Question

我成功编译了最初在AIX上编译的64位Linux上的一些UDF

Linux正在运行IBM DB2 v9.5

这是有时为某些值返回的函数

#include <cstdlib>
#include <cstdio>
#include <iostream>
#include "cbudf.hpp"


using namespace std;

extern "C"
void SQL_API_FN cbzdt(SQLUDF_VARCHAR_FBD *pArg,
                      SQLUDF_VARCHAR *pResult,
                      SQLUDF_NULLIND *pArgInd,
                      SQLUDF_NULLIND *pResultInd,
                      SQLUDF_CHAR *pSqlState,
                      SQLUDF_VARCHAR *pFuncName,
                      SQLUDF_VARCHAR *pSpecName,
                      SQLUDF_VARCHAR *pMsgTxt)
{
short i;
unsigned long x = 0;

cout << "entered cbzdt";

if (pArg->length < CM_UDF_ZDT_SIZE){
   *pResultInd = CM_UDF_IS_NULL_IND;
   cout << "pArg length is:" << pArg->length << " and that is smaller than CM_UDF_ZDT_SIZE: " << CM_UDF_ZDT_SIZE;
   }
else
   {
   cout << "entered else";
   for (i = 0; i < 4; i++)
      x = (x << 8) | pArg->data[i];
   if (x & 0xff000000){
      *pResultInd = CM_UDF_IS_NULL_IND;
      cout << "x mask is true.. making null";
      }
   else
      {
      cout << "entered nested else";
      *pResultInd = CM_UDF_NOT_NULL_IND;
      CtDate zDate(x);
//      strcpy(pResult, (const char *)(zDate.asString("%Y-%m-%d")));
      strncpy(pResult, (const char *)(zDate.asString("%Y-%m-%d")),11);
      if (pResult[0] == '3'){             // if year is 3000
         pResult[9] = '1';               // fix day
         cout << "pResult =1";
         }
      }
   }
}

这是数据：

COID_                      TIMESTAMP_                 CHANGE_DATE                
-------------------------- -------------------------- ---------------------------
1814              2010-02-17-15.44.22.995784 x'0025780F000000000804011B'
1814              2010-02-17-15.44.55.583734 x'0025780F000000000804011B'
1814              2010-02-17-15.46.09.929014 x'0025780F000000000804011B'
1324              2009-12-02-21.31.13.058337 x'0025769F000000000804011B'
1324              2009-12-02-21.32.47.387150 x'0025769F000000000804011B'
1324              2009-12-02-21.34.03.020405 x'0025769F000000000804011B'
1324              2009-12-08-19.54.13.492488 x'002576A1000000000804011B'
1324              2009-12-08-19.55.22.041794 x'002576A1000000000804011B'
1358              2009-12-09-15.14.05.593887 x'002576A1000000000804011B'



After function applied:

COID_                      TIMESTAMP_                 3         
-------------------------- -------------------------- ----------
1814              2010-02-17-15.44.22.995784 2011-01-05
1814              2010-02-17-15.44.55.583734 2011-01-05
1814              2010-02-17-15.46.09.929014 2011-01-05
1324              2009-12-02-21.31.13.058337 -         
1324              2009-12-02-21.32.47.387150 -         
1324              2009-12-02-21.34.03.020405 -         
1324              2009-12-08-19.54.13.492488 -         
1324              2009-12-08-19.55.22.041794 -         
1358              2009-12-09-15.14.05.593887 -

有人有什么想法吗？

我的猜测是这个if语句：

if (x & 0xff000000){
          *pResultInd = CM_UDF_IS_NULL_IND;

在某些值上返回true。它可能是与64位相关的东西吗？

此外，couts不会在DB2输出中打印

由于

Answer 1

将二进制zDate的一部分转换为可用日期的规则是什么？您可能在SQL中提出了等效的转换函数，并完全避免使用外部编译的UDF。 HEX函数将二进制值转换为十六进制字符串，然后您可以使用SUBSTR迭代以构建整数值。

至于cout消息，我不认为cout在外部UDF中会有多大用处。您可以更好地将调试信息保存到本地文件中。

将DB2 UDF从AIX迁移到Linux（RHEL）

1 个答案: