Question

我有一个简单的应用程序，用户必须输入用户名和密码才能访问第二个屏幕。要做到这一点，我使用sqlite 的 * PY

class LoginPage(Screen):
    def findAccount(self):
        username = self.ids["login_in"].text
        password = self.ids["pass_in"].text
        c.execute('''SELECT * FROM customer WHERE name = ? AND password = ?;
                  ''', (username, password))
        records = c.fetchone() #returns (1, Bob, qwerty, bob@gmail.com, 071234556)
        return records
    def verify(self):
        try:
            records = LoginPage.findAccount(self)
            if self.ids["login_in"].text == records[1] and self.ids["pass_in"].text == records[2]:
                self.manager.current = "user"
        except TypeError:
            print("Something is not correct")

class UserInfoPage(Screen):
    def show_info(self):
        pass
class ScreenManagement(ScreenManager):
    pass

kv_file = Builder.load_file("login.kv")
class LoginApp(App):
    def build(self):
        return kv_file

*的Kv

ScreenManager:
    LoginPage:
    UserInfoPage:

<LoginPage>:
    orientation: "vertical"
    name: "login"
    BoxLayout:
        TextInput:
            id: login_in
        TextInput:
            id: pass_in
            password: True
        Button:
            text: "GO"
            on_release:
                root.findAccount()
                root.verify()
<UserInfoPage>:
    name: "user"
    BoxLayout:
        Label:
            id: the_id
            text: 'id: '
        Label:
            text: 'name: '
        Label:
            text: 'email: '
        Label:
            text: 'phone number: '

使用＆＃39; show_info（）＆＃39;方法我想在＆＃39; UserInfoPage＆＃39;上显示用户的信息。屏幕。要做到这一点，我需要来自变量＆＃39;记录的信息。来自＆＃39; findAccount＆＃39;帐户方法。我试着这样做：

def show_info(self):
    information = LoginPage.findAccount(self)
    user_id = information[0]
    self.ids["the_id"].text = self.ids["the_id"].text + str(user_id)

但它会引发错误，因为＆＃39; findAccount（）＆＃39;方法链接到第一个屏幕。如何保存我在第一个屏幕上获得的信息并将其传输到第二个屏幕？谢谢。

Answer 1

首先，我们必须纠正某些错误：

没有必要在findAccount()中致电.kv，因为verify()正在.py中调用它。
LoginPage.findAccount(self)语句不正确，您必须直接使用self来调用它，即正确的是self.findAccount()。
避免使用try，除非没有必要，否则在您的情况下，只需验证records不是None。

要创建新方法不需要添加信息，更好的方法是创建属性并通过屏幕分配，以获得适当的屏幕，我们使用get_screen() ScreenManager方法}}

<强> *。PY

class LoginPage(Screen):
    def findAccount(self):
        username = self.ids["login_in"].text
        password = self.ids["pass_in"].text
        c.execute('''SELECT * FROM customer WHERE name = ? AND password = ?;
                  ''', (username, password))
        records = c.fetchone() #returns (1, Bob, qwerty, bob@gmail.com, 071234556)
        return records

    def verify(self):
        records = self.findAccount()
        if records is not None:
            _id, name, password, email, phone_number = records
            if self.ids["login_in"].text == name and self.ids["pass_in"].text == password:
                user_screen = self.manager.get_screen("user")
                user_screen._id = str(_id)
                user_screen._name = name
                user_screen._email = email
                user_screen._phone_number = phone_number

                self.manager.current = "user"
            else:
                print("Something is not correct")

class UserInfoPage(Screen):
    pass

class ScreenManagement(ScreenManager):
    pass

<强> *。KV

ScreenManager:
    LoginPage:
    UserInfoPage:

<LoginPage>:
    orientation: "vertical"
    name: "login"
    BoxLayout:
        TextInput:
            id: login_in
        TextInput:
            id: pass_in
            password: True
        Button:
            text: "GO"
            on_release:
                root.verify()

<UserInfoPage>:
    name: "user"

    _id: ""
    _name: ""
    _email: ""
    _phone_number: ""
    BoxLayout:
        Label:
            text: 'id: ' + root._id
        Label:
            text: 'name: ' + root._name
        Label:
            text: 'email: ' + root._email
        Label:
            text: 'phone number: ' + root._phone_number

Kivy：在屏幕之间传输信息

1 个答案: