如何在QSCINTILA中实现Sublimitext逐层折叠功能

我正在努力实现 fold_by_level QScintilla组件上的Sublimitetext3功能,但我不太清楚如何实现,到目前为止,我已经想出了以下代码:

import sys
import re
import math

from PyQt5.Qt import *  # noqa

from PyQt5.Qsci import QsciScintilla
from PyQt5 import Qsci
from PyQt5.Qsci import QsciLexerCPP


class Foo(QsciScintilla):

    def __init__(self, parent=None):
        super().__init__(parent)

        # http://www.scintilla.org/ScintillaDoc.html#Folding
        self.setFolding(QsciScintilla.BoxedTreeFoldStyle)

        # Indentation
        self.setIndentationsUseTabs(False)
        self.setIndentationWidth(4)
        self.setBackspaceUnindents(True)
        self.setIndentationGuides(True)

        # Set the default font
        self.font = QFont()
        self.font.setFamily('Consolas')
        self.font.setFixedPitch(True)
        self.font.setPointSize(10)
        self.setFont(self.font)
        self.setMarginsFont(self.font)

        # Margin 0 is used for line numbers
        fontmetrics = QFontMetrics(self.font)
        self.setMarginsFont(self.font)
        self.setMarginWidth(0, fontmetrics.width("000") + 6)
        self.setMarginLineNumbers(0, True)
        self.setMarginsBackgroundColor(QColor("#cccccc"))

        # Indentation
        self.setIndentationsUseTabs(False)
        self.setIndentationWidth(4)
        self.setBackspaceUnindents(True)

        lexer = QsciLexerCPP()
        lexer.setFoldAtElse(True)
        lexer.setFoldComments(True)
        lexer.setFoldCompact(False)
        lexer.setFoldPreprocessor(True)
        self.setLexer(lexer)

        QShortcut(QKeySequence("Ctrl+K, Ctrl+J"), self,
                  lambda level=-1: self.fold_by_level(level))
        QShortcut(QKeySequence("Ctrl+K, Ctrl+1"), self,
                  lambda level=1: self.fold_by_level(level))
        QShortcut(QKeySequence("Ctrl+K, Ctrl+2"), self,
                  lambda level=2: self.fold_by_level(level))
        QShortcut(QKeySequence("Ctrl+K, Ctrl+3"), self,
                  lambda level=3: self.fold_by_level(level))
        QShortcut(QKeySequence("Ctrl+K, Ctrl+4"), self,
                  lambda level=4: self.fold_by_level(level))
        QShortcut(QKeySequence("Ctrl+K, Ctrl+5"), self,
                  lambda level=5: self.fold_by_level(level))

    def fold_by_level(self, lvl):
        if lvl < 0:
            self.foldAll(True)
        else:
            for i in range(self.lines()):
                level = self.SendScintilla(
                    QsciScintilla.SCI_GETFOLDLEVEL, i) & QsciScintilla.SC_FOLDLEVELNUMBERMASK
                level -= 0x400
                print(f"line={i+1}, level={level}")
                if lvl == level:
                    self.foldLine(i)


def main():
    app = QApplication(sys.argv)
    ex = Foo()
    ex.setText("""\
#include <iostream>
using namespace std;

void Function0() {
    cout << "Function0";
}

void Function1() {
    cout << "Function1";
}

void Function2() {
    cout << "Function2";
}

void Function3() {
    cout << "Function3";
}


int main(void) {
    if (1) {
        if (1) {
            if (1) {
                if (1) {
                    int yay;
                }
            }
        }
    }

    if (1) {
        if (1) {
            if (1) {
                if (1) {
                    int yay2;
                }
            }
        }
    }

    return 0;
}\
    """)
    ex.resize(800, 600)
    ex.show()
    sys.exit(app.exec_())


if __name__ == "__main__":
    main()

我关注的文档是 https://www.scintilla.org/ScintillaDoc.html#Folding 和 http://pyqt.sourceforge.net/Docs/QScintilla2/classQsciScintilla.html 。

正如我所说 按级别折叠\u 该特性的行为与Sublimitext完全一样,但我不确定ST的特性实现细节。无论如何,让我在Sublimitext上测试一些基本序列后发布一些屏幕截图,这可以澄清我在这里要实现的目标:

序列1: {ctrl+k, ctrl+5}, {ctrl+k, ctrl+j} {ctrl+k, ctrl+4}, {ctrl+k, ctrl+j} {ctrl+k, ctrl+3}, {ctrl+k, ctrl+j} {ctrl+k, ctrl+2}, {ctrl+k, ctrl+j} {ctrl+k, ctrl+1}, {ctrl+k, ctrl+j}

序列2: {ctrl+k, ctrl+5}, {ctrl+k, ctrl+4}, {ctrl+k, ctrl+3}, {ctrl+k, ctrl+2}, {ctrl+k, ctrl+1}

我确信Sublimitext行为有更多的内部细节,但如果我的示例在测试序列后的行为与这些快照上发布的行为完全相同,那么可以说该功能已经变得非常方便使用了。