问题描述
我的python cgi如何返回特定的http状态代码，例如403或418？
How can my python cgi return a specific http status code, such as 403 or 418?
我尝试了显而易见的（打印状态：403 Forbidden），但它不起作用。
I tried the obvious (print "Status:403 Forbidden") but it doesn't work.
推荐答案
print 'Status: 403 Forbidden'
print
适合我。您确实需要第二次打印，因为您需要一个双换行来结束HTTP响应标头。否则，您的网络服务器可能会抱怨您没有向它发送一套完整的标题。
Works for me. You do need the second print though, as you need a double-newline to end the HTTP response headers. Otherwise your web server may complain you aren't sending it a complete set of headers.
sys.stdout('Status: 403 Forbidden\r\n\r\n')
在技术上可能更正确，根据RFC（假设您的CGI脚本未在Windows上以文本模式运行）。然而，两个行结尾似乎无处不在。
may be technically more correct, according to RFC (assuming that your CGI script isn't running in text mode on Windows). However both line endings seem to work everywhere.
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