本文介绍了OnClick(视图v)抛出空指针异常。有人可以帮忙吗?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
单击(查看v)抛出空指针异常。一直试图将它分类几个小时。但无济于事。如果有人能帮助我解决这个问题,我会很高兴。
谢谢你的时间。
click(view v) is throwing a null pointer exception. have been trying to sort it for hours. but to no avail. I would appriciate if anyone could help me sort this issue.
Thanks for you time.
import android.support.v7.app.ActionBarActivity;
import android.support.v7.app.ActionBar;
import android.support.v4.app.Fragment;
import android.os.Bundle;
import android.util.Log;
import android.view.LayoutInflater;
import android.app.Activity;
import android.view.Menu;
import android.view.MenuItem;
import android.view.View;
import android.view.ViewGroup;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
import android.os.Build;
public class MainActivity extends ActionBarActivity
{
ankita db1 = null;
@Override
protected void onCreate(Bundle savedInstanceState)
{
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
if (savedInstanceState == null)
{
getSupportFragmentManager().beginTransaction().add(R.id.container, new PlaceholderFragment()).commit();
//add();
}
//TextView txt1 = (TextView)findViewById(R.id.userName);
//TextView txt2 = (TextView)findViewById(R.id.password);
db1 = new ankita(this);
Log.e("error", "before db create");
db1.CreateDb();
Log.e("error","after db create");
add();
//db1.insertrecord();
}
public void add()
{
Log.e("hello", "entered db insert");
// TODO Auto-generated method stub
Button addb= (Button) findViewById(R.id.add);
try
{
addb.setOnClickListener(new View.OnClickListener()
{
@Override
public void onClick(View v)
{
Log.e("hello", "entered db click");
// TODO Auto-generated method stub
TextView txt1 = (TextView) findViewById(R.id.userName);
EditText txt = (EditText) findViewById(R.id.userName);
TextView txt2 = (TextView) findViewById(R.id.password);
Log.e("hello", "entered db insert1");
String uname = (String) txt1.getText();
Log.e("hello", "entered db insert2");
String pass = (String) txt2.getText();
Log.e("hello", "entered db insert3");
db1.insertrecord(uname,pass);
}
});}
catch(Exception e)
{
System.out.println("helloji"+e);
}
}
修正了代码块并减少了制表
code block corrected and tabulation reduced
推荐答案
TextView txt1 = (TextView) findViewById(R.id.userName);
EditText txt = (EditText) findViewById(R.id.userName);
两者都使用相同的ID。可能
Both use the same id. Probably
EditText txt
为空。使用正确的ID来修复它。
is null. use the correct id to fix it.
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