Question

我正在尝试在MySQL中创建一个UDF函数。

我想根据输入参数返回7位十进制/双精度。

我在MySQL中安装了以下函数：

CREATE FUNCTION postrank_udf RETURNS REAL SONAME 'postrank.so'

我的功能选择postrank（120,0,1444763428）的正确结果应为13.8604775。但是，它现在返回14，除非我做postrank（120,0,1444763428.0000000）。下面是我的UDF功能。有什么想法吗？

   #ifdef STANDARD
   /* STANDARD is defined, don't use any mysql functions */
   #include <stdlib.h>
   #include <stdio.h>
   #include <string.h>
   #ifdef __WIN__
   typedef unsigned __int64 ulonglong;/* Microsofts 64 bit types */
   typedef __int64 longlong;
   #else
   typedef unsigned long long ulonglong;
   typedef long long longlong;
   #endif /*__WIN__*/
   #else
   #include <my_global.h>
   #include <my_sys.h>
   #if defined(MYSQL_SERVER)
   #include <m_string.h>/* To get strmov() */
   #else
   /* when compiled as standalone */
   #include <string.h>
   #define strmov(a,b) stpcpy(a,b)
   #define bzero(a,b) memset(a,0,b)
   #define memcpy_fixed(a,b,c) memcpy(a,b,c)
   #endif
   #endif
   #include <mysql.h>
   #include <ctype.h>

   #ifdef HAVE_DLOPEN

   #if !defined(HAVE_GETHOSTBYADDR_R) || !defined(HAVE_SOLARIS_STYLE_GETHOST)
   static pthread_mutex_t LOCK_hostname;
   #endif

   #include <math.h>

   my_bool postrank_init(UDF_INIT *initid, UDF_ARGS *args, char *message);
   void postrank_deinit(UDF_INIT *initid __attribute__((unused)));
   double postrank(UDF_INIT* initid, UDF_ARGS* args __attribute__((unused)),
                     char* is_null __attribute__((unused)), char* error __attribute__((unused)));

   my_bool postrank_init(UDF_INIT *initid, UDF_ARGS *args, char *message)
   {
     if(!(args->arg_count == 3)) {
       strcpy(message, "Expected three arguments");
       return 1;
     }

     args->arg_type[0] = REAL_RESULT;
     args->arg_type[1] = REAL_RESULT;
     args->arg_type[2] = REAL_RESULT;

     return 0;
   }

   void postrank_deinit(UDF_INIT *initid __attribute__((unused)))
   {

   }

   double postrank(UDF_INIT* initid, UDF_ARGS* args __attribute__((unused)),
                     char* is_null __attribute__((unused)), char* error __attribute__((unused)))
   {
     double ups = *((double *)(args->args[0]));
     double downs = *((double *)(args->args[1]));
     double d = *((double *)(args->args[2]));

       double final = 0.00;
       double my_sign = 1.0;

       if(ups - downs > 0){
        my_sign = 1.0;
       }else if(ups - downs < 0){
        my_sign = -1.0;
       }else{
        my_sign = 0.0;
       }

       int myups = (int) ups;
       int mydowns = (int) downs;

       int my_max = 1;
       if(abs(myups-mydowns) > 1){
        my_max = abs(myups-mydowns);
       }

    double dbmy_max = (double) my_max;

    double result = log10(dbmy_max) * my_sign + ((d - 1444000000.0) / 64800.0);

    final = round( 10000000.0 * result ) / 10000000.0;

    return final;
   }

   #endif /* HAVE_DLOPEN */

Answer 1

您的代码没问题，请尝试这个小调试代码，然后打印您所期望的内容。

#include <usual.h> 
int main(  )
{
  //  int ups = *((int *)(args->args[0]));
  //  int downs = *((int *)(args->args[1]));
  //  double d = *((double *)(args->args[2])); 
  int ups = 120;
  int downs = 0;
  double d = 1444763428;
  double debugd = 0.00; 
  double my_sign = 1.0; 
  if ( ups - downs > 0 )
  {
    my_sign = 1.0;
  }
  else if ( ups - downs < 0 )
  {
    my_sign = -1.0;
  }
  else
  {
    my_sign = 0.0;
  } 
  int my_max = 1;
  if ( abs( ups - downs ) > 1 )
  {
    my_max = abs( ups - downs );
  } 
  double dbmy_max = ( double ) my_max; 
  double result =
    log10( dbmy_max ) * my_sign + ( d - 1444000000.0 ) / 64800.0; 
  //  return round(10000000.0 * result) / 10000000.0;
  debugd = round( 10000000.0 * result ) / 10000000.0;
  printf( "\n debugd == %5.7f ", debugd ); 
}

MySQL UDF只返回整数 - 想要双倍

1 个答案: