python - PyQt5 ComboBox - 如何在不影响下拉列表的情况下设置 CurrentText 的颜色？

Question

以下代码段正确设置了 ComboBox 下拉列表中各个条目的颜色。 However, when an item is selected and transferred to the CurrentText field, all of the entries in the dropdown change to the color of CurrentText.如何转换条目的颜色以显示为 CurrentText 而不影响下拉列表？

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

class ComboDemo(QWidget):

    def __init__(self):
        super().__init__()
        self.initUI()

    def initUI(self):

        def combo_changed():
            for color in ('red', 'green', 'blue'):
                if color == cb.currentText():
                    cb.setStyleSheet('color: {}'.format(color))

        grid = QGridLayout()
        cb = QComboBox()
        grid.addWidget(cb, 0, 0)
        model = cb.model()
        for color in ('red', 'green', 'blue'):
            entry = QStandardItem(color)
            entry.setForeground(QColor(color))
            model.appendRow(entry)

        cb.currentIndexChanged.connect(combo_changed)

        self.setLayout(grid)
        self.show()

app = QApplication(sys.argv)
c = ComboDemo()
app.exec_()

Answer 1

你必须使用QComboBox:editable:

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

class ComboDemo(QWidget):

    def __init__(self):
        super().__init__()
        self.initUI()

    def initUI(self):

        def combo_changed():
            for color in ('red', 'green', 'blue'):
                if color == cb.currentText():
                    cb.setStyleSheet("QComboBox:editable{{ color: {} }}".format(color))

        grid = QGridLayout()
        cb = QComboBox()
        grid.addWidget(cb, 0, 0)
        model = cb.model()
        for color in ('red', 'green', 'blue'):
            entry = QStandardItem(color)
            entry.setForeground(QColor(color))
            model.appendRow(entry)

        cb.currentIndexChanged.connect(combo_changed)
        self.setLayout(grid)
        self.show()

app = QApplication(sys.argv)
c = ComboDemo()
app.exec_()