Question

我试图将一些简单的信息从SQLite数据库显示到QTableView中。我跟着one answer from SO，就其自身而言，它正在发挥作用。当我尝试将相同的代码实现到我的GUI（只是一个带有QTableView对象的简单主窗口）时，它什么都没显示。这是代码：

from PyQt4 import QtCore, QtGui
from gui import Ui_MainWindow
from dialog import Ui_Dialog
from PyQt4.QtSql import QSqlQueryModel,QSqlDatabase,QSqlQuery

import sys

class Glavni(QtGui.QMainWindow):
    def __init__(self, parent=None):
        super(Glavni, self).__init__()
        self.ui=Ui_MainWindow()
        self.ui.setupUi(self)
        self.show()

        #QtCore.QObject.connect(self.ui.actionRegistar, QtCore.SIGNAL("triggered()"), self.popup)
        db = QSqlDatabase.addDatabase("QSQLITE")
        db.setDatabaseName("baza.db")
        db.open()

        projectModel = QSqlQueryModel()
        projectModel.setQuery("select name from people",db)

        projectView = QtGui.QTableView()
        projectView.setModel(projectModel)
        projectView.show()

    def popup(self):
        dialog = QtGui.QDialog()
        dialog.show()

class Dialog(QtGui.QDialog):
    def __init__(self,parent=None):
        super(Dialog,self).__init__()
        self.ui = Ui_Dialog()
        self.ui.setupUi(self)
        self.show()

def main():
    app = QtGui.QApplication(sys.argv)
    prozor = Glavni()
    sys.exit(app.exec_())

if __name__ == '__main__':
    main()

我做错了什么？ Q4View在QT Designer中的名称是lista，如果它是相关的。谢谢。

Answer 1

projectView = QtGui.QTableView() #THIS part is wrong if the GUI is designed through Designer
projectView.setModel(projectModel)
projectView.show()

正确的代码与此类似：

projectView = self.ui.myList #or some other name, which is the SAME AS that object name in   Designer
projectView.setModel(projectModel)
projectView.show()

它有效;）

基于PyQt4 QTableView模型

1 个答案: