I've been playing with macro code for a day trying to figure out how to translate this bit of VBA code into an OO Basic macro without notable success. I know there has to be an easy way to do this but I'm not finding it...
VBA code:
sub VBA_Original
For i = Sheets.Count To 1 Step -1
X = Application.Match(Sheets(i).Name, Worksheets("source info").Range("I2:I17"), 0)
If IsError(X) Then
Application.DisplayAlerts = False
Sheets(i).Delete
Application.DisplayAlerts = True
End If
Next
end sub
Any pointers would be appreciated.
Thanks!
Last edited by stickman68 on Mon Feb 19, 2018 10:50 pm, edited 1 time in total.
LibreOffice 6.0.1 on Windows 10/ LibreOffice 6.0.1 with MacOS 10.13 / LibreOffice 6.0.1 on Linux Mint 18
"Playing for a day" is a total waste of time. You could also let your cat walk over the keyboard until she wrote a poem by mere chance.
We are not the humanoid macro recorders for VBA snippets
3 options:
- learn how to program
- do your stuff manually
- stick with Excel
Please, edit this topic's initial post and add "[Solved]" to the subject line if your problem has been solved.
Ubuntu 18.04 with LibreOffice 6.0, latest OpenOffice and LibreOffice
Thanks for all the excellent help with your solution. I don't know how I didn't see it myself.
If you had bothered to look at my other post you would see that I can write a macro in OO Basic. I'm just missing something here.
LibreOffice 6.0.1 on Windows 10/ LibreOffice 6.0.1 with MacOS 10.13 / LibreOffice 6.0.1 on Linux Mint 18
sx = ThisComponent.getSheets()
a() = sx.getElementNames() 'simple array of names
rg = sx.getByName("source info").getCellRangeByName("I2:I17")
b() = rg.getDataArray() 'nested array of row values
uba = uBound(a())
ubb = uBound(b())
for i = 0 to uba
for j = 0 to ubb
if a(i) = b(j)(0) then exit for
next j
if j > ubb then sx.removeByName(a(i))
next i
The tricky part is that the DataArray b() is a nested array of the rows in one column. When you inspect the variable, you see that each element in b is another array with one element (because there is only one column in each row) and b(j)(0) addresses the first element of element b(j) which is the cell value or string.
When the loop runs until the last element without calling the "exit for" clause, j is by one greater than the uBound of array b() which means that the name a(i) was not found in b() in which case we remove the element named a(i) from the sheets collection sx.
Please, edit this topic's initial post and add "[Solved]" to the subject line if your problem has been solved.
Ubuntu 18.04 with LibreOffice 6.0, latest OpenOffice and LibreOffice
While numSheets > numRows
for i = 0 to ubaNames
for j = 0 to ubaRows
if aNames(i) = aRows(j)(0) then
exit for
end if
next j
if j > ubaRows then
print aNames(i)
objSheets.removeByName(aNames(i))
end if
next i
numSheets = objSheets.Count
wend
LibreOffice 6.0.1 on Windows 10/ LibreOffice 6.0.1 with MacOS 10.13 / LibreOffice 6.0.1 on Linux Mint 18
stickman68 wrote:Any idea why it exits the sub when it hits the objSheets.removeByName(aNames(i))
Only you can find out. The Basic debugging tools are the same as in VBA. I guess it is because NOT(numSheets > numRows)
Please, edit this topic's initial post and add "[Solved]" to the subject line if your problem has been solved.
Ubuntu 18.04 with LibreOffice 6.0, latest OpenOffice and LibreOffice