为什么不能将结果提取到listView小部件中?

时间:2019-06-23 10:25:51

标签: python sqlite pyqt python-3.7

我试图通过单击单选按钮从数据库表中获取记录。但是,这些项目没有显示在我的listview小部件中,并且在单击单选按钮后界面崩溃。请帮助我。

我的代码:

def working(self,action):
    txt=(action.text())
    cricket=Ui_MainWindow.connect('cricket.db')
    sql="SELECT player from statistics WHERE ctg='BAT';";
    sql1="SELECT player from statistics WHERE ctg='BOWL';";
    sql2="SELECT player from statistics WHERE ctg='AR';";
    sql3="SELECT player from statistics WHERE ctg='WK';";
    curse=cricket.cursor()
    #i=random.randint(1,30)
    if(radioButton.isChecked()):
            try:
                curse.execute(sql)
                rec=curse.fetchall()
                listWidget.additems(rec)
            except:
                print("error")
    elif(radioButton_2.isChecked()):
         try:
             curse.execute(sql1)
             rec=curse.fetchall()
             listWidget.additems(rec)
         except:
                print("error")
    elif(radioButton_3.isChecked()):
        try:
            curse.execute(sql2)
            rec=curse.fetchall()
            listWidget.additems(rec)
        except:
            print("error")
    elif(radioButton_4.isChecked()):
        try:
            curse.execute(sql3)
            rec=curse.fetchall()
            listWidget.additems(rec)
        except:
             print("error")

我尝试在命令提示符下运行sql命令,并且记录在那里显示而没有任何错误。

编辑:我也已在我的setupUi方法中添加了这些

  self.radioButton.toggled.connect(self.working)
  self.radioButton_2.toggled.connect(self.working)
  self.radioButton_3.toggled.connect(self.working)
  self.radioButton_4.toggled.connect(self.working)

0 个答案:

没有答案