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Auto-incrementing a number field
Message
From
22/12/1998 12:31:25
 
General information
Forum:
Visual FoxPro
Category:
Coding, syntax & commands
Miscellaneous
Thread ID:
00168371
Message ID:
00169708
Views:
12
>>>>>>Sorry Dan, must have hit the enter kit by accident on my last response.
>>>>>>
>>>>>>Any way, here is my suggestion:
>>>>>>
>>>>>>In your form init method:
>>>>>>
>>>>>>SELECT ;
>>>>>>MAX(id_field) ;
>>>>>>FROM table ;
>>>>>>INTO ARRAY aHighestID
>>>>>>
>>>>>>IF _TALLY > 0
>>>>>> THISFORM.object.value = aHighestID[1] + 1
>>>>>>ELSE && No records on the table
>>>>>> THISFORM.object.value = 1
>>>>>>ENDIF
>>>>>>
>>>>>>This should do it for you.
>>>>>>
>>>>>>Good luck
>>>>>
>>>>>Many thanks Juan to you and others who helped me with this...
>>>>>although there were many very good syggestions, your example sure helped me to finally resolve this issue!
>>>>>
>>>>>This is why I think the UT is the greatest thing since sliced bread!!
>>>>>
>>>>>Thanks to all,
>>>>>Dan
>>>>
>>>>Dan, I'm sorry to interrupt here, but this way is highly suspicious in multi-user environment.
>>>
>>>I agree 100% with you Edward, however, this is one is those rare instances where this short and simole functions will work since it is NOT going to be used in a multi-user environment, should work fine.
>>>
>>>Any other app I would normally have used something like UniqueId()....
>>>
>>>Dan
>>
>>I see. Anyway, you don't need in SELECT here. You can just:
>>Select MyTable
>>Set Order To Tag ID
>>GO BOTTOM
>>nNewid=MyTable.Id+1
>
>I thought of useing something like above, but i wasnt sure if it would pick up any deleted records before the table had been packed and thus show a false number that didnt represent the actual number of records in the table.
>I havent tried your code yet, but would this be the case?
>
>Dan

If SET DELETED ON then GO BOTTOM will pickup the last non-deleted record in index sequence, and SET DELETED OFF will allow to pick up the last record regardless deleted status, i.e. you have a choice depending on what you want to achieve.
Edward Pikman
Independent Consultant
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