cgi 下载文件解决方法 _CGI

cgi 下载文件
web是这样写的：
<html>
<head>
<title>Test</title>
</head>
<body>
<form action="cgi-bin/PostTest.cgi" method="POST">
<input type="file" name="file" value="" />
<input type="submit" value="Submit" />
</form>
</body>
</html>

我想点击Submit之后PostTest.cgi 可以接受到上传的文件，
不知道用c 怎么处理。可以使用libcgi这个库。

我的代码是这样的：可是无法实现。

C/C++ code


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

int main()
{
    cgi_init();
    cgi_process_form();
    cgi_init_headers();

    puts(""
        "<!DOCTYPE html PUBLIC '-//W3C//DTD HTML 4.01 Transitional//EN'>"
        "<html>"
        "<head>"
        "       "
         " <meta http-equiv='content-type' content='text/html; charset=ISO-8859-1'>"
        "       "
         "   <meta name='author' content='Rafael Steil'>"
          "    <title>LIBCGI Examples</title>"
           "   </head>"
           "  <body text='#000000' bgcolor='#ffffff' link='#0000ee' vlink='#551a8b' alink='#0000ee'>"
        "");
     if (cgi_param("name"))
     {
                printf("name: %s<br>", cgi_param("name"));
     }
     else
     {
                puts("name: Empty<br>");
     }

     if (cgi_param("number"))
     {
                printf("number: %s<br>", cgi_param("number"));
     }
     else
     {
                puts("number: Empty<br>");
     }
 
     //char **lines;
     //unsigned int total, i;
    // printf("Content-Type:   application/xml\n\n");
     printf("Content-Disposition:   attachment;   filename=DOWNLOAD.xml\n\n");
    printf("file name = : %s<br>", cgi_param("file"));
    FILE *fp = fopen(cgi_param("file"), "rb");

     if(fp != NULL)
     {
        printf("Open file [%s] Successfully<br>", cgi_param("file"));
     }
     else
     {
        printf("Open file [%s] Erroy<br>", cgi_param("file"));
     }



     puts(""
        "</body>"
        "</html>"
        "");

        cgi_end();

        return 0;

}

------解决方案--------------------
cgi.h 和 cgi.cpp 文件能给我看看吗》？
------解决方案--------------------
cgi程序如果在vs6.0里面的话，服务器用apache运行cgi程序需要配置哪些东西呀？