Question

以下代码会创建一个QTableView。双击其项目将使用委派的QComboBox设置它。

问题：
单击ComboBox时，其下拉菜单会显示瞬间，然后折叠回其展开状态。

enter image description here

如果使用comboBox将combox.setEditable(True)设置为可编辑，下拉菜单将根据需要保持打开状态。但随后combobox的项目变得可编辑。并不是所需要的。由于combobox的项目只能选择。

如何解决自我崩溃的combobox行为？

P.S。我注意到，当ComboBox设置为可编辑且其下拉菜单展开时，模型的data()会不断被调用...自我崩溃行为可能是由模型？

import sys
from PyQt4.QtCore import *
from PyQt4.QtGui import *

class ComboDelegate(QItemDelegate):
    comboItems=['Combo_Zero', 'Combo_One','Combo_Two']
    def createEditor(self, parent, option, proxyModelIndex):
        combo = QComboBox(parent)
        combo.addItems(self.comboItems)
        # combo.setEditable(True)
        self.connect(combo, SIGNAL("currentIndexChanged(int)"), self, SLOT("currentIndexChanged()"))
        return combo

    def setModelData(self, combo, model, index):
        comboIndex=combo.currentIndex()
        text=self.comboItems[comboIndex]        
        model.setData(index, text)
        print '\t\t\t ...setModelData() 1', text

    @pyqtSlot()
    def currentIndexChanged(self): 
        self.commitData.emit(self.sender())

class MyModel(QAbstractTableModel):
    def __init__(self, parent=None, *args):
        QAbstractTableModel.__init__(self, parent, *args)
        self.items=['Data_Item01','Data_Item02','Data_Item03']

    def rowCount(self, parent=QModelIndex()):
        return len(self.items)
    def columnCount(self, parent=QModelIndex()):
        return 1

    def data(self, index, role):        
        if not index.isValid(): return QVariant()

        row=index.row()
        item=self.items[row]

        if row>len(self.items): return QVariant()

        if role == Qt.DisplayRole:
            print ' << >> MyModel.data() returning ...', item
            return QVariant(item) 

    def flags(self, index):
        return Qt.ItemIsEditable | Qt.ItemIsEnabled

    def setData(self, index, text):
        self.items[index.row()]=text

if __name__ == '__main__':
    app = QApplication(sys.argv)

    model = MyModel()
    tableView = QTableView()
    tableView.setModel(model)

    delegate = ComboDelegate()

    tableView.setItemDelegate(delegate)
    tableView.resizeRowsToContents()

    tableView.show()
    sys.exit(app.exec_())

Answer 1

您的原始代码在PyQt5中运行，combobox保持打开状态。但是用户必须单击以打开编辑器，然后单击以打开combobox。为避免这种情况，我在代码中将QComboBox替换为QlistWidget。另外我设置了editorGeometry：

import sys
# from PyQt4.QtCore import *
# from PyQt4.QtGui import *

from PyQt5.QtCore import *
from PyQt5.QtWidgets import *

class ComboDelegate(QItemDelegate):
    editorItems=['Combo_Zero', 'Combo_One','Combo_Two']
    height = 25
    width = 200
    def createEditor(self, parent, option, index):
        editor = QListWidget(parent)
        # editor.addItems(self.editorItems)
        # editor.setEditable(True)
        editor.currentItemChanged.connect(self.currentItemChanged)
        return editor

    def setEditorData(self,editor,index):
        z = 0
        for item in self.editorItems:
            ai = QListWidgetItem(item)
            editor.addItem(ai)
            if item == index.data():
                editor.setCurrentItem(editor.item(z))
            z += 1
        editor.setGeometry(0,index.row()*self.height,self.width,self.height*len(self.editorItems))

    def setModelData(self, editor, model, index):
        editorIndex=editor.currentIndex()
        text=editor.currentItem().text() 
        model.setData(index, text)
        # print '\t\t\t ...setModelData() 1', text

    @pyqtSlot()
    def currentItemChanged(self): 
        self.commitData.emit(self.sender())

class MyModel(QAbstractTableModel):
    def __init__(self, parent=None, *args):
        QAbstractTableModel.__init__(self, parent, *args)
        self.items=['Data_Item01','Data_Item02','Data_Item03']

    def rowCount(self, parent=QModelIndex()):
        return len(self.items)
    def columnCount(self, parent=QModelIndex()):
        return 1

    def data(self, index, role):        
        if not index.isValid(): return QVariant()

        row=index.row()
        item=self.items[row]

        if row>len(self.items): return QVariant()

        if role == Qt.DisplayRole:
            # print ' << >> MyModel.data() returning ...', item
            return QVariant(item) 

    def flags(self, index):
        return Qt.ItemIsEditable | Qt.ItemIsEnabled

    def setData(self, index, text):
        self.items[index.row()]=text

if __name__ == '__main__':
    app = QApplication(sys.argv)

    model = MyModel()
    tableView = QTableView()
    tableView.setModel(model)

    delegate = ComboDelegate()

    tableView.setItemDelegate(delegate)
    for i in range(0,tableView.model().rowCount()):
        tableView.setRowHeight(i,tableView.itemDelegate().height)
    for i in range(0,tableView.model().columnCount()):
        tableView.setColumnWidth(i,tableView.itemDelegate().width)

    tableView.show()
    sys.exit(app.exec_())

Answer 2

以下是尝试将QComboBox替换为与QToolButton和QMenu相关联的QAction。

外观与QComboBox大致相同，并且具有设置多个QActions已设置的附加功能（目前无法拥有多个组合框的项目）选中）。

enter image description here

import sys
from PyQt4.QtCore import *
from PyQt4.QtGui import *

class ComboDelegate(QItemDelegate):
    comboItems=['Combo_Zero', 'Combo_One','Combo_Two']
    def __init__(self, parent):
        QItemDelegate.__init__(self, parent=None)
        self.actionEmitted=None

    def createEditor(self, parent, option, index):
        if not index.isValid(): return

        model=index.model()
        itemName=model.data(index, Qt.DisplayRole)

        toolButton=QToolButton(parent)
        toolButton.setText( itemName.toString() )

        toolButton.setPopupMode(QToolButton.InstantPopup)        

        menu=QMenu(parent)

        action1=QAction('Action 01', menu, checkable=True)
        action2=QAction('Action 02', menu, checkable=True)
        action3=QAction('Action 03', menu, checkable=True)

        action1.setObjectName('Action 01')
        action2.setObjectName('Action 02')
        action3.setObjectName('Action 03')       

        action1.setChecked(True)
        action3.setChecked(True)

        self.connect(action1, SIGNAL("triggered(bool)"), self, SLOT("actionTriggered()"))
        self.connect(action2, SIGNAL("triggered(bool)"), self, SLOT("actionTriggered()"))
        self.connect(action3, SIGNAL("triggered(bool)"), self, SLOT("actionTriggered()"))

        menu.addAction(action1)
        menu.addAction(action2)
        menu.addAction(action3)

        toolButton.setMenu(menu) 

        return toolButton

    def setModelData(self, toolButton, model, index):
        print '\t\t\t ...setModelData() 1', toolButton, model, index

        if not self.actionEmitted: return

        menu=toolButton.menu()
        for action in menu.actions():
            actionName=action.objectName()
            actionStatus=action.isChecked()            
            if actionStatus:
                model.setData(index, actionName)
                print '####', actionName, actionStatus

    @pyqtSlot()
    def actionTriggered(self): 
        self.actionEmitted=self.sender()        
        self.commitData.emit( self.actionEmitted )
        print 'actionTriggered.....', self.actionEmitted


class MyModel(QAbstractTableModel):
    def __init__(self, parent=None, *args):
        QAbstractTableModel.__init__(self, parent, *args)
        self.items=['Data_Item01','Data_Item02','Data_Item03']

    def rowCount(self, parent=QModelIndex()):
        return len(self.items)
    def columnCount(self, parent=QModelIndex()):
        return 1

    def data(self, index, role):        
        if not index.isValid(): return QVariant()

        row=index.row()
        item=self.items[row]

        if row>len(self.items): return QVariant()

        if role == Qt.DisplayRole:
            print ' << >> MyModel.data() returning ...', item
            return QVariant(item) 

    def flags(self, index):
        return Qt.ItemIsEditable | Qt.ItemIsEnabled

    def setData(self, index, itemName):
        self.items[index.row()]=itemName

if __name__ == '__main__':
    app = QApplication(sys.argv)

    combo=QComboBox()
    combo.addItems(['Combo_Zero', 'Combo_One','Combo_Two'])

    model = MyModel()
    tableView = QTableView()
    tableView.setModel(model)

    delegate = ComboDelegate(tableView)

    tableView.setItemDelegate(delegate)
    tableView.resizeRowsToContents()

    tableView.show()
    sys.exit(app.exec_())

如何使用QComboBox作为QTableView的委托

2 个答案: