>>>>Hi, Is is possible to use an sql statement(s) to count and number duplicates? The original table looks something like this, showing an excerpt where there are three entries for one date:
>>>
>>>

>>>SELECT * 
>>>    FROM ch5 
>>>    WHERE mdate IN (
>>>        SELECT mdate FROM ch5 ;
>>>        GROUP BY mdate ;
>>>        HAVING COUNT(mdate) > 1)
>>>

>>>
>>>>

>>>>Mdate            Vs
>>>>05/28/2013     v1
>>>>05/28/2013     v1
>>>>05/28/2013     v1
>>>>

>>>>
>>>>The following is a start in that it will give me a table that shows how many duplicates exist for each date.
>>>>
>>>>

>>>>SELECT mdate,COUNT(mdate) FROM ch5 ;
>>>>    INTO TABLE dups;
>>>>    GROUP BY mdate ;
>>>>    HAVING COUNT(mdate) > 1
>>>>

>>>>
>>>>Here is a part of the report that the above produces:
>>>>
>>>>

>>>>mdate         cnt_mdate
>>>>05/28/2013   3
>>>>08/28/2013   2
>>>>

>>>>
>>>>And by using set relationship and a do while loop, I can count and label the duplicates. But I was just wondering if an SQL statement(s) could do it all? The final output that I need is::
>>>>
>>>>

>>>>Mdate            cnt/date    Vs
>>>>05/28/2013     1              v1
>>>>05/28/2013     2
>>>>05/28/2013     3
>>>>

>>>>
>>>>Thanks,
>>>>Steve
>>
>>Thank you, I tried your code:
>>
>>

>>SELECT * 	;
>>    FROM ch5	;
>>    WHERE mdate IN (	;
>>        SELECT mdate FROM ch5 ;
>>        GROUP BY mdate ;
>>        HAVING COUNT(mdate) > 1)
>>

>>
>>but I get an error: file ch5.dbf does not exist
>
>Why would it? ch5 is a table isn't it? You used it in your example. Sorry, but this is the way to do it. You must have done something else wrong.
>
>Walter,

You are Correct. I had made a mistake. You code now gives a print out of all duplicates (with counts > 1); but it does not number the members of each group of duplicates, or otherwise mark off the appearance of the first of each of the duplicates?
Thank you,
Steve