Question

当我启动我的应用程序时出现此错误，我该如何解决？

12112-12112 / com.javacodegeeks.androidqrcodeexample E / AndroidRuntime：致命异议：主要 java.lang.RuntimeException：无法实例化活动ComponentInfo {com.javacodegeeks.androidqrcodeexample / com.javacodegeeks.androidqrcodeexample.Login}：显示java.lang.NullPointerException

这是我的.xml文件：

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
    package="com.javacodegeeks.androidqrcodeexample"
    android:versionCode="1"
    android:versionName="1.0" >

    <uses-sdk
        android:minSdkVersion="8"
        android:targetSdkVersion="19" />

    <application
        android:allowBackup="true"
        android:icon="@drawable/ic_launcher"
        android:label="@string/app_name" >
        <activity
            android:name="com.javacodegeeks.androidqrcodeexample.AndroidBarcodeQrExample"
            android:label="@string/app_name" >
        </activity>
        <activity
            android:name="com.javacodegeeks.androidqrcodeexample.Login"
            android:label="@string/app_name" >
            <intent-filter>
                <action android:name="android.intent.action.MAIN" />

                <category android:name="android.intent.category.LAUNCHER" />
            </intent-filter>
        </activity>
    </application>

</manifest>

在这里我的活动：

package com.javacodegeeks.androidqrcodeexample;

/**
 * Created by Andre on 14/06/15.
 */

import android.app.Activity;
import android.app.AlertDialog;
import android.content.ActivityNotFoundException;
import android.content.DialogInterface;
import android.content.Intent;
import android.net.Uri;
import android.os.Bundle;
import android.view.View;
import android.widget.EditText;
import android.widget.Toast;


public class Login  extends Activity {

    EditText username = (EditText)findViewById(R.id.editText);
    EditText password = (EditText)findViewById(R.id.editText2);

    public void onCreate(Bundle savedInstanceState) {

        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_login);
    }

    public void login(View view) {
        if (username.getText().toString().equals("admin") && password.getText().toString().equals("admin")) {
            Toast toast = Toast.makeText(this, "CORRECT PW", Toast.LENGTH_LONG);
            toast.show();
            //correcct password
            Intent i = new Intent(getApplicationContext(), AndroidBarcodeQrExample.class);
            i.putExtra("Disco", username.getText().toString());
            startActivity(i);
        } else {
            Toast toast = Toast.makeText(this, "UNCORRECT PW", Toast.LENGTH_LONG);
            toast.show();
            //wrong password
        }
    }

}

Answer 1

移动线条：

EditText username = (EditText)findViewById(R.id.editText);
EditText password = (EditText)findViewById(R.id.editText2);

在onCreate

内

EditText username;
EditText password;

public void onCreate(Bundle savedInstanceState) {

    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_login);
    username = (EditText)findViewById(R.id.editText);
    password = (EditText)findViewById(R.id.editText2);
}

无法实例化活动ComponentInfo启动我的应用程序

1 个答案: