On 19-Nov-08, at 8:42 AM, Ken Ramsay wrote:
>
> Thanks guys,
>
> MySQL defines the field as a DATETIME , so the return type is time_t.
No; you are using an expression, not the column, so the result type
is indeed BIGINT, as Markus surmised (his explanation was correct, as
far as I can determ
Quoting Ken Ramsay <[EMAIL PROTECTED]>:
>
> Thanks guys,
>
> MySQL defines the field as a DATETIME , so the return type is time_t. As
> far as I could find out, there is no helper funtion to convert this into
> secs from 1 Jan 1970 , so I got around it by using the difftime() call
> and subtractin
Thanks guys,
MySQL defines the field as a DATETIME , so the return type is time_t. As
far as I could find out, there is no helper funtion to convert this into
secs from 1 Jan 1970 , so I got around it by using the difftime() call
and subtracting the returned value from 1/1/1970. difftime() retur
