在我的Android代码中
我想像json那样获取联系人的姓名，电子邮件和电话号码，然后想要显示。
这是我的代码：
Join但是我得到了nullpointer异常错误：
SELECT *
FROM t1 AS a
inner join t2 AS b
on a.id =b.id
and a.desc != b.desc
错误：
public class MainActivity extends Activity {
public TextView outputText;
String[] phoneNumber;
String[] email;
String name;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
outputText = (TextView) findViewById(R.id.textView1);
try {
//fetchContacts();
outputText.setText(fetchContacts());
} catch (JSONException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
public String fetchContacts() throws JSONException {
Uri CONTENT_URI = ContactsContract.Contacts.CONTENT_URI;
String _ID = ContactsContract.Contacts._ID;
String DISPLAY_NAME = ContactsContract.Contacts.DISPLAY_NAME;
String HAS_PHONE_NUMBER = ContactsContract.Contacts.HAS_PHONE_NUMBER;
Uri PhoneCONTENT_URI = ContactsContract.CommonDataKinds.Phone.CONTENT_URI;
String Phone_CONTACT_ID = ContactsContract.CommonDataKinds.Phone.CONTACT_ID;
String NUMBER = ContactsContract.CommonDataKinds.Phone.NUMBER;
Uri EmailCONTENT_URI = ContactsContract.CommonDataKinds.Email.CONTENT_URI;
String EmailCONTACT_ID = ContactsContract.CommonDataKinds.Email.CONTACT_ID;
String DATA = ContactsContract.CommonDataKinds.Email.DATA;
ContentResolver contentResolver = getContentResolver();
Cursor cursor = contentResolver.query(CONTENT_URI, null, null, null,
null);
List<Contact> contacts = new ArrayList<Contact>();
Gson gson = new Gson();
// Loop for every contact in the phone
if (cursor.getCount() > 0) {
while (cursor.moveToNext()) {
String contact_id = cursor
.getString(cursor.getColumnIndex(_ID));
name = cursor.getString(cursor.getColumnIndex(DISPLAY_NAME));
int hasPhoneNumber = Integer.parseInt(cursor.getString(cursor
.getColumnIndex(HAS_PHONE_NUMBER)));
if (hasPhoneNumber > 0) {
int p = 0;
// Query and loop for every phone number of the contact
Cursor phoneCursor = contentResolver.query(
PhoneCONTENT_URI, null, Phone_CONTACT_ID + " = ?",
new String[] { contact_id }, null);
while (phoneCursor.moveToNext()) {
phoneNumber[p] = phoneCursor.getString(phoneCursor
.getColumnIndex(NUMBER));
p++;
}
phoneCursor.close();
int q = 0;
// Query and loop for every email of the contact
Cursor emailCursor = contentResolver.query(
EmailCONTENT_URI, null, EmailCONTACT_ID + " = ?",
new String[] { contact_id }, null);
while (emailCursor.moveToNext()) {
email[q] = emailCursor.getString(emailCursor
.getColumnIndex(DATA));
q++;
}
emailCursor.close();
contacts.add(new Contact(name, phoneNumber, email));
}
}
}
return gson.toJson(contacts);
}
}
这里我将phonenumber和email存储为数组。是否存在数组初始化问题？任何想法的人？
答案 0 :(得分：1)
您可以使用这些方法从联系人列表中找到联系人
class FetchDeviceContact extends AsyncTask<Void, Integer, String>
{
protected void onPreExecute (){
Constant.showProgressDialog(AddDeviceContactScreeen.this);
}
protected String doInBackground(Void...arg0) {
arrayList.clear();
ContentResolver cr = AddDeviceContactScreeen.this.getContentResolver();
Cursor cur = AddDeviceContactScreeen.this.getContentResolver().query(ContactsContract.Contacts.CONTENT_URI, null, null, null, null);
while (cur.moveToNext()) {
String id = cur.getString(cur.getColumnIndex(ContactsContract.Data._ID));
String name = cur.getString(cur.getColumnIndex(ContactsContract.Data.DISPLAY_NAME));
String photoUri = cur.getString(cur.getColumnIndex(ContactsContract.Data.PHOTO_THUMBNAIL_URI));
Bitmap my_btmp = BitmapFactory.decodeResource(getResources(), R.drawable.ic_arrow_up_blue);
String email = null;
String phoneNo = null;
Cursor phonecur = AddDeviceContactScreeen.this.getContentResolver().query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null,
ContactsContract.CommonDataKinds.Phone.CONTACT_ID + " = ?",
new String[]{id}, null);
if (photoUri != null) {
Uri my_contact_Uri = Uri.parse(photoUri);
try {
my_btmp = MediaStore.Images.Media.getBitmap(AddDeviceContactScreeen.this.getContentResolver(), my_contact_Uri);
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
if (phonecur.getCount() > 0) {
while (phonecur.moveToNext()) {
phoneNo = phonecur.getString(phonecur.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));
}
}
Cursor emailCur = cr.query(
ContactsContract.CommonDataKinds.Email.CONTENT_URI, null,
ContactsContract.CommonDataKinds.Email.CONTACT_ID + " = ?",
new String[]{id}, null);
while (emailCur.moveToNext()) {
// to get the contact names
// = cur1.getString(cur1.getColumnIndex(ContactsContract.CommonDataKinds.Email.DATA)
email = emailCur.getString(emailCur.getColumnIndex(ContactsContract.CommonDataKinds.Email.DATA));
if (email != null) {
System.out.println("Email============== :" + email);
}
}
emailCur.close();
ContactBean bean = new ContactBean();
bean.setName(name);
bean.setEmail(email);
bean.setImage(my_btmp);
bean.setPhone_number(phoneNo);
if (phoneNo == null || email == null) {
} else {
arrayList.add(bean);
}
}
return "";
}
protected void onProgressUpdate(Integer...a){
}
protected void onPostExecute(String result) {
Constant.cancelDialog();
}
}
new FetchDeviceContact().execute();
为它创建 getter 和 setter 类： -
public class ContactBean {
String name;
String email;
Bitmap image;
String phone_number;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
public Bitmap getImage() {
return image;
}
public void setImage(Bitmap image) {
this.image = image;
}
public String getPhone_number() {
return phone_number;
}
public void setPhone_number(String phone_number) {
this.phone_number = phone_number;
}
}
答案 1 :(得分：0)
下面：
phoneNumber[p] = phoneCursor.getString(phoneCursor.getColumnIndex(NUMBER));
NullPointerException为phoneNumber而未初始化，因此导致 null。
使用Cursor. getCount()将其初始化为：
phoneNumber = new String[phoneCursor.getCount()];
phoneCursor.moveToFirst();
while (phoneCursor.moveToNext()) {
//..your code here..
}
答案 2 :(得分：0)
我认为你应该解决：
BasicCredentials credentials=new BasicCredentials(userName,passWord);
JiraClient jira=new JiraClient("https://myproject.info",credentials);
Issue.SearchResult sr = jira.searchIssues("Your project Name(in my case TC)", 100);
for (Issue i : sr.issues) {
System.out.println(i.getSummary());
System.out.println(i.getKey());
etc ...
}