Re : Excel VBA Array to be expressed as a variation of Partial
Dimension

1. Enters an Excel worksheet.

2. Assume that,
Dim Xrange as Variant

3. To read into Xrange,
Xrange = ActiveSheet.Range(A1:A10)

4. And then, Xrange contains 10 elements.

5. Programming logic follows that it would be required to examine the
contents of Xrange but merely the second half of the stated array
namely, Xrange(6) to Xrange(10).

6. To exemplify the query conceptually, would it be veritable to
express as follows,
(Xrange(I,J), I = 6 to 10, J=1)
or, Alternatively ?

7. Please share your experience. Regards.

for i = 6 to 10
msgbox "xRange(" & i & ",1)= " & xRange(i,1)
Next

--
Regards,
Tom Ogilvy


" wrote:

Re : Excel VBA Array to be expressed as a variation of Partial
Dimension

1. Enters an Excel worksheet.

2. Assume that,
Dim Xrange as Variant

3. To read into Xrange,
Xrange = ActiveSheet.Range(A1:A10)

4. And then, Xrange contains 10 elements.

5. Programming logic follows that it would be required to examine the
contents of Xrange but merely the second half of the stated array
namely, Xrange(6) to Xrange(10).

6. To exemplify the query conceptually, would it be veritable to
express as follows,
(Xrange(I,J), I = 6 to 10, J=1)
or, Alternatively ?

7. Please share your experience. Regards.

wrote:
Re : Excel VBA Array to be expressed as a variation of Partial
Dimension

1. Enters an Excel worksheet.

2. Assume that,
Dim Xrange as Variant

3. To read into Xrange,
Xrange = ActiveSheet.Range(A1:A10)

4. And then, Xrange contains 10 elements.

5. Programming logic follows that it would be required to examine the
contents of Xrange but merely the second half of the stated array
namely, Xrange(6) to Xrange(10).

6. To exemplify the query conceptually, would it be veritable to
express as follows,
(Xrange(I,J), I = 6 to 10, J=1)
or, Alternatively ?

7. Please share your experience. Regards.

If the functions in the freely downloadable file at
http://home.pacbell.net/beban are available to your workbook

PartialXRange = SubArray(XRange, , , 6, 10)

Alan Beban

Mr. Tom Ogilvy and Mr. Alan Beban,

Thank you very much indeed.