Question

我正在尝试执行以下操作：

打开包含带有URL（GET-Requests）列表的CSV文件
阅读CSV文件并将条目写入列表
打开每个网址并阅读答案
将答案写回新的CSV文件

我收到以下错误：

Traceback (most recent call last):
  File "C:\Users\l.buinui\Desktop\request2.py", line 16, in <module>
    req = urllib2.urlopen(url)
  File "C:\Python27\lib\urllib2.py", line 127, in urlopen
    return _opener.open(url, data, timeout)
  File "C:\Python27\lib\urllib2.py", line 404, in open
    response = self._open(req, data)
  File "C:\Python27\lib\urllib2.py", line 427, in _open
    'unknown_open', req)
  File "C:\Python27\lib\urllib2.py", line 382, in _call_chain
    result = func(*args)
  File "C:\Python27\lib\urllib2.py", line 1247, in unknown_open
    raise URLError('unknown url type: %s' % type)
URLError: <urlopen error unknown url type: ['http>

以下是我正在使用的代码：

import urllib2
import urllib
import csv

# Open and read the source file and write entries to a list called link_list
source_file=open("source_new.csv", "rb")
source = csv.reader(source_file, delimiter=";")
link_list = [row for row in source]
source_file.close()

# Create an output file which contains the answer of the GET-Request
out=open("output.csv", "wb")
output = csv.writer(out, delimiter=";")
for row in link_list:
    url = str(row)
    req = urllib2.urlopen(url)
    output.writerow(req.read())

out.close()

那里出了什么问题？

提前感谢任何提示。

干杯

Answer 1

使用row变量会将一个list元素（它只包含一个元素，url）传递给urlopen，但是传递row [0]会传递包含url的字符串。

csv.reader返回它读取的每一行的列表，无论该行中有多少项。

Answer 2

现在正在运作。如果我在循环中直接引用row[0]，则没有问题。

import urllib2
import urllib
import csv

# Open and read the source file and write entries to a list called link_list
source_file=open("source.csv", "rb")
source = csv.reader(source_file, delimiter=";")
link_list = [row for row in source]
source_file.close()

# Create an output file which contains the answer of the GET-Request
out=open("output.csv", "wb")
output = csv.writer(out)
for row in link_list:
    req = urllib2.urlopen(row[0])
    answer = req.read()
    output.writerow([answer])


out.close()

使用urllib2.urlopen（）的未知网址类型

2 个答案: