在python

Question

我试图设置一个循环来跟踪多个ordernums，并为增加值的每个新订单分配一个nw ordernum。

这里是完整的代码：

*这是使用Klaus D.建议的固定代码

import tkinter
from tkinter import *
from tkinter import ttk
from tkinter import messagebox
from itertools import count



class ImprintPlusApp:

    def __init__(self, master):
        self.counter = count(1)

        master.title("Imprint Plus Manager")

        self.frame_header = ttk.Frame(master)
        self.frame_header.pack()

        ttk.Label(self.frame_header, text = "Bienvenidos a Imprint Plus Manager")

        self.frame_crearorden = ttk.Frame(master)
        self.frame_crearorden.pack()

        ttk.Label(self.frame_crearorden, text = "Nombre").grid(row = 0, column = 0, padx = 5)
        ttk.Label(self.frame_crearorden, text = "Email").grid(row = 2, column = 0, padx = 5)
        ttk.Label(self.frame_crearorden, text = "Numero Cel/Tel").grid(row = 4, column = 0, padx = 5)
        ttk.Label(self.frame_crearorden, text = "Información Adicional").grid(row = 6, column = 0, padx = 5)

        self.entry_name = ttk.Entry(self.frame_crearorden, width = 24)
        self.entry_email = ttk.Entry(self.frame_crearorden, width = 24)
        self.entry_numtc = ttk.Entry(self.frame_crearorden, width = 24)
        self.entry_addinf = Text(self.frame_crearorden, width = 50, height = 10)

        self.entry_name.grid(row = 0, column = 1, padx = 5)
        self.entry_email.grid(row = 2, column = 1, padx = 5)
        self.entry_numtc.grid(row = 4, column = 1, padx = 5)
        self.entry_addinf.grid(row = 7, column = 0, columnspan = 2, padx = 5)

        ttk.Button(self.frame_crearorden, text = "Submit", command = self.submit).grid(row = 8, column = 1,columnspan = 1, padx = 5)
        ttk.Button(self.frame_crearorden, text = "Clear", command = self.clear).grid(row = 8, columnspan = 1, padx = 5)


    def submit(self):
        result = next(self.counter)
        orderResult = str(result)
        print ("Nombre: {}".format(self.entry_name.get()))
        print ("Email: {}".format(self.entry_email.get()))
        print ("Num Cel/Tel: {}".format(self.entry_numtc.get()))
        print ("Información Adicional: {}".format(self.entry_addinf.get(1.0, "end")))
        self.clear()


        messagebox.showinfo(title = "Orden #"+ orderResult, message = "Orden Guardada")

    def clear(self):
        self.entry_name.delete(0, "end")
        self.entry_email.delete(0, "end")
        self.entry_numtc.delete(0, "end")
        self.entry_addinf.delete(1.0, "end")


def main():

    root = Tk()
    app = ImprintPlusApp(root)
    root.mainloop()

if __name__ == '__main__':
    main() 

Answer 1

您可以使用itertools.count()。当你调用next()方法时，它会创建一个只返回一个数字的生成器：

from itertools import count

counter = count(1)
for row in rows:
    row_id = counter.next()

在这个例子中，通过行的每次迭代，您将获得row_id的另一个值，从给定的参数（1）开始：1，2，{{ 1}}，...

生成器也可以直接在3循环中使用：

for

这将快速打印一个数字列表。

在课堂上，最好在for number in count(1): print(number) 中创建计数器，并在需要时调用__init__()：

next()