我想要做的是从邮政编码中检索城市和州.这是我到目前为止所拥有的:

def find_city(zip_code):
    zip_code = str(zip_code)
    url = 'http://www.unitedstateszipcodes.org/' + zip_code
    source_code = requests.get(url)
    plain_text = source_code.text
    index = plain_text.find(">")
    soup = BeautifulSoup(plain_text, "lxml")
    stuff = soup.findAll('div', {'class': 'col-xs-12 col-sm-6 col-md-12'})

我也尝试使用id ="zip-links",但这不起作用.但事情就是这样:当我跑步时,print(plain_text)我得到以下内容:

Forbidden
You don't have permission to access /80123
on this server.

所以我想我的问题是:是否有更好的方法从邮政编码获得城市和州？或者有一个原因,unitedstateszipcodes.gov不合作.毕竟,很容易看到源,标签和文本.谢谢

您需要添加用户代理:

headers = {"User-agent":"Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/47.0.2526.80 Safari/537.36"}
def find_city(zip_code):
    zip_code = str(zip_code)
    url = 'http://www.unitedstateszipcodes.org/' + zip_code
    source_code = requests.get(url,headers=headers)

完成后,响应为200,您将获得源:

In [8]:  url = 'http://www.unitedstateszipcodes.org/54115'

In [9]: headers = {"User-agent":"Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/47.0.2526.80 Safari/537.36"}

In [10]:  url = 'http://www.unitedstateszipcodes.org/54115'
In [11]: source_code = requests.get(url,headers=headers)
In [12]: source_code.status_code
Out[12]: 200

如果您想要详细信息,可以轻松解析:

In [59]:  soup = BeautifulSoup(plain_text, "lxml")

In [60]: soup.find('div', id='zip-links').h3.text
Out[60]: 'ZIP Code: 54115'

In [61]: soup.find('div', id='zip-links').h3.next_sibling.strip()
Out[61]: 'De Pere, WI 54115'

In [62]:  url = 'http://www.unitedstateszipcodes.org/90210'

In [63]: source_code = requests.get(url,headers=headers).text

In [64]:  soup = BeautifulSoup(source_code, "lxml")

In [65]: soup.find('div', id='zip-links').h3.text
Out[66]: 'ZIP Code: 90210'

In [70]: soup.find('div', id='zip-links').h3.next_sibling.strip()
Out[70]: 'Beverly Hills, CA 90210'

您还可以将每个结果存储在数据库中,并首先尝试在数据库中进行查找.