如何在QScintilla中实现SublimeText逐级功能

Question

我正在尝试fold_by_level在QScintilla组件上实现SublimeText3功能,但我不知道如何做到这一点,到目前为止我已经提出了这个代码:

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()

我遵循的文档是https://www.scintilla.org/ScintillaDoc.html#Folding和http://pyqt.sourceforge.net/Docs/QScintilla2/classQsciScintilla.html.

正如我所说,该fold_by_level功能的目的与SublimeText完全相同,但我不确定ST的功能实现细节.在任何情况下,让我在SublimeText上测试一些基本序列之后发布一些截图,这些序列可以澄清我在这里想要实现的内容:

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SublimeText行为有更多的内部细节,但是如果我的示例在测试序列之后表现得与在这些镜头上发布完全相同,那么您可以说该功能已经变得非常方便使用.

Answer 1

您的示例的问题主要是由 QsciScintilla API 中的一些不良命名引起的。FoldLine和FoldAll方法确实应该被调用and，因为它们实际上撤消了之前的状态。这意味着，例如，如果两个连续的行具有相同的折叠级别，则调用两次将不会导致净变化。toggleFoldLinetoggleFoldAllfoldLine

在下面的实现中，我使用了更明确的 Scintilla 消息，以便只有真正需要折叠的行才会受到影响。我还更改了键盘快捷键以匹配 SublimeText 中的默认设置：

class Foo(QsciScintilla):
    def __init__(self, parent=None):
        ...
        QShortcut(QKeySequence("Ctrl+K, Ctrl+J"), self, self.fold_by_level)
        QShortcut(QKeySequence("Ctrl+K, Ctrl+0"), self, self.fold_by_level)
        ...

    def fold_by_level(self, level=0):
        SCI = self.SendScintilla
        if level:
            level += 0x400
            MASK = QsciScintilla.SC_FOLDLEVELNUMBERMASK
            for line in range(self.lines()):
                foldlevel = SCI(QsciScintilla.SCI_GETFOLDLEVEL, line) & MASK
                print('line=%i, level=%i' % (line + 1, foldlevel), end='')
                if foldlevel == level:
                    line = SCI(QsciScintilla.SCI_GETFOLDPARENT, line)
                    if SCI(QsciScintilla.SCI_GETFOLDEXPANDED, line):
                        print(', foldline:', line + 1, end='')
                        SCI(QsciScintilla.SCI_FOLDLINE, line,
                            QsciScintilla.SC_FOLDACTION_CONTRACT)
                print()
        else:
            SCI(QsciScintilla.SCI_FOLDALL, QsciScintilla.SC_FOLDACTION_EXPAND)