如何将404个错误代码分开与Python中的200分

Question

这是我的代码，我要做的是分开400个响应状态代码子域。如您所见，我在子域中提供了三个URL 索引列表。具有不同的状态代码

现在我要做的是将404个错误状态代码分开。

statuscode = []
statuscode.append(400) 
statuscode.append(403)
statuscode.append(404)
 subdomains = []
    subdomains.append("https://teyrtguhigkfjn.s3.amazonaws.com/")
    subdomains.append("http://google.com")
    subdomains.append("https://lasdfgfldsakjas.s3.amazonaws.com/")

    for x in subdomains:
        url =  x

        try:
            req = requests.get(url)
            req1 = str(req.status_code) + " " + str(url) + 'n'
            req2 = str(req.status_code)
            req3 = str(url)
            print "n" + str(req1)
            except requests.exceptions.RequestException as e:
            print "Can't make the request to this Subdomain " + str(url) + 'n'

    if statuscode in str(req1):
        print "nTrying to Collect the URL's whose status is 400, 400, 403"
print str(req2) + ' ' + str(req3)

，但我没有成功。请查看问题，我想我知道在str(req1(中的statuscode的这一行中的问题在哪里。我的猜测

希望您能解决问题。

谢谢

Answer 1

import requests
statuscode = [400, 403, 404]
subdomains = ["https://teyrtguhigkfjn.s3.amazonaws.com/", "http://google.com", "https://lasdfgfldsakjas.s3.amazonaws.com/"]
for url in subdomains:
    try:
        req = requests.get(url)
        req1 = str(req.status_code) + " " + str(url) + 'n'
        req2 = str(req.status_code)
        print "n" + str(req1)
        if req.status_code in statuscode:             #----->Update
            print "nTrying to Collect the URL's whose status is 400, 400, 403"
            print str(req2) + ' ' + url
    except requests.exceptions.RequestException as e:
        print "Can't make the request to this Subdomain " + str(url) + 'n'