python - 在 QComboBox 的 View QTableView 中单击信号

Question

我需要一个带有按钮的 QListView 用于 QComboBox，但是因为编写委托(delegate)是一个巨大的痛苦。我选择了 QTableView，因为每一行总是有相同的按钮。我唯一的问题是我似乎无法从 QComboBox 中捕捉到 clicked 信号。

我附上一个mwe来说明我的意思:

from PyQt5.QtCore import QModelIndex, Qt, QAbstractTableModel, QVariant
from PyQt5.QtWidgets import QApplication, QComboBox, QTableView, QWidget, QVBoxLayout


class Model(QAbstractTableModel):
    def data(self, index, role=Qt.DisplayRole):
        if role == Qt.DisplayRole:
            col = index.column()
            if col == 0:
                return str(index.row())
            elif col == 1:
                return '✎'
            elif col == 2:
                return '✘'
        return QVariant()

    def columnCount(self, parent=QModelIndex()):
        return 3

    def rowCount(self, parent=QModelIndex()):
        return 5


if __name__ == '__main__':
    import sys

    app = QApplication(sys.argv)
    model = Model()

    main = QWidget()
    layout = QVBoxLayout(main)

    view = QTableView()
    view.clicked.connect(lambda _: print('Click Table'))  # Works fine
    view.setModel(model)
    layout.addWidget(view)

    combo = QComboBox()
    combo.setModel(model)
    combo.setView(QTableView())
    combo.view().clicked.connect(lambda _: print('Click Combo'))  # Does'nt show
    layout.addWidget(combo)

    main.resize(500, 300)
    main.show()
    sys.exit(app.exec_())

Answer 1

您可以使用以下信号触发组合框中的点击:

view = QTableView()
view.clicked.connect(lambda: print('Click Table'))  # Works fine
view.setModel(model)
layout.addWidget(view)

combo = QComboBox()
combo.setModel(Model())
combo.setView(QTableView())

combo.activated.connect(lambda: print('Click Combo'))  # Works fine too
layout.addWidget(combo)

或者，如果您只需要检测单击或键盘输入何时更改所选项目:

combo.currentIndexChanged.connect(lambda: print('Click Combo'))