我可以检索联系人ID,但稍后我希望根据联系人ID单独检索电话号码。下面的代码返回电话号码的空结果。 (我希望以后能够一起检索姓名和电话号码并填写视图,但我只是想让电话号码先工作)。
在我的onCreate中我有这段代码
String phoneNum = getPhoneNumber(myID);
TextView phoneTextView = (TextView) findViewById(R.id.textViewPhone);
phoneTextView.setText(phoneNum);
这是getPhoneNumber()
的方法 protected String getPhoneNumber(String id) {
ArrayList<String> phones = new ArrayList<String>();
Cursor cursor = getContentResolver().query(
ContactsContract.CommonDataKinds.Phone.CONTENT_URI,
null,
ContactsContract.CommonDataKinds.Phone.CONTACT_ID + " = ?",
new String[]{id}, null);
while (cursor.moveToNext()) {
phones.add(cursor.getString(cursor.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER)));
}
cursor.close();
String phoneNum;
phoneNum = phones.get(0);
return phoneNum;
}//end getPhoneNumber();
}
这会产生错误java.lang.IndexOutOfBoundsException:索引0无效,大小为0,我计划为其创建一些错误处理。但是,我仍然确定我有前面代码中的ID,所以我不知道为什么ArrayList返回null。如果您想查看该代码,它也在我的onCreate中:
Cursor cursor = getContentResolver().query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null, null, null, null);
if (cursor.getCount() != 0) {
int numContacts = cursor.getCount();
ArrayList<String> idList = new ArrayList<>();
Random rand = new Random();
int randomNum = rand.nextInt(numContacts);
while (cursor.moveToNext()) {
String id = cursor.getString(cursor.getColumnIndex(ContactsContract.Contacts._ID));
idList.add(id);
}
myID = idList.get(randomNum);
String myString = Integer.toString(randomNum);
TextView myTextView = (TextView) findViewById(R.id.textViewID);
myTextView.setText(myString);
if (myID != null) {
myTextView.setText(myID);
} else {
myTextView.setText("Try Again!");
}
} else {
Toast.makeText(getApplicationContext(), "Your have no contacts.", Toast.LENGTH_SHORT).show();
}
cursor.close();
答案 0 :(得分:1)
// You can fetch the Contact Number and Email With Following Methods.
String phone = getPhoneNumber(ContactId);
String email = getEmail("" + ContactId);
private String getPhoneNumber(long id) {
String phone = null;
Cursor phonesCursor = null;
phonesCursor = queryPhoneNumbers(id);
if (phonesCursor == null || phonesCursor.getCount() == 0) {
// No valid number
//signalError();
return null;
} else if (phonesCursor.getCount() == 1) {
// only one number, call it.
phone = phonesCursor.getString(phonesCursor
.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));
} else {
phonesCursor.moveToPosition(-1);
while (phonesCursor.moveToNext()) {
// Found super primary, call it.
phone = phonesCursor.getString(phonesCursor
.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));
break;
}
}
return phone;
}
private Cursor queryPhoneNumbers(long contactId) {
ContentResolver cr = getContentResolver();
Uri baseUri = ContentUris.withAppendedId(ContactsContract.Contacts.CONTENT_URI,
contactId);
Uri dataUri = Uri.withAppendedPath(baseUri,
ContactsContract.Contacts.Data.CONTENT_DIRECTORY);
Cursor c = cr.query(dataUri, new String[]{ContactsContract.CommonDataKinds.Phone._ID, ContactsContract.CommonDataKinds.Phone.NUMBER,
ContactsContract.CommonDataKinds.Phone.IS_SUPER_PRIMARY, ContactsContract.RawContacts.ACCOUNT_TYPE,
ContactsContract.CommonDataKinds.Phone.TYPE,
ContactsContract.CommonDataKinds.Phone.LABEL},
ContactsContract.Data.MIMETYPE + "=?",
new String[]{ContactsContract.CommonDataKinds.Phone.CONTENT_ITEM_TYPE}, null);
if (c != null && c.moveToFirst()) {
return c;
}
return null;
}
private String getEmail(String id) {
String email = "";
ContentResolver cr = getContentResolver();
Cursor emailCur = cr.query(
ContactsContract.CommonDataKinds.Email.CONTENT_URI,
null,
ContactsContract.CommonDataKinds.Email.CONTACT_ID + " = ?",
new String[]{id}, null);
while (emailCur.moveToNext()) {
// This would allow you get several email addresses
// if the email addresses were stored in an array
email = emailCur.getString(
emailCur.getColumnIndex(ContactsContract.CommonDataKinds.Email.DATA));
// String emailType = emailCur.getString(
// emailCur.getColumnIndex(ContactsContract.CommonDataKinds.Email.TYPE));
}
emailCur.close();
return email;
}
答案 1 :(得分:0)
我无法根据联系人ID成功检索代码 - 但这篇帖子给了我一个线索:Retrieving a phone number with ContactsContract in Android - function doesn't work
我已修改原始代码以在DISPLAY_NAME上进行查询,现在情况正常。这是我用来检索电话号码的方法:
private String retrieveContactNumber(String contactName) {
Log.d(TAG, "Contact Name: " + contactName);
Cursor cursorPhone = getContentResolver().query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI,
new String[]{ContactsContract.CommonDataKinds.Phone.NUMBER},
ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME + " = ? AND " +
ContactsContract.CommonDataKinds.Phone.TYPE + " = " +
ContactsContract.CommonDataKinds.Phone.TYPE_MOBILE,
new String[]{contactName},
null);
if (cursorPhone.moveToFirst()) {
contactNumber = cursorPhone.getString(cursorPhone.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));
}
cursorPhone.close();
Log.d(TAG, "Contact Phone Number: " + contactNumber);
return contactNumber;
}