Why might HttpOpenRequest fail with error 122?

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轻奢々
轻奢々 2020-12-20 02:52

The following code

fRequestHandle = HttpOpenRequestA(
                   fConnectHandle, 
                   \"POST\", url.c_str(), 
                   NULL,         


        
3条回答
  •  不知归路
    2020-12-20 03:22

    EDIT:

    Looks like this guy had the same problem, the URL is too long.

    http://social.msdn.microsoft.com/Forums/en-US/windowsmobiledev/thread/68612c89-bbce-4d88-926d-5d76771be944

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