Question

如果我向前搜索字母x（按钮Next），一切正常，但是一旦我改变方向（按钮Previous），就会发生这种情况：

QsciScintilla.findFirst()不会移动选择。即，首先按下按钮Previous不会执行任何操作；
QsciScintilla.findNext()以2为步长，因此跳过了一个字符。

我正在考虑将ATM的查找逻辑从C ++转换为Python，从而潜在地解决了这个问题，但是很高兴知道我犯了一些新手错误，从而避免了所有额外的工作...

代码如下：

from PyQt5.QtWidgets import *
from PyQt5.QtCore import *
from PyQt5.QtGui import *
from PyQt5.Qsci import *
import sys


class FindAndReplace(QWidget):
    def __init__(self, *arg, **kwarg):
        super(self.__class__, self).__init__(*arg, **kwarg)

        rows = QVBoxLayout()

        self.editor = QsciScintilla()
        self.editor.setText(f'{"x"*40}\n{"y"*40}\n{"z"*40}\n')
        rows.addWidget(self.editor)

        self.text_to_find = ''
        self.state_ = tuple()

        self.find = QLineEdit()
        self.find_previous = QPushButton('&Previous')
        self.find_next = QPushButton('&Next')
        self.find_lbl = QLabel('&Find')
        self.find_lbl.setBuddy(self.find)
        row = QHBoxLayout()
        for w in (self.find_lbl, self.find, self.find_previous, self.find_next):
            row.addWidget(w)
        rows.addLayout(row)

        self.re = QCheckBox('&Regular expressions')
        self.cs = QCheckBox('&Case sensitive')
        self.wo = QCheckBox('Whole &words')
        self.wrap = QCheckBox('Wrap aroun&d')
        self.show_ = QCheckBox('&Unfold folded text')
        self.posix = QCheckBox('POSI&X-compatible RE')
        row = QHBoxLayout()
        for w in (self.re, self.cs, self.wo, self.wrap, self.show_, self.posix):
            row.addWidget(w)
        rows.addLayout(row)

        self.setLayout(rows)

        self.find_previous.clicked.connect(lambda: self.findText(forward = False))
        self.find_next.clicked.connect(lambda: self.findText(forward = True))

    def findText(self, forward):
        text_to_find = self.find.text()
        state_ = ( \
            self.re.isChecked(), self.cs.isChecked(),
            self.wo.isChecked(), self.wrap.isChecked(),
            forward, -1, -1,
            self.show_.isChecked(), self.posix.isChecked(),
        )

        if text_to_find != self.text_to_find or state_ != self.state_:
            self.text_to_find = text_to_find
            self.state_ = state_
            # search with new conditions.
            self.editor.findFirst(text_to_find, *state_)
        else:
            # search with previously set conditions.
            self.editor.findNext()


if __name__ == '__main__':
    app = QApplication(sys.argv)
    FindAndReplace().show()
    sys.exit(app.exec_())

Answer 1

findNext函数对我来说似乎不可靠。如果我使用getSelection在line中明确输入index和findFirst，并避免完全使用findNext，那么一切都会按预期进行：

def findText(self, forward):
    text_to_find = self.find.text()

    if forward:
        line, index = self.editor.getSelection()[2:]
    else:
        line, index = self.editor.getSelection()[:2]

    state_ = (
        self.re.isChecked(), self.cs.isChecked(),
        self.wo.isChecked(), self.wrap.isChecked(),
        forward, line, index,
        self.show_.isChecked(), self.posix.isChecked(),
        )

    self.text_to_find = text_to_find
    self.state_ = state_
    self.editor.findFirst(text_to_find, *state_)

查看最新的源代码（ qsciscintilla.cpp ，第1853行），我看到了：

// Finally adjust the start position so that we don't find the same one again.
if (findState.forward)
    findState.startpos = targend;
else if ((findState.startpos = targstart - 1) < 0)
    findState.startpos = 0;

我可能误解了代码的意图，但是为什么在这里减去一？ AFAICS，这将在向后搜索时产生一个错误的错误。

使用QsciScintilla.findNext的向后搜索无法按预期进行

1 个答案: