无法读取带有URL的csv以在python中对其进行web刮取

我是python新手，所以我尝试了VisualStudio和Windows7

import csv
from bs4 import BeautifulSoup 
import requests 

contents = []
with open('websupplies.csv','r') as csvf: # Open file in read mode
   urls = csv.reader(csvf)

   for url in urls:
      contents.append(url) # Add each url to list contents


for url in contents:  # Parse through each url in the list.
   page = requests.get(url).content
   soup = BeautifulSoup(page, "html.parser")

   price = soup.find('span', attrs={'itemprop':'price'})
   availability = soup.find('div', attrs={'class':'product-availability'})

https://www.websupplies.gr/epeksergastis-intel-core-i5-8400-9mb-2-80ghz-bx80684i58400
https://www.websupplies.gr/epeksergastis-intel-celeron-g3930-2mb-2-90ghz-bx80677g3930
https://www.websupplies.gr/epeksergastis-amd-a6-9500-bristol-ridge-dual-core-3-5ghz-socket-am4-65w-ad9500agabbox

但我得到-找不到..'的连接适配器['a url']'

为什么?

csv的结构如下所示

import csv
from bs4 import BeautifulSoup 
import requests 

contents = []
with open('websupplies.csv','r') as csvf: # Open file in read mode
   urls = csv.reader(csvf)

   for url in urls:
      contents.append(url) # Add each url to list contents


for url in contents:  # Parse through each url in the list.
   page = requests.get(url).content
   soup = BeautifulSoup(page, "html.parser")

   price = soup.find('span', attrs={'itemprop':'price'})
   availability = soup.find('div', attrs={'class':'product-availability'})

https://www.websupplies.gr/epeksergastis-intel-core-i5-8400-9mb-2-80ghz-bx80684i58400
https://www.websupplies.gr/epeksergastis-intel-celeron-g3930-2mb-2-90ghz-bx80677g3930
https://www.websupplies.gr/epeksergastis-amd-a6-9500-bristol-ridge-dual-core-3-5ghz-socket-am4-65w-ad9500agabbox

---

他们最后没有一个半列

有问题，它说请求需要http方案，也许这就是问题所在？当您读取文件中的行时，还必须删除/n。您的文件是一个简单的URL列表。它不是真正的CSV

CSV读取器将每一行读入自己的列表。因此，加载数据的结构将是：

[
  ["https://www.websupplies.gr/epeksergastis-intel-core-i5-8400-9mb-2-80ghz-bx80684i58400"],
  ["https://www.websupplies.gr/epeksergastis-intel-celeron-g3930-2mb-2-90ghz-bx80677g3930"],
  ["https://www.websupplies.gr/epeksergastis-amd-a6-9500-bristol-ridge-dual-core-3-5ghz-socket-am4-65w-ad9500agabbox"],
]

解决此问题的一种方法是使用

url[0]

作为

请求的参数。get

，但真正正确的解决方法是根本不使用CSV。由于每行只有一条数据，您可以直接读取数据并将其传递给请求：

with open('websupplies.csv','r') as csvf: # Open file in read mode 
   for line in csvf:
      contents.append(line.strip('\n')) # Add each url to list contents

---

你能展示你的CSV的结构吗？可能是@Mad my URL的重复，他们有http协议。在你的回答中有一个问题，看起来像是一条评论。我知道，但我没有足够的声誉，我会写一条评论。我没有以txt文件的形式回复你是说？