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Use a variable to identify a variable

Posted on 2004-08-03
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Last Modified: 2010-05-02
Is there a way to use a variable to identify a variable-- for example:

Value1ID as integer
Value2ID as integer
Value3ID as integer

For x=1 to 3
    Value & x & ID = 123*x
Next x
0
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Question by:kevman123
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14 Comments
 
LVL 44

Expert Comment

by:Arthur_Wood
ID: 11705237
use an array:

Dim ValueID(3) as Integer

For x=1 to 3
    ValueID(x) = 123*x
Next x


AW
0
 
LVL 6

Expert Comment

by:bkthompson2112
ID: 11705267
Hi kevman123,

Use an array:

Dim ValueID(3) as Integer

For x = 1 to 3
  ValueID(x-1) = 123*x
Next x

bkt
0
 
LVL 6

Expert Comment

by:bkthompson2112
ID: 11705275
Gah!  Sorry AW.
Too slow.
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Author Comment

by:kevman123
ID: 11705281
That's the problem, it is already an array- sorry I forgot to put that-

Value1ID(3) as integer
Value2ID(3) as integer
Value3ID(3) as integer

For x=1 to 3
    Value & x & ID(x) = 123*x
Next x
0
 
LVL 6

Accepted Solution

by:
bkthompson2112 earned 1000 total points
ID: 11705341
Use a two-dimensional array

Dim ValueID(3)(3) as integer
0
 

Author Comment

by:kevman123
ID: 11705342
I guess I could use a 2 dimensional array- however, I would like to know if my original question is possible, not just a workaround.
0
 

Author Comment

by:kevman123
ID: 11705348
We're on the same wavelength!
0
 
LVL 6

Expert Comment

by:bkthompson2112
ID: 11705550
Using arrays is the only way I know of doing that in VB.
0
 

Author Comment

by:kevman123
ID: 11705570
Ok- I'll hold out for a little bit and see if anyone has an answer for it- otherwise I'll just give you the points since you gave the most feedback-
0
 
LVL 44

Assisted Solution

by:Arthur_Wood
Arthur_Wood earned 1000 total points
ID: 11705603
No, your original idea will not work.  You could however define a Class to hold an array of Items, and then create an array of instances of your class - which is functionally identical to a two-dimensional array.  But that is NOT the same thing as the idea you asked about in the original question.  The problem with:

Value & x & ID(x)

is that there is no 'variable' defined in you code for 'Value", and by using the & operator, the resultant entity would be a STRING, and not the NAME of a variable - and in any case, the Name of a variable and the VARIABLE itself are not the same thing.

Consider the famous painting of a PIPE, which is labeld "This is NOT a PIPE" (it is not a PIPE, because it is a PAINTING of a PIPE and thus, as such, is nothing more that a REPRESENTATION of a PIPE and not the PIPE itself)

Similarly, if you were to be arrested and then convicterd, the authorities would not throw your NAME in jail, they would throw YOU - your PHYSICAL self - in jail.

AW
0
 

Author Comment

by:kevman123
ID: 11705679
I believe a split is in order here- thanks for the quick responses guys!
0
 

Author Comment

by:kevman123
ID: 11705695
Well, I should have reversed the assisted and accepted answers- anyhow- thanks again-
0
 
LVL 44

Expert Comment

by:Arthur_Wood
ID: 11708751
as long as the points were assigned the same, Accepted and Assisted are irrrelevant.  And as for the points, with just 9,900,000 more, I can get a Tall Mocha Latte at Starbucks for only $4.50  LOL.

AW
0
 
LVL 6

Expert Comment

by:bkthompson2112
ID: 11708842
AW, agreed.

>ust 9,900,000 more...   ROFL
0

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