如何在python的字典中的键中存储多个值？

Question

Maybe my question title is different from the question content.

statuscode = []
statuscode.append(200) 

for x in find_from_sublister(hostname):
    x2 = x.strip()
    url = "http://"+ x2
    try:
        req = requests.get(url)
        req1 = str(req.status_code) + " " + str(url) + '\n'
        req2 = str(req.status_code)
        req3 = str(url)
        dict = {req2 : req3}

        print " \n " + str(req1)

    except requests.exceptions.RequestException as e:
        print "Can't make the request to this Subdomain " + str(url) + '\n'

for keys, values in dict.iteritems():
    print "finding the url's whose status code is 200"

    if keys == statuscode[0]:
        print values
    else:
        print "po"

I am using the code to do some following stuffs.

1. It will find the subdomains from the Sub-lister (Locally)
2. Then it will go to find the status code of each subdomain which was find by the sublister, with the help of for loop . for x in find_from_sublister(hostname):

Note: find_from_sublister(hostname) is a function for finding the subdomain from the sublister.

3. Then print the status code with the URL. Here print " \\n " + str(req1)

[All goes well, but the problem starts here ]

Now what I want is to seperate the URLs which have 200 status code.

so I heard then it can happen by using the dictionary in python. So, I try to use the dictionary. As you can see dict = {req2 : req3}

Now I also make a list of index which have value 200

Then I compare the keys to list of index. here keys == statuscode[0]

and if they match then it should print all the URL's which have 200 status code.

But the result I am getting is below,

finding the url's whose status code is 200 
po 

You see the value po its else statment value,

else:
   print "po"

Now the problem is Why I am getting this else value why not the URL's which have status code 200?

Hope I explained you clearly. And waiting for someone who talk to me on this.

Thanks

Note: I am Using Python 2.7 ver.

Answer 1

In your case, you don't even need the dictionary.

I've tried to clean up the code as much as possible, including more descriptive variable names (also, it's a bad idea to override builtin names like dict ).

urls_returning_200 = []

for url in find_from_sublister(hostname):
    url = "http://" + url.strip()
    try:
        response = requests.get(url)
        if response.status_code == 200:
            urls_returning_200.append(url)
        print " \n {} {}\n".format(response.status_code, url)
    except requests.exceptions.RequestException as e:
        print "Can't make the request to this Subdomain {}\n".format(url)

print "finding the url's whose status code is 200"
print urls_returning_200