Question

我无法运行代码来读取Tastekid的API（python），它说错误：403（请求禁止）但是我可以在浏览器中访问相同的URL。用钥匙也试过了。

请在下面查询

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

在向网址

发出请求之前，您需要添加适当的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)

此处设置标题中的User-Agent键，以便告诉服务器我们是从浏览器而不是程序发出请求

<强>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

带标题
```
$ python test.py 
everything is fine
```

无法访问Tastekid的API。它说错误：403（请求被禁止）

1 个答案: