【发布时间】：2021-09-05 05:18:23
【问题描述】：
我正在使用表格来控制绘图的可见性和颜色。我想要一个复选框来切换可见性和一个下拉菜单来选择颜色。为此，我有如下内容。感觉就像拥有一个持久的编辑器可以防止使用复选框。
这个例子有点做作（模型/视图的设置方式），但说明了在编辑器打开时复选框如何不起作用。
我怎样才能拥有一个可以与可见组合框一起使用的复选框？使用两列更好吗？
import sys
from PyQt5 import QtWidgets, QtCore
class ComboDelegate(QtWidgets.QItemDelegate):
def __init__(self, parent):
super().__init__(parent=parent)
def createEditor(self, parent, option, index):
combo = QtWidgets.QComboBox(parent)
li = []
li.append("Red")
li.append("Green")
li.append("Blue")
li.append("Yellow")
li.append("Purple")
li.append("Orange")
combo.addItems(li)
combo.currentIndexChanged.connect(self.currentIndexChanged)
return combo
def setEditorData(self, editor, index):
editor.blockSignals(True)
data = index.model().data(index)
if data:
idx = int(data)
else:
idx = 0
editor.setCurrentIndex(0)
editor.blockSignals(False)
def setModelData(self, editor, model, index):
model.setData(index, editor.currentIndex())
@QtCore.pyqtSlot()
def currentIndexChanged(self):
self.commitData.emit(self.sender())
class PersistentEditorTableView(QtWidgets.QTableView):
def __init__(self, *args, **kwargs):
super().__init__(*args, **kwargs)
QtCore.pyqtSlot('QVariant', 'QVariant')
def data_changed(self, top_left, bottom_right):
for row in range(len(self.model().tableData)):
self.openPersistentEditor(self.model().index(row, 0))
class TableModel(QtCore.QAbstractTableModel):
def __init__(self, parent=None):
super(TableModel, self).__init__(parent)
self.tableData = [[1, 2, 3], [1, 2, 3], [1, 2, 3]]
self.checks = {}
def columnCount(self, *args):
return 3
def rowCount(self, *args):
return 3
def checkState(self, index):
if index in self.checks.keys():
return self.checks[index]
else:
return QtCore.Qt.Unchecked
def data(self, index, role=QtCore.Qt.DisplayRole):
row = index.row()
col = index.column()
if role == QtCore.Qt.DisplayRole:
return '{0}'.format(self.tableData[row][col])
elif role == QtCore.Qt.CheckStateRole and col == 0:
return self.checkState(QtCore.QPersistentModelIndex(index))
return None
def setData(self, index, value, role=QtCore.Qt.EditRole):
if not index.isValid():
return False
if role == QtCore.Qt.CheckStateRole:
self.checks[QtCore.QPersistentModelIndex(index)] = value
self.dataChanged.emit(index, index)
return True
return False
def flags(self, index):
fl = QtCore.QAbstractTableModel.flags(self, index)
if index.column() == 0:
fl |= QtCore.Qt.ItemIsEditable | QtCore.Qt.ItemIsUserCheckable
return fl
if __name__ == "__main__":
app = QtWidgets.QApplication(sys.argv)
view = PersistentEditorTableView()
view.setItemDelegateForColumn(0, ComboDelegate(view))
model = TableModel()
view.setModel(model)
model.dataChanged.connect(view.data_changed)
model.layoutChanged.connect(view.data_changed)
index = model.createIndex(0, 0)
persistet_index = QtCore.QPersistentModelIndex(index)
model.checks[persistet_index] = QtCore.Qt.Checked
view.data_changed(index, index)
view.show()
sys.exit(app.exec_())
【问题讨论】：
标签： python pyqt pyqt5 qtableview