Re: [HACKERS] how can i get the binary format of timestamp?

[[EMAIL PROTECTED]]

Hello,Tom！
   Thanks for your answer.
   To invoid missing something of the timestamp, i just copy the tm2timestamp 
func into my program.(Of course, some other related functions are copied, too), 
result is still refusing me.

What i need is: 41A7 7DBA D400 0000 stands for "2006-03-30 18:18:18" ( 41A7 
7DBA D400 0000 is got from the Copy to File)
When using tm2timestamp, result is:
if define HAVE_INT64_TIMESTAMP
result is:  4E66 E642 0030 C274
if not define HAVE_INT64_TIMESTAMP
{
  if define INT64_IS_BUSTED result is: 6972 2142 0000 00DB
  if not define INT64_IS_BUSTED result is: 4221 7269 DB00 0000
}

So, maybe  I forget to define something??

/******************************************************************************************************
* my codes 
*
*/
my program is listed:
time_t time = 1143713898;
struct tm* tmField = gmtime(&time);    // I know that struct tm dislikes the 
struct tm in "time.h"
tmField->tm_year += 1900;
tmField->tm_mon += 1;                  // So tmFiled is same as pg_tm
int tzp = 8;
char buffer[256];

timestamp timestampField ;
tm2timestamp(tmField,0,&tzp,&timestampField);
timestamp_send(timestampField,buffer);   // I changed some codes here, to get 
the binary format of timestamp 

/*********************************************************************************************************/
     

so ,what switch should i turn on?

======= 2006-04-18 21:47:42 You Wrote：=======

>"[EMAIL PROTECTED]" <[EMAIL PROTECTED]> writes:
>> timestamp is defined as int64 or float8, when it is defined as int64, it 
>> looks like timeval, first 32 bits for seconds,second 32 bits for usec.
>
>No, it's seconds times 1000000, not times 2^32.
>
>                       regards, tom lane
>
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