Attn: Barry Davidson


Posted by Anthony on July 20, 2001 5:57 AM

Hey Barry! If you don't mind, I have one more question with regard to the solution you provided for me. Here it goes: You explained to me yesterday that if I divided my number in question by 1000, I'd achieve the desired results; in which case I did. However, what I neglected to mention was that I need the number appear as follows:

My report contains the number 1120----1120/1000 = 1.120

I need the result to read 1.1204

Is this possible?

Thanks Again

Anthony

Posted by Aladin Akyurek on July 20, 2001 6:01 AM

Anthony -- Why is that decimal 4? Does it mean that every entry should get a 4 as last digit in the decimals part?

PS. I'll come back to you very soon on the other thing.

Aladin

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Posted by Anthony on July 20, 2001 6:25 AM

Thanks Aladin for the immediate response! To answer your question, the last number does not have to end in a 4. 1120/1000 = 1.120----On my original document that is generated from another database, the number reads 1.1204----another e.g. 1280/1000 = 1.280---this number reads 1.2849.

When I extract the original report to excel, it does not provide me with the data from the original report. I've tried saving the file many different ways without obtaining the desired results. Any suggestions?
Also, thanks again for continuing to work on that project for me.

Regards

Anthony

Posted by Aladin Akyurek on July 20, 2001 6:55 AM

Anthony,

I'll propose something, but that is still guesswork. The best thing to do is to provide those numbers from the database as they exactly look (If they consist of unequal numbers of digits, we need to know that) and how each one should raed. In the meantime, you might try the following:

=A1/(10^(LEN(A1)-1))

Aladin

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Posted by Barrie Davidson on July 20, 2001 7:45 AM

Anthony, how does the number end up in your spreadsheet as 1120 when it is 1.1204 in the original document? Is a digit being dropped?

Barrie



Posted by Anthony on July 20, 2001 8:41 AM

Thanks for the help. It worked out just fine.

Anthony