PythonScript get filenames in both views

Michael Vincent · Feb 5, 2022, 3:07 AM

PythonScript experts:

Is there a way to get the name of the “active” file in both views? That is, say I have split into 2 Scintilla views - left and right - and there is a file in each. From an application and notepad object view, there is only 1 active file - the one with focus and I can get that with notepad.getCurrentFilename(). But the file in the other Scintilla view that is “viewable” but not “active” in terms of focus - how do I get it’s name?

I see there is :

but they don’t seem to be implemented in PythonScript. Also, not sure how cycling through all opened files will be able to tell me which one is the “viewable” (but not “focus” “active”) in the secondary view.

Cheers.

Alan Kilborn · Feb 5, 2022, 3:07 AM

@michael-vincent said in PythonScript get filenames in both views:

Is there a way to get the name of the “active” file in both views?

for (filename, bufferID, index, view) in notepad.getFiles():
    if index == notepad.getCurrentDocIndex(0) and view == 0:
        print('view0 file:', filename)
    elif index == notepad.getCurrentDocIndex(1) and view == 1:
        print('view1 file:', filename)

Alan Kilborn · Feb 6, 2022, 2:42 AM

The above question and code leads to making a nice function like this:

def notepad_getCurrentFilename(view_to_query=None):
    if view_to_query == None:
        return notepad.getCurrentFilename()
    for (filename, bufferID, index, view) in notepad.getFiles():
        if view == view_to_query and index == notepad.getCurrentDocIndex(view_to_query):
            return filename
    return ''

and some code to exercise it:

print('notepad_getCurrentFilename():', notepad_getCurrentFilename())
print('notepad_getCurrentFilename(0):', notepad_getCurrentFilename(0))
print('notepad_getCurrentFilename(1):', notepad_getCurrentFilename(1))
current_view = notepad.getCurrentView()
print('notepad_getCurrentFilename(current_view):', notepad_getCurrentFilename(current_view))
other_view = 1 if current_view == 0 else 0
print('notepad_getCurrentFilename(other_view):', notepad_getCurrentFilename(other_view))

Some cautions/surprises:

If only one view is open, it could be view1 (aka editor2 aka “secondary”) not view0 (aka editor1 aka “primary”)!

Note: “aka” = also known as

If a view is not open and you ask for the getCurrentDocIndex() on that view, it will return an “interesting” thing:

On 32-bit Notepad++, 4294967295L is returned. As an unsigned number, this is FFFFFFFF in hex, leading to a possible signed interpretation of -1 in 32-bit land.
On 64-bit Notepad++, 18446744073709551615 is returned. As an unsigned number, this is FFFFFFFFFFFFFFFF in hex, leading to a possible signed interpretation of -1 in 64-bit land.

I’m not sure why PythonScript can’t just give us a -1 meaning that the view specified is not open.

A possible workaround is to see if the return value is greater than 4294967294 and if so deduce that the view is not open:

view = 1
print('view{v} is{n} open'.format(v=view, n=' not' if notepad.getCurrentDocIndex(view) > 4294967294 else ''))

Or one could run the for loop on notepad.getFiles() and remember which of the views is encountered:

view_is_open = [ False, False ]
for (filename, bufferID, index, view) in notepad.getFiles(): view_is_open[view] = True
print('view_is_open:', view_is_open)

We could turn that into a function:

def notepad_isViewOpen(view_to_test):
    view_is_open = [ False, False ]
    for (filename, bufferID, index, view) in notepad.getFiles(): view_is_open[view] = True
    return True if view_is_open[view_to_test] else False

and some code to exercise it:

print('notepad_isViewOpen(0):', notepad_isViewOpen(0))
print('notepad_isViewOpen(1):', notepad_isViewOpen(1))

Alan Kilborn · Dec 1, 2022, 1:29 PM

@Alan-Kilborn said:

Regarding:

If a view is not open and you ask for the getCurrentDocIndex() on that view, it will return an “interesting” thing

See:

https://github.com/bruderstein/PythonScript/issues/255