b.push(a[o]);
   }
}

console.log(b);

使用Beautifulsoup进行python任务中的链接

9 个答案:

导入需求模块

访问网站

检索所有锚标签