如何使用python在域内请求职业页面url

Question

import requests
s = ['jobs','careers','opportunities']
u = ['yahoo.com','statestreet.com']
f = []

for i in u:
    for j in s:
        w = "http://{}/{}".format(i,j)
        print w
        r = requests.get(w)
        print r.status_code
            if(r.status_code == 200):
                f.append(w)
                break
        print f

此代码适用于大多数网站，但不适用于www.surveymonkey.com.

Answer 1

您没有说明什么不起作用，或者您希望代码做什么。但是，代码中最后的if语句中的缩进似乎不正确

纠正这一点（并编辑变量名，使其更具描述性，如下所示）至少可以像我预期的那样运行：

import requests

paths = ['jobs', 'careers', 'opportunities']
domains = ['yahoo.com', 'statestreet.com', 'surveymonkey.com']
results = []

for domain in domains:
    for path in paths:
        url = "http://{}/{}".format(domain, path)
        print url
        r = requests.get(url)
        print r.status_code
        if r.status_code == requests.codes.OK:
            results.append(url)
            break
print results