Question

我有一个python程序，用于访问数据库以查找需要在屏幕上显示的内容。它的目的是显示何时可以提取某人的订单。我有所有代码来获取数据并将其显示在窗口上。但是，我需要每60秒重新查询一次数据库，因为有些订单会被拿走，并且需要从列表中消失，还有一些需要添加。我只是不知道该怎么做，因为一旦调用了app.mainloop（），它就需要人与窗口的交互才能使事情发生。任何帮助将不胜感激...很抱歉长期困扰！

Answer 1

以下是我汇总的一个示例，向您展示了如何使用after()检查这么多秒来更新跟踪器的一些基础知识。

如果您有任何疑问，请告诉我。

import tkinter as tk


class Example(tk.Tk):
    def __init__(self):
        tk.Tk.__init__(self)
        self.geometry("600x400")
        self.current_ticket_number = 1
        self.data = [[97, "Mike"], [98, "Kaite"], [99, "Tom"]]

        self.display_frame = tk.Frame(self)
        self.display_frame.grid(row=2, column=0, columnspan=3, sticky="nsew")

        self.lbl1 = tk.Label(self, text="Next ticket number: {}".format(self.current_ticket_number))
        self.lbl1.grid(row=0, column=0)
        self.lbl2 = tk.Label(self, text="Customer Name: ".format(self.current_ticket_number))
        self.lbl2.grid(row=0, column=1)
        self.entry1 = tk.Entry(self)
        self.entry1.grid(row=0, column=2)

        tk.Button(self, text="Refresh List", command=self.refresh).grid(row=1, column=0, pady=5)
        tk.Button(self, text="Submit new ticket", command=self.new_ticket).grid(row=1, column=1, pady=5)

        self.timed_refresh()

    def new_ticket(self):
        x = self.entry1.get().strip()
        if x != "":
            self.data.append([self.current_ticket_number, x])
            #self.refresh() # you could do self.refresh() here if you want to update as soon as you create a ticket
            #I left it out though so you can see how after() works below.
            if self.current_ticket_number >= 99:
                self.current_ticket_number = 1
            else:
                self.current_ticket_number += 1

    def refresh(self):
        self.display_frame.destroy()
        self.display_frame = tk.Frame(self)
        self.display_frame.grid(row=2, column=0, columnspan=3, sticky="nsew")
        for ndex, item in enumerate(self.data):
            tk.Label(self.display_frame, text=r"Order #{} is ready for {}.".format(item[0], item[1])).grid(row=ndex, column=1)
            tk.Button(self.display_frame, text=r"Remove Ticket".format(item[0], item[1]), command=lambda x=ndex: self.remove_ticket(x)).grid(row=ndex, column=0)

    def remove_ticket(self, ndex):
        self.data.pop(ndex)
        self.refresh()

    def timed_refresh(self):
        #this after statement is set for every 6 seconds
        self.after(6000, self.timed_refresh)
        self.refresh()

if __name__ == "__main__":
    Example().mainloop()

Answer 2

这就是我最后得到的。似乎在做我需要做的事。感谢您的所有帮助，一切就绪。

class Application(tk.Frame):
    def __init__(self,master=None):
    self.createWidgets()

    def createWidgets(self):
        tk.Frame.__init__(self)
        self.pack()
        for b in range(0,int(len(myItems)/2)):
            #print (myItems[b])
            self.btn = tk.Button(self)
            self.btn["text"] = myItems[b,0]
            # self.btn["command"] = (lambda tck=b, binst=btn : pickUp(tck, binst))
            # self.btn["command"] =  lambda ticketNo=myItems[b,1] : self.pickUp(ticketNo)
            self.btn.pack(fill='x')

    def pp(self) :
        #print('Im in pp')
        self.destroy()
        getArrowDataAndUpdateSQLite()
        myItems = getDisplayData()
        app.createWidgets()
        app.master.after(30000, self.pp)

app = Application()
app.master.title('Customer Order Status')
app.master.after(30000,app.pp)
app.mainloop()

如何每60秒刷新python tkinter窗口的内容

2 个答案: