模块请求，如何得到简单的打印URL和响应？

Question

任何人都可以解释一下，我如何才能得到简单的url和回复结果？我尝试了这么多次，但什么也没有，因为现在我可以打印，就像：

50
0.4110674999999999
........, [<Response [200]>], [<Response [200]>], [<Response [200]>]]
[......, ['http://example.com.com/catalogue/page-48.html'], ['http://example.com.com/catalogue/page-49.html'], ['http://example.com.com/catalogue/page-50.html']]

我需要像

<Response [200]>
https://example.com/

非常感谢。

Ps。另外，为什么在安装模块grequests之后，我在控制台上得到了这条消息？

C:\P3\lib\site-packages\grequests.py:22: MonkeyPatchWarning: Monkey-patching ssl after ssl has already been imported may lead to errors, including RecursionError on Python 3.6. It may also silently lead to incorrect behaviour on Python 3.7. Please monkey-patch earlier. See https://github.com/gevent/gevent/issues/1016. Modules that had direct imports (NOT patched): ['urllib3.util.ssl_ (C:\\P3\\lib\\site-packages\\urllib3\\util\\ssl_.py)', 'urllib3.util (C:\\P3\\lib\\site-packages\\urllib3\\util\\__init__.py)'].
  curious_george.patch_all(thread=False, select=False)

我怎么才能修好它？卸载完整的python，安装补丁还是什么？谢谢!

import grequests
from bs4 import BeautifulSoup
import time

def get_urls():
    urls = []
    for x in range(1,51):
        urls.append(f'http://books.toscrape.com/catalogue/page-{x}.html')
    return urls

def get_data(urls):
    reqs = [grequests.get(link) for link in urls]
    resp = grequests.map(reqs)
    return resp


if __name__ == '__main__':
    start = time.perf_counter()
    urls = get_urls()
    url = len(get_urls())
    resp = get_data(urls)
    respo = len(get_data(urls))
    fin = time.perf_counter() - start

    resp_list = resp
    chunked_resp = list()
    chunk_size = respo
    urls_list = urls
    chunked_url = list()
    chunk_size = url

    print(urls)
    print(url)
    print(resp)
    print(respo)
    print(fin)  

    resp_list = resp
    chunked_resp = list()
        chunk_size = 1


    for i in range(0, len(resp_list), chunk_size):
        chunked_resp.append(resp_list[i:i+chunk_size])
    print(chunked_resp) 

    urls_list = urls
    chunked_url = list()
        chunk_size = 1


    for i in range(0, len(urls_list), chunk_size):
        chunked_url.append(urls_list[i:i+chunk_size])
    print(chunked_url)      

Answer 1

好的，我只能为打印url找到一个解决方案：

def get_data(urls):
    reqs = [grequests.get(link) for link in urls]
    resp = grequests.get(reqs)
    return resp
    
    
if __name__ == '__main__':
    start = time.perf_counter()
    urls = get_urls()
    resp = get_data(urls)
    resp = '\n'.join(resp)
    url = '\n'.join(urls)

http://books.toscrape.com/catalogue/page-48.html
http://books.toscrape.com/catalogue/page-49.html
http://books.toscrape.com/catalogue/page-50.html

     resp = '\n'.join(resp)
TypeError: can only join an iterable

但是对于resp，我得到了TypeError:只能加入一个可迭代的

Ps。我开始学习python最多一个月..。:(