如何在Windows上手动启动fastCGI应用程序？

Question

我已经配置了一个Web服务器，通过命名管道使用'远程'fastCGI应用程序（它实际上在同一个Windows主机上）。我现在试图找出如何启动fastCGI应用程序来使用这个管道，但我不确定这应该如何完成。其他操作系统似乎有spawn-fcgi实用程序用于执行此操作，但对于Windows似乎没有任何类似的操作。

这是我的APP：

#include <stdio.h> 
#include "fcgi_stdio.h" 

int main(int argc, char ** argv) 
{ 
    while (FCGI_Accept() >= 0) { 
    printf("Content-type: text/html\r\n" 
     "\r\n" 
     "<title>Web Services Interface Module</title>" 
     "<h1>Web Services Interface Module</h1>\n"); 
    } 
    return(0); 
} 

出于兴趣我使用的深渊Web服务器，虽然我希望不会对答案的轴承。

问候

Answer 1

的FCGI界面不会让你这样做，而是使用FCGX接口。 调用FCGX_Open_Socket以在特定端口上侦听例如9345或命名管道。

FCGX_OpenSocket(":9345", 500); 

然后你不需要使用像spawn_fcgi这样的工具来启动你的应用程序。在fcgiapp.c

Answer 2

/* 
*---------------------------------------------------------------------- 
* 
* FCGX_OpenSocket -- 
* 
* Create a FastCGI listen socket. 
* 
* path is the Unix domain socket (named pipe for WinNT), or a colon 
* followed by a port number. e.g. "/tmp/fastcgi/mysocket", ":5000" 
* 
* backlog is the listen queue depth used in the listen() call. 
* 
* Returns the socket's file descriptor or -1 on error. 
* 
*---------------------------------------------------------------------- 
*/ 
DLLAPI int FCGX_OpenSocket(const char *path, int backlog); 

默认情况下libfcgi从标准输入读取。所以重新打开stdin句柄作为管道。

dup2(FCGX_OpenSocket("pipe name", 5),0); 

Answer 3

变化FCGX_Init

int FCGX_Init(void) 
{ 
    char *p; 

    int listen_socket; 

    if (libInitialized) { 
     return 0; 
    } 

    if (OS_LibInit(NULL) == -1) { 
     return OS_Errno ? OS_Errno : -9997; 
    } 

    /*sureone socket*/ 
     /* 9010 is your listen port*/ 
    listen_socket = FCGX_OpenSocket(":9010", 400); 
    if(listen_socket < 0) exit(1); 
    printf("FCGX_InitRequest...\n"); 
    FCGX_InitRequest(&the_request, listen_socket, 0); 
    /*end sureone*/ 

    //FCGX_InitRequest(&the_request, FCGI_LISTENSOCK_FILENO, 0); 



    p = getenv("FCGI_WEB_SERVER_ADDRS"); 
    webServerAddressList = p ? StringCopy(p) : NULL; 

    libInitialized = 1; 
    return 0; 
} 

Answer 4

我拿出刚在Windows工作对我来说代码：

int main() 
{ 
    char **initialEnv = environ; //Keep track of initial environment 
    int count = 0;   
    int listenSocket; 


    //It's ugly, but in Windows we need to initialize the 
    //socket library. We can do it by calling 
    //libfcgi's OS_LibInit() function 
    OS_LibInit(NULL); 

    //Open a socket. Here, we use localhost:9000 
    listenSocket = FCGX_OpenSocket("localhost:9000", 5); 
    if (listenSocket < 0) { 
     exit(1); 
    } 

    FCGX_Request request; 
    FCGX_Init(); 
    FCGX_InitRequest(&request, listenSocket, 0); 


    while (FCGX_Accept_r(&request) >= 0) { 
     //Init I/O streams wrapper as well as set the new environment 
     FCGI_stdin->stdio_stream = NULL; 
     FCGI_stdin->fcgx_stream = request.in; 
     FCGI_stdout->stdio_stream = NULL; 
     FCGI_stdout->fcgx_stream = request.out; 
     FCGI_stderr->stdio_stream = NULL; 
     FCGI_stderr->fcgx_stream = request.err; 
     environ = request.envp; 

     //Funny stuff 
     char *contentLength = getenv("CONTENT_LENGTH"); 
     int len; 

     printf("Content-type: text/html\r\n" 
       "\r\n" 
       "<title>FastCGI echo</title>" 
       "<h1>FastCGI echo</h1>\n" 
       "Request number %d, Process ID: %d<p>\n", ++count, getpid()); 

     if (contentLength != NULL) { 
      len = strtol(contentLength, NULL, 10); 
     } else { 
      len = 0; 
     } 

     if (len <= 0) { 
      printf("No data from standard input.<p>\n"); 
     } 
     else { 
      int i, ch; 

      printf("Standard input:<br>\n<pre>\n"); 
      for (i = 0; i < len; i++) { 
       if ((ch = getchar()) < 0) { 
        printf("Error: Not enough bytes received on standard input<p>\n"); 
        break; 
       } 
       putchar(ch); 
      } 
      printf("\n</pre><p>\n"); 
     } 

    } /* while */  

    return 0; 
}