使用 asyncio/aiohttp 返回 404 响应时遇到问题

Question

import time
import asyncio
import aiohttp
async def is_name_available(s, name):
async with s.get("https://twitter.com/%s" % name) as res:
if res.raise_for_status == 404:
print('%s is available!' % name)
return name
async def check_all_names(names):
async with aiohttp.ClientSession(raise_for_status=True) as s:
tasks = []
for name in names:
task = asyncio.create_task(is_name_available(s, name))
tasks.append(task)
return await asyncio.gather(*tasks)
def main():    
with open('names.txt') as in_file, open('available.txt', 'w') as out_file:        
names = [name.strip() for name in in_file]
start_time = time.time()
results = asyncio.get_event_loop().run_until_complete(check_all_names(names))
results = [i for i in results if i]
out_file.write('n'.join(results))
print(f'[ <? ] Checked {len(names)} words in {round(time.time()-start_time, 2)} second(s)')
if __name__ == '__main__':
main()

我似乎无法弄清楚如何使用我从我的另一个项目中使用的这种 asyncio/aiohttp 结构is_name_available只返回 404'd 链接。我是python的初学者，任何帮助都值得赞赏。

Answer 1

此行不正确：

if res.raise_for_status == 404:

raise_for_status是一个方法，所以你应该调用它，而不是将其与数字进行比较(数字将始终返回 false(。在您的情况下，您不想首先调用raise_for_status，因为您不想在遇到 404 时引发异常，而是检测到它。要检测 404，您可以简单地编写：

if res.status == 404:

另请注意，您不想指定raise_for_status=True，因为它会在if有机会运行之前引发 404 异常。