本文介绍了android中的java.lang.nullpointerexception的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧! 问题描述 29岁程序员,3月因学历无情被辞! 嗨所有:) 当我尝试午餐登录活动时我有这个例外 01-12 21:04:28.994:E / AndroidRuntime(1442):java.lang.RuntimeException:无法启动活动ComponentInfo {com.msagroup.sbhmg / com.msagroup.sbhmg.LogIn}:java.lang。 NullPointerException 这里是我的.XML文件:hi all :)I have this exception when I try to lunch log in activity in android01-12 21:04:28.994: E/AndroidRuntime(1442): java.lang.RuntimeException: Unable to start activity ComponentInfo{com.msagroup.sbhmg/com.msagroup.sbhmg.LogIn}: java.lang.NullPointerExceptionhere is my .XML file:<?xml version="1.0" encoding="utf-8"?><RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="match_parent" android:layout_height="match_parent" > <EditText android:id="@+id/etmail" android:layout_width="match_parent" android:layout_height="wrap_content" android:layout_alignParentLeft="true" android:layout_alignParentTop="true" android:layout_marginTop="41dp" android:inputType="textEmailAddress" android:text="@string/email" > <requestFocus /> </EditText> <EditText android:id="@+id/etpassword" android:layout_width="match_parent" android:layout_height="wrap_content" android:layout_alignParentLeft="true" android:layout_below="@+id/etmail" android:layout_marginTop="16dp" android:inputType="textPassword" android:text="@string/password"/> <Button android:id="@+id/btnlogin" android:layout_width="match_parent" android:layout_height="wrap_content" android:layout_alignParentLeft="true" android:layout_below="@+id/etpassword" android:layout_marginTop="16dp" android:text="@string/login" /> <TextView android:id="@+id/tvcomment" android:layout_width="match_parent" android:layout_height="wrap_content" android:layout_alignParentLeft="true" android:layout_centerVertical="true" android:text="@string/empty" /></RelativeLayout> 和.Java文件:and .Java file:package com.msagroup.sbhmg;import android.app.Activity;import android.content.Intent;import android.os.Bundle;import android.view.View;import android.view.View.OnClickListener;import android.widget.Button;import android.widget.EditText;import android.widget.TextView;public class LogIn extends Activity {@Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.login); Button _gotousermain = (Button) findViewById(R.id.btngotoenglishmain); EditText _etmail = (EditText) findViewById(R.id.etmail); EditText _etpassword = (EditText) findViewById(R.id.etmail); final String _email = _etmail.getText().toString(); final String _pass = _etpassword.getText().toString(); final TextView _tvcomm = (TextView) findViewById(R.id.tvcomment); _gotousermain.setOnClickListener(new OnClickListener() {public void onClick(View v) {if(_email == "msay@live.com" && _pass == "123"){startActivity(new Intent(LogIn.this, ArMain.class));}else{if(_email == "msay@live.com" && _pass == "123"){startActivity(new Intent(LogIn.this, ArMain.class));}else{_tvcomm.setText("Invalid Info. !");}}}});}} 谢谢大家:)thank you all :)推荐答案我发现了问题 这是一个愚蠢的问题:\ IDbtngotoenglishmain at Button _gotousermain =(Button )findViewById(R.id.btngotoenglishmain); 在xml文件中不存在 谢谢大家:)I found the problemit was a stupid one :\the ID "btngotoenglishmain"atButton _gotousermain = (Button) findViewById(R.id.btngotoenglishmain);is not exists at the xml filethank you all :) 这篇关于android中的java.lang.nullpointerexception的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持! 上岸,阿里云!