2016-12-01 21:37 GMT+01:00 Igor Tandetnik <i...@tandetnik.org>:
> On 12/1/2016 1:57 PM, Cecil Westerhof wrote:
>>
>> At the moment I have the following code:
>> SELECT totalUsed, COUNT(*) AS Count
>> FROM tips
>> GROUP BY totalUsed
>>
>> This shows the total number of records for every value of totalUsed.
>> Would it be possible to get the total number of records also. (Sum of
>> all the Count's.)
>
>
> SELECT totalUsed, COUNT(*) AS Count FROM tips GROUP BY totalUsed
> union all
> SELECT null, COUNT(*) FROM tips;
Works likes a charm and is significant faster as the solution from
Nomad. But from him I took:
SELECT 'Total:', COUNT(*) FROM tips
Thanks, both.
--
Cecil Westerhof
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