csv - 如何将 CSV 列表中的 url 传递到 python GET 请求中

Question

我有一个 CSV 文件，其中包含 Google 扩展程序 ID 列表。
我正在编写一段代码，用于读取扩展 ID、添加网上商店 url，然后执行基本的获取请求:

import csv
import requests

with open('small.csv', 'rb') as f:
    reader = csv.reader(f)
    for row in reader:
        urls = "https://chrome.google.com/webstore/detail/" + row[0]
        print urls
        r = requests.get([urls])

运行此代码会产生以下回溯:

Traceback (most recent call last):
  File "C:\Users\tom\Dropbox\Python\panya\test.py", line 9, in <module>
    r = requests.get([urls])
  File "C:\Python27\lib\site-packages\requests\api.py", line 69, in get
    return request('get', url, params=params, **kwargs)
  File "C:\Python27\lib\site-packages\requests\api.py", line 50, in request
    response = session.request(method=method, url=url, **kwargs)
  File "C:\Python27\lib\site-packages\requests\sessions.py", line 465, in request
    resp = self.send(prep, **send_kwargs)
  File "C:\Python27\lib\site-packages\requests\sessions.py", line 567, in send
    adapter = self.get_adapter(url=request.url)
  File "C:\Python27\lib\site-packages\requests\sessions.py", line 641, in get_adapter
    raise InvalidSchema("No connection adapters were found for '%s'" % url)
InvalidSchema: No connection adapters were found for '['https://chrome.google.com/webstore/detail/blpcfgokakmgnkcojhhkbfbldkacnbeo']'

如何修改代码，以便它接受列表中的网址，并发出 GET 请求？

Answer 1

requests.get 需要一个字符串，但您正在创建并传递一个列表 [urls]

r = requests.get([urls])

将其更改为仅

r = requests.get(urls)

它应该可以工作。