Am 27.03.2015 um 05:28 schrieb Andrew Bernard:
Well, answering my own question here. The following works, but now I
feel the need to ask is the best way to do this sort of operation?
Yes, I think so. Why wouldn’t layer = 3 or = 2 suffice for
LedgerLineSpanner also?
Yours, Simon
\version "2.19.17"
treble = \relative c'' {
\override Stem.layer = #3
e'
\change Staff = "bass"
\override NoteHead.layer = #3
<e,,,, g,>-\markup {
\with-dimensions #'(0 . 3) #'(0 . 0)
\with-color #white
\filled-box #'(-1 . 2.5) #'(3 . 7) #0
}
\change Staff = "treble"
g'' c
c1
}
bass = \relative c {
\clef bass
<c, f c,>1 ~
<c f c,>
}
\score {
\new PianoStaff
<<
\new Staff = "treble" \with {
}
{ \treble }
\new Staff = "bass" \with {
\override LedgerLineSpanner.layer = #10
}
{ \bass }
>>
\layout { }
}
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