> dim x as string > dim y as integer > x="y" > > I have the name of the variable in a string. How > should i use the string variable to access the value > of the integer variable.
you should consider using naming conventions. dim strX as string dim intY as integer strX= trim(str(inty)) > dim x(1 to 3,1 to 10) > > Assuming the arrayname alone is available how do i > find the no of dimensions of the array and the bounds > of the 2nd and 3rd dimensions.... > > Thanks a lot... dim lngUpper as long dim lngLower as long lngUpper1 = ubound(x(), 1) '1st lngUpper2 = ubound(x(), 2) '2nd lngUpper3 = ubound(x(), 3) '3rd lngLower1 - lbound(x(), 1) '1st lngLower2 - lbound(x(), 2) '2nd lngLower3 - lbound(x(), 3) '3rd etc ------------------------ Yahoo! Groups Sponsor --------------------~--> Make a clean sweep of pop-up ads. Yahoo! Companion Toolbar. Now with Pop-Up Blocker. Get it for free! http://us.click.yahoo.com/L5YrjA/eSIIAA/yQLSAA/k7folB/TM --------------------------------------------------------------------~-> '// ======================================================= Rules : http://ReliableAnswers.com/List/Rules.asp Home : http://groups.yahoo.com/group/vbHelp/ ======================================================= Post : [EMAIL PROTECTED] Join : [EMAIL PROTECTED] Leave : [EMAIL PROTECTED] '// ======================================================= Yahoo! Groups Links <*> To visit your group on the web, go to: http://groups.yahoo.com/group/vbhelp/ <*> To unsubscribe from this group, send an email to: [EMAIL PROTECTED] <*> Your use of Yahoo! Groups is subject to: http://docs.yahoo.com/info/terms/
