如何在Python中错误输入时显示弹出窗口？

我正在尝试编写一个tkinter gui来读取用户的输入。 如果输入不以“”开头https://www.youtube.com/watch?v=“然后会显示一个弹出窗口，通知用户链接无效，用户应该可以重试。在输入的url有效之前，应一直执行此操作。如果输入有效，将调用另一个函数（我还没有编写）

但是，当url无效时，弹出窗口显示，但它永远不会返回到输入，并始终显示弹出窗口，直到我终止程序

这是我的代码，非常感谢您的帮助：

import tkinter as tk
from tkinter import messagebox


class App:
def __init__(self, root):
    # setting title
    root.title("YouTube Downloader")
    # setting window size

    width = 600
    height = 500
    screenwidth = root.winfo_screenwidth()
    screenheight = root.winfo_screenheight()
    alignstr = '%dx%d+%d+%d' % (width, height, (screenwidth - width) / 2, (screenheight - height) / 2)
    root.geometry(alignstr)
    root.resizable(width=False, height=False)

    self.LineEdit = tk.Entry(root)
    self.LineEdit["borderwidth"] = "2px"
    self.LineEdit.place(x=250, y=50, width=289, height=30)

    LineLabel = tk.Label(root)
    LineLabel["text"] = "Enter YouTube URL to analyze:"
    LineLabel.place(x=60, y=50, width=177, height=30)

    GoButton = tk.Button(root)
    GoButton["text"] = "Analyze"
    GoButton.place(x=170, y=130, width=195, height=58)
    GoButton["command"] = self.DisplayInput

    def DisplayInput(self):
        isvalid = False
        while not isvalid:
            result = self.LineEdit.get()
            line = str(result).lower()
            if not line.startswith("https://www.youtube.com/watch?v="):
                messagebox.showerror("Error!", "The URL doesn't seems to be a valid YouTube URL.\n\nPlease try again.")
            else:
                isvalid = True
        print(result)


if __name__ == "__main__":
    root = tk.Tk()
    app = App(root)
    root.mainloop()

---

而不是在DisplayInput函数中使用while循环。只要使用if语句，每当用户输入的url不是以您想要的url开头时，就会打开错误消息框。这将仅在用户按analyze时打开错误消息框

import tkinter as tk
from tkinter import messagebox


class App:
    def __init__(self, root):
        # setting title
        root.title("YouTube Downloader")
        # setting window size

        width = 600
        height = 500
        screenwidth = root.winfo_screenwidth()
        screenheight = root.winfo_screenheight()
        alignstr = '%dx%d+%d+%d' % (width, height, (screenwidth - width) / (screenheight - height) / 2)
        root.geometry(alignstr)
        root.resizable(width=False, height=False)

        self.LineEdit = tk.Entry(root)
        self.LineEdit["borderwidth"] = "2px"
        self.LineEdit.place(x=250, y=50, width=289, height=30)

        #I added this line
        self.isvalid = False

        LineLabel = tk.Label(root)
        LineLabel["text"] = "Enter YouTube URL to analyze:"
        LineLabel.place(x=60, y=50, width=177, height=30)

        GoButton = tk.Button(root)
        GoButton["text"] = "Analyze"
        GoButton.place(x=170, y=130, width=195, height=58)

        def DisplayInput():
            # I changed this part
            result = self.LineEdit.get()
            line = str(result).lower()
            if not line.startswith("https://www.youtube.com/watch?v="):
                messagebox.showerror("Error!", "The URL doesn't seems to be a valid YouTube URL.\n\nPlease try again.")
            else:
                self.isvalid = True
            print(result)
        GoButton["command"] = DisplayInput


if __name__ == "__main__":
    root = tk.Tk()
    app = App(root)
    root.mainloop()

---

更新：我设法解决了这个问题

我对“DisplayInput”功能做了一个小改动：

    def DisplayInput(self):
    result = self.LineEdit.get()
    line = str(result).lower()
    if not line.startswith("https://www.youtube.com/watch?v="):
        messagebox.showerror("Error!", "The URL doesn't seems to be a valid YouTube URL.\n\nPlease try again.")
        # I added the line below
        root.mainloop()
    print(result)

---

@谢谢你的回答。但是，我得到了以下错误：Tkinter回调回溯异常（最近一次调用是最后一次）：在调用返回self.func（*args）文件“C:/Users/s/Google Drive/Studies/Python/My Games/YouTube Downloader/1.py”第41行的DisplayInput GoButton[“command”]=DisplayInput名称错误：名称“DisplayInput”未定义您运行的代码与我发布的代码完全相同吗？我运行了代码，没有收到任何错误。我复制/粘贴了。。。奇怪的是为什么它不起作用