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February 26th, 2002, 01:58 AM
#1
Please help w/ VB question..
Okay, I need assistance with a VB program.. Let's say in the most basic program we have to take the amount of an item and divide it by the price, to get the unit price..
I need to take "16 Oz" from a textbox which would be the amount variable divide it by let's say "16" from the amount to get the individual price per unit.. Okay.. Easy enough... using the Var(txtAmount.text) code.. I can extract the "16" from the "16 Oz" and I get an even 1 as my output to the unit price.. But now I need to extract the "Oz" from the string to output "1 Oz" How would I do that? I've tried all different things, but I keep receiving errors. I was hoping some knowledgable individual could give me some ideas... Thank you.. Also.. If you could, please respond within the next 3 1/2 hours.. I have a deadline.. Thank you..
=]
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February 26th, 2002, 02:19 AM
#2
i can't tell if you're dealing with a database or not...sounds like you might be so i'll answer as if you were...
firstly...you shouldn't be mixing numbers and text strings...makes life too difficult...
i'd keep qty = integer, unit/wt = string then you can do all your manipulation with out this issue...
but if you're stuck with it...
just assign the oz part of the string to a var and then concatenate that to the unit price cal var
myvalue = varUnitPrice & " " & varUnitWt
and if unit wt never changes....you can just
myvalue = varUnitPrice & " oz."
again...this is based on a guess that you're using vba and are working with a db...if not...hope it still applies or maybe point you rightly...
I used to be With IT. But then they changed what IT was. Now what I'm with isn't IT, and what's IT seems scary and weird." - Abe Simpson
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February 26th, 2002, 02:25 AM
#3
the same method should work in most programs with vb, to get specifics look in the vb help file or reference book under concatenate.
If you give me the code, I could probably do it myself fairly quickly
Preliminary operational tests were inconclusive (the dang thing blew up)
\"Ask not what the kernel can do for you, ask what you can do for the kernel!\"
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February 26th, 2002, 04:08 AM
#4
The code looks something like this..
Dim size As String
Dim price As Single
Dim unit As Single
Dim weight As String
Private Sub Label1_Click()
End Sub
Private Sub cmdCalc_Click()
size = Val(txtSize.Text)
price = txtPrice.Text
weight = Str$(txtSize.Text)
unit = price / size
txtUnit.Text = unit & " " & weight
/// Here I need to get it to print out.. The "Oz" part of the "17 Oz" without the "17" but I'm not sure if the language can differentiate between 1 and 7 as ASCII characters, or if it can actually eliminate the numbers from the string and throw in the "Oz"; I still receive an error from this.. I can't quite figure this one out.. I can get it to work fine.. But as an exercize.. I am trying this out.. I can't put in "txtUnit.Text = Unit & " Oz."" because if it were any other unit of measurement it would not work accurately.. If you can figure this out, that would be great.. I'll keep checking back.. Thank you.
=]
///
End Sub
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February 26th, 2002, 04:35 AM
#5
Member
why not have Oz as a label?
\"Not all humans are reasonably good.\" -John Locke
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February 26th, 2002, 05:21 AM
#6
how about
weight = right(varWholeString, 2)
or
weight = right(varWholeString, 3)
2 or 3 depending on whether you use oz. or jus oz (no <.>)
in this ex the last 2 or 3 chars can be eval'ed whatever they are...
I used to be With IT. But then they changed what IT was. Now what I'm with isn't IT, and what's IT seems scary and weird." - Abe Simpson
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February 26th, 2002, 05:27 AM
#7
VB
If you have a text box that contain 16 OZ, you can use the code
SDK as String
SDK = Right(txtAmount.text, 2)
' SDK will = OZ
This code will give you the last 2 caracters from the right. You can also write
SDK = Left(txtAmount.text, 2)
'SDK will = 16
To get the 2 first caracters from the left and
SDK = Mid(txtAmount.text, 2, 4)
SDK will = ' OZ"
The command MID will put the 4 caracters from the second caracters on in of your txtAmount.text in the variable SDK.
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February 26th, 2002, 05:41 PM
#8
Thanks for the help.. =]
Okay..
SDK, your method worked perfectly!
Zigar, thanks for your help, too!
Kezil, you too!
Mrln, I would if I could.. But let's say they entered mL instead of Oz.. I would be stuck with mL still.. But thank you for the help.
=]
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