在python中的函数内创建一个自动递增循环

Question

我正在尝试建立一个循环来跟踪订单编号的数量，并为每个新订单分配一个 nw 订单编号以增加值。

完整代码如下：

*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

您可以为此使用itertools.count()。它创建了一个生成器，当您调用它时，它只返回一个接一个的数字，它是 next() 方法：

from itertools import count

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

在此示例中，每次遍历行时，您都会从给定参数（1）开始获得 row_id 的另一个值：1、2、3、...

生成器也可以直接用于for 循环：

for number in count(1):
    print(number)

这将快速打印出一个数字列表。

在您的课堂上，最好在__init__() 创建计数器并在需要时调用next()：

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

    …(more code)…

def submit(self):
    my_id = self.counter.next()

    …(more code)…