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Find numbers that contain and add to a diff field
Message
De
24/06/1999 11:00:39
Paul Frost
Instem Computer Systems Ltd
Stone, Royaume Uni
 
 
À
24/06/1999 09:29:22
Brad Stickley
Parker Pneumatic Div. North America
Wake Forest, Caroline du Nord, États-Unis
Information générale
Forum:
Visual FoxPro
Catégorie:
Codage, syntaxe et commandes
Divers
Thread ID:
00233452
Message ID:
00233513
Vues:
32
>>>>>I have a table with part numbers that show a length for the part numbers in another field.
>>>>>
>>>>>I need to find all of the numbers that contain "2479" and replace the length for these numbers with a value that is 4 more than the current value.
>>>>>
>>>>>Can anyone help me with a quick code to do this?
>>>>>
>>>>>Thanks in advance.
>>>>>
>>>>>Brad.
>>>>
>>>>Is this field a numeric field or char field?
>>>>
>>>>Jeff
>>>
>>>The part number is a character field and the length is numeric.
>>
>>Simplitic way
>>
>>select table1
>>locate
>>scan for at('2479',partnofield)0
>> replace lenfield with lenfield+4
>>endscan
>>
>>
>>HTH
>>Jeff
>
>Thanks Jeff that did it!
>
>I do very little programming and am working on learning more. Stuff like this is very helpful.
>
>Brad.

even more simplistic

select table1
replace lenfield with lenfield+4 all for like( '*2479*, partnofield)

Paul
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