无法访问 Tastekid 的 API。它说错误：403（请求被禁止）

Question

我无法运行代码来读取 Tastekid 的 API (python)，它显示错误：403（禁止请求）但是我可以访问与浏览器中的 url 相同的内容。 钥匙也试过了。

请在查询下方找到

from urllib2 import Request, urlopen, URLError
req = Request('http://www.tastekid.com/api/similar?q=red+hot+chili+peppers%2C+pulp+fiction')
try:
    response = urlopen(req)
except URLError as e:
    if hasattr(e, 'reason'):
        print 'We failed to reach a server.'
        print 'Reason: ', e.reason
    elif hasattr(e, 'code'):
        print 'The server couldn\'t fulfill the request.'
        print 'Error code: ', e.code
else:
    'everything is fine'

Answer 1

在向 url 发出请求之前，您需要添加正确的headers

headers = { 'User-Agent' : "Mozilla/4.0 (compatible; MSIE 5.5; Windows NT)" }    
req = Request('http://www.tastekid.com/api/similar?q=red+hot+chili+peppers%2C+pulp+fiction', headers = headers)

---

这里设置了标头中的User-Agent 键，以便告诉服务器我们正在从浏览器而不是程序发出请求

测试

• 文件名test.py

  from urllib2 import Request, urlopen, URLError
  
  #Changes made in the below two line
  
  headers = { 'User-Agent' : 'Mozilla/4.0 (compatible; MSIE 5.5; Windows NT)' }
  req = Request('http://www.tastekid.com/api/similar?q=red+hot+chili+peppers%2C+pulp+fiction', headers = headers)
  
  
  try:
      response = urlopen(req)
  except URLError as e:
      if hasattr(e, 'reason'):
          print 'We failed to reach a server.'
          print 'Reason: ', e.reason
      elif hasattr(e, 'code'):
          print 'The server couldn\'t fulfill the request.'
          print 'Error code: ', e.code
  else:
      print 'everything is fine'
  

• 没有标题

  $ python test.py 
  We failed to reach a server.
  Reason:  Forbidde
  

• 带有标题

  $ python test.py 
  everything is fine