从包含python中URL列表的文本文件中打印文本文件中URL的内容

Question

我有一个包含 URL 列表的文本文件，我愿意将 URL 的内容打印到另一个文本文件中，并将 URL 作为标题。我已经使用这个项目文件https://pypi.org/project/Wikipedia-API/ 来提取内容，但是我必须一个接一个地输入链接，我不想深入研究，因为我的列表很大，每个文本文件至少有 3000 个链接。

谁能帮我解决这个问题，将不胜感激。

编辑：

我用下面的方法试过了，但是输出的txt文件里没有内容。

import urllib
import datetime as dt
from datetime import datetime

import time

linklist = []
with open ("test.txt", 'r', encoding = 'utf=8') as wikitxt :
         #content = wikitxt.read().splitlines()
         for i in wikitxt:
                  linklist.append (i)

output = open('Wikipedia_content.txt', 'w', encoding='utf-8')

startTime = time.time()
endTime = time.time()
runTime = endTime - startTime
print("Runtime is %3f seconds" % runTime)

这是我使用过的txt文件https://pastebin.com/Y4bwsHGB，这是我需要使用的文本文件：https://pastebin.com/SXDAu8jV。

提前致谢。

问题：

Traceback (most recent call last):


 File "C:/Users/suva_/Desktop/Project specification/data/test2.py", line 13, in <module>
    output_file.write((urlopen(link).read()))
  File "D:\Python 36\lib\urllib\request.py", line 228, in urlopen
    return opener.open(url, data, timeout)
  File "D:\Python 36\lib\urllib\request.py", line 531, in open
    response = self._open(req, data)
  File "D:\Python 36\lib\urllib\request.py", line 554, in _open
    'unknown_open', req)
  File "D:\Python 36\lib\urllib\request.py", line 509, in _call_chain
    result = func(*args)
  File "D:\Python 36\lib\urllib\request.py", line 1389, in unknown_open
    raise URLError('unknown url type: %s' % type)
urllib.error.URLError: <urlopen error unknown url type: https>

最终修复：

import urllib
import datetime as dt
from datetime import datetime
import requests
import time
import re
import html2text

startTime = time.time()
def text_opener():
    linklist=[]
    with open ("test.txt", 'r', encoding = 'utf=8') as wikitxt :

         #content = wikitxt.read().splitlines()
        for i in wikitxt:
            try:
                linklist.append(i.strip())
            except UnicodeEncodeError as enror:
                linklist.append  ("")

    return linklist

linklist = text_opener() # put the content in a list and then opened the text

'''
This is a string of characters which I wanted to remove from the URL content

rejectedChar = list('!"#$%&\'()*+,-./:;<=>?@[\\]^_`{|}~0123456789')
rejectedChar.append("\t")
special="\t" 
regexWords = r"[\w']+"

'''


'''STOPWORDS LIST WHICH CONTAINS A BUNCH OF WORDS WHICH I DON"T NEED TO BE PRINTED--- ONLY FOR LARGE FILES
#stopwords = []
#with open('stopwords.txt', 'r', encoding='utf-8') as inFile:
 #   for i in inFile:
  #      stopwords.append(i.strip())
'''
content = ""
count = 0

for i in linklist:
    print(count,"   ",i.encode('utf-8'))
    count+=1
    try:
        f = urllib.request.urlopen(i).read()
        content+=str(f)
    except Exception as e:
        continue
#print((linklist[0:4000]).encode('utf-8'))

#combinedstops= rejectedChar+stopwords # combining them together

#for item in combinedstops:
    #content=content.replace(item,"") # now this items are removed from the 
#content

def output_file (content):
    with open('June_wikipedia_content.txt', 'w', encoding = 'utf-8') as output:
              output.write(str(content))

##    try:
##        output_file (content)
##    except UnicodeEncodeError as enror:
##        print ("Got lost in the game")
#sky=open("sky.txt",'w')
#sky.write(str(content))
output_file (content)

#print("hahahahahaha",stopwords)

#for i in content:
  #       i = re.findall(regexWords, i)
    #     i = [i for i in i if i in stopwords]


startTime = time.time()
endTime = time.time()
runTime = endTime - startTime
print("Runtime is %3f seconds" % runTime)

Answer 1

您可以使用以下函数打开文本文件并将所有链接存储在一个列表中：

with open('links.txt') as f:
    content = f.read().splitlines()

变量content 是一个列表，其中每个元素都包含与URL 关联的字符串。这仅在您的links.txt 具有逐行排列的 URL 时才有效，即：

www.google.co.in
www.wikipedia.co.in
www.youtube.co.in 

一旦你得到这个列表，你就可以用一个简单的 for 循环遍历它，然后做你想做的事情。

如果您想要更详细的答案，我建议发布链接的示例文本文件。

编辑：

这可行，但它会将整个数据转储到文件中。数据格式不正确。这是你需要的吗？

from urllib.request import urlopen
with open('links.txt') as f:
    content = f.read().splitlines()

with open('Wikipedia_content.txt', 'w') as output_file:
for link in content :
    output_file.write(link)
    output_file.write((urlopen(link).read()))