在我的代码中,我试图进行AlertDialog登录,但是一旦按下按钮就无法关闭它,有人知道吗?我尝试从构建器创建AlertDialog对象,但是它不起作用。这是我的代码:
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答案 0 :(得分:0)
您应该致电builder.create()
而不是致电builder.show()
。它会向您返回一个对话框实例,然后将可以调用show()
和dismiss()
方法。
答案 1 :(得分:0)
您的函数应如下所示:
model.add(LSTM(10, return_sequences = False, input_shape = (8, 8)))
然后您可以像这样调用它:
public AlertDialog createSigninDialog (){
final AlertDialog.Builder builder = new AlertDialog.Builder(MainActivity.this);
LayoutInflater inflater = MainActivity.this.getLayoutInflater();
View view = inflater.inflate(R.layout.dialog_signin, null);
builder.setView(view);
final AlertDialog dialog = builder.create();
signin = (Button)view.findViewById(R.id.singing);
user = (EditText)view.findViewById(R.id.user_input);
password = (EditText)view.findViewById(R.id.password_input);
signin.setOnClickListener(
new View.OnClickListener() {
@Override
public void onClick(View v) {
if (user.getText().toString().equals("1234567890") && password.getText().toString().equals("1234")){
Toast.makeText(MainActivity.this, "Login", Toast.LENGTH_SHORT).show();
//Dismiss here
dialog.dismiss();
} else {
Toast.makeText(MainActivity.this, "Datos incorrectos", Toast.LENGTH_SHORT).show();
//Dismiss here
dialog.dismiss();
}
}
}
);
return dialog;
}