Question

初学者，我正在创建一个非常简单的“登录”样式程序，目前我使用if语句来查看用户输入的内容是否与我已声明的内容相匹配，如果他们做得对，他们会得到“登录“消息，如果没有，则程序终止。

但是，我希望这样，用户在终止之前最多可以尝试3次获取凭据。我该怎么做呢？

import tkinter as tk

class NumberBox():
    def __init__(self, root,  display_number):
        self.number_label = tk.Label(root, text=display_number)
        self.number_label.pack()

    def changeNumber(self, display_number):
        self.number_label.config(text=display_number)

class NumberBoxesList():    
    def __init__(self, root,  start_number = 5):
        self.number = start_number
        self.root = root
        self.boxes = []
        tk.Button(root, text="Add Class", command= self.addBox).pack()
        tk.Button(root, text="Number UP", command=lambda: self.change_number("add")).pack()
        tk.Button(root, text="Number DOWN", command=lambda: self.change_number("subtract")).pack()

    def addBox(self):
        self.boxes.append(NumberBox(self.root,  self.number))

    def change_number(self,  operation):
        if operation == "add":
            self.number += 1
        if operation == "subtract":
            self.number -= 1
        for box in self.boxes:
            box.changeNumber(self.number)

root = tk.Tk()
boxList = NumberBoxesList(root)

for _ in range(5):
    boxList.addBox()

root.mainloop()

非常感谢

Answer 1

我也会使用这样的循环：

    List<String> usernames = accessClass.setUsernames();
    List<String> passwords = accessClass.setPasswords();

    while(attempts > 0 && !loggedIn) {

        System.out.print("Please enter your username: ");

        user1 = wordScan.next();
        int attempts = 3;
        boolean loggedIn = false;

        attempts = attempts -1;
        if(usernames.contains(user1)){

            System.out.print("Now enter your password: ");

            pass1=wordScan.next();

            if(passwords.contains(pass1)){

                System.out.println("Logged in.");
                System.out.println("Your username is "+user1);
                loggedIn = true;
                random.setSeed(System.currentTimeMillis());
                numGen = random.nextInt(899999)+100000;

                System.out.println("Your access code is: "+numGen);
            }
        }
    }

    if(!loggedIn) {
        System.out.println("Invalid Username or password... Terminating.");
        System.exit(0);
    }

    wordScan.close();
}

Answer 2

如果你想最多循环3次，这样的事情可以解决问题：

for (int i = 0; i<3; i++)
{
 ... your code
 // when username + password match : use break; at the end in order not to do the remaining loops
}

Answer 3

for或while循环应该足够了，如果你的用户和密码数组是链接的（user1有pass1，user2有pass2等等），你可以获得列表中的索引使用indexOf并比较两者：

if(usernames.contains(user1)){
    System.out.print("Now enter your password: ");
    for(int i = 0; i < 3; i++){
        pass1 = wordScan.next();
        if(passwords.contains(pass1) 
                && usernames.indexOf(user1) == passwords.indexOf(pass1)){
            // correct
            break;
        } else if (i == 2){
            // incorrect and end of try
        }
    }
}

Answer 4

试试这段代码：

INSERT ... ON CONFLICT DO NOTHING/UPDATE

我如何激发重试循环

4 个答案: