android – 获取空指针异常？

2019-10-03 12:25:11 阅读：181 来源： 互联网

我试图将一个数组中的联系人名称和他们的类型放在另一个数组中,但是无法通过空指针异常来完成.这是我的代码.我已经指出了我得到空指针异常的行.请帮忙..提前致谢.

package application.test;
import android.app.Activity;
import android.content.ContentResolver;
import android.database.Cursor;
import android.database.SQLException;
import android.os.Bundle;
import android.provider.ContactsContract;
import android.provider.ContactsContract.CommonDataKinds.Phone;  
import android.provider.ContactsContract.Contacts.Data;
import android.util.Log;
import android.view.LayoutInflater;
import android.view.ViewGroup.LayoutParams;
import android.widget.LinearLayout;
import android.widget.ListView;
import android.widget.RelativeLayout;

public final class TestActivity extends Activity {
String[] name;
String[] phoneType;
ListView lv;
ListViewAdapter lva;


    public static final String TAG = "ContactManager";
@Override
public void onCreate(Bundle savedInstanceState)
{
    Log.v(TAG, "Activity State: onCreate()");
    super.onCreate(savedInstanceState);
    LayoutParams params = new RelativeLayout.LayoutParams(LayoutParams.FILL_PARENT,LayoutParams.WRAP_CONTENT);        
    LinearLayout mainLayout=new LinearLayout(this);
    mainLayout.setOrientation(LinearLayout.VERTICAL);               
    LayoutInflater layoutInflater = getLayoutInflater();        
    mainLayout.addView(layoutInflater.inflate(R.layout.main,null));
    mainLayout.addView(layoutInflater.inflate(R.layout.extra,null));

    this.addContentView(mainLayout, params);

      lv = (ListView)findViewById(android.R.id.list);
     lva = new ListViewAdapter(this,name,phoneType); 
    lv.setAdapter(lva);
    testGetContacts();
}


private void testGetContacts() { 

        ContentResolver cr = getContentResolver();

        String[] projection = new String[] { Data._ID,
                ContactsContract.Contacts.DISPLAY_NAME, Phone.TYPE}; 

        Cursor cur = cr.query(ContactsContract.Data.CONTENT_URI,
                projection, null, null, null); 


        if (cur != null && cur.moveToFirst()) { 

        try {

            int indexID =  cur.getColumnIndexOrThrow(ContactsContract.Contacts._ID);
            int indexName = cur.getColumnIndexOrThrow(ContactsContract.Contacts.DISPLAY_NAME);
             int indexPhoneType = cur.getColumnIndexOrThrow(Phone.TYPE);

          while (cur.moveToNext()) {
             int  i=1;
              String id = cur.getString(indexID);    
 //HERE LIES NULL POINTER EXCEPTION   name[i] = cur.getString(indexName);  
 //HERE TOO              phoneType[i] =  cur.getString(indexPhoneType);

             i++;


              System.out.println(id + "\n");
              System.out.println(name + "\n");
              System.out.println(phoneType + "\n");


          }


        } catch (SQLException sqle) {
           //handling exception       
        } finally { 
         if (!cur.isClosed()) {
             cur.close();
         }     
     }

        }

}
}

解决方法:

在使用之前初始化String []名称.
你可以这样做：

name=new String[cur.getCount()];
String s="";
while (cur.moveToNext()) {

     int  i=1;
     String id = cur.getString(indexID);    
     name[i] = cur.getString(indexName);  
     phoneType[i] =  cur.getString(indexPhoneType);         

     //System.out.println(id + "\n");
     //System.out.println(name + "\n");
     //System.out.println(phoneType + "\n");

     s=s+"id="+id+" name="+name[i]+" phoneType="+phoneType[i]+"\n";
     i++;
}
Toast.makeText(getApplicationContext(),i+" - "+s).show();

编辑：

在布局文件夹中创建一个xml文件.

main.xml中

<LinearLayout
    xmlns:android="http://schemas.android.com/apk/res/android"
    android:layout_width="fill_parent"
    android:layout_height="fill_parent" android:orientation="vertical" >    

    <ListView
        android:layout_width="fill_parent"
        android:layout_height="fill_parent"
        android:id="@+id/listview"
        android:cacheColorHint="#0000"
        />
</LinearLayout>

现在在你的TestActivity.class中：

public final class TestActivity extends Activity {

String[] name;
String[] phoneType;
ListView lv;
String s[];
public static final String TAG = "ContactManager";

@Override
public void onCreate(Bundle savedInstanceState)
{
    super.onCreate(savedInstanceState);
    setContentView(R.id.main);        

    testGetContacts();

    lv = (ListView)findViewById(R.id.listview);
    ArrayAdapter<String> sa=new ArrayAdapter<String>(context, android.R.layout.simple_list_item_1,s);
    lv.setAdapter(sa);        
}


private void testGetContacts() { 

    ContentResolver cr = getContentResolver();    
    String[] projection = new String[] { Data._ID,
                ContactsContract.Contacts.DISPLAY_NAME, Phone.TYPE};     
    Cursor cur = cr.query(ContactsContract.Data.CONTENT_URI,
                projection, null, null, null);     

    if (cur != null && cur.moveToFirst()) { 

        try {

            int indexID =  cur.getColumnIndexOrThrow(ContactsContract.Contacts._ID);
            int indexName = cur.getColumnIndexOrThrow(ContactsContract.Contacts.DISPLAY_NAME);
             int indexPhoneType = cur.getColumnIndexOrThrow(Phone.TYPE);

          name=new String[cur.getCount()];
          s=new String[cur.getCount()];

          while (cur.moveToNext()) {

               int  i=1;
               String id = cur.getString(indexID);    
               name[i-1] = cur.getString(indexName);  
               phoneType[i-1] =  cur.getString(indexPhoneType);       


              String temp="id="+id+"-name="+name[i-1]+"-phoneType="+phoneType[i-1];
              s[i-1]=temp;
              i++;
}

}

标签：android,nullpointerexception
来源： https://codeday.me/bug/20191003/1849415.html

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android – 获取空指针异常？