我正在嘗試爲安全目的輸入用戶名和密碼以進入我的應用程序。我試圖設置用戶名和密碼作爲對象,但它給了我上述錯誤。EditText中的EditText(android.content.Context)無法應用於(android.view.View)
任何線索?
我使用過Android Studio
package com.example.xfrehner1.jacksonattendance;
import android.content.Intent;
import android.support.v7.app.AlertDialog;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.view.View;
import android.widget.EditText;
public class LogIn extends AppCompatActivity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_log_in);
}
/*Creates objects for the username and password*/
EditText username = new EditText(findViewById(R.id.userEditText));//it highlights this red
EditText password = new EditText(findViewById(R.id.passEditText));//this too
//AlertDialog wrongUP = new AlertDialog.Builder(LogIn.this).create();
//wrongUP.setTitle("Error");
public void enterUPButton(View v){
Intent afterLog= new Intent(this, firstScreen.class);
if(username.getText().toString().equals("admin") && password.getText().toString().equals("Password1234")){
startActivity(afterLog);
}else{
System.exit(0);
}
}
}
初始化你的'views'裏面的oncreate,同時在類 –
'EditText password = new EditText(findViewById(R.id.passEditText))中持有全局引用'你是怎麼想出來的?你是在哪裏找到那個東西的? – njzk2