我有一个联系人在移动像名称=“ABC”。 phone number =“123456789”type =“work”google number =“987654321”type =“work”。现在,当我更新号码“123456789”的联系人时,首先获取该联系人的ID,然后更新与phone.type =“工作”的联系人。但问题是,当我更新的联系人,然后联系人将更新的数字,如电话号码和谷歌number.So如何更新只有手机的联系号码,但没有任何其他帐户与此ID加入。我写的代码如下:?当多个号码在一个联系人下更新唯一的手机联系号码联系人
public Long getID(String number) {
Uri uri = Uri.withAppendedPath(PhoneLookup.CONTENT_FILTER_URI,
Uri.encode(number));
Cursor c = getContentResolver().query(uri,
new String[] { PhoneLookup._ID}, null, null, null);
while (c.moveToNext()) {
return c.getLong(c.getColumnIndex(PhoneLookup._ID));
}
return null;
}
public int gettype(String number) {
Uri uri = Uri.withAppendedPath(PhoneLookup.CONTENT_FILTER_URI,
Uri.encode(number));
Cursor c = getContentResolver().query(uri,
new String[] { PhoneLookup.TYPE }, null, null, null);
while (c.moveToNext()) {
return c.getInt(c.getColumnIndex(PhoneLookup.TYPE));
}
return 0;
}
Long id = getID(delnumber);
int contact_type= gettype(delnumber);
String selectPhone = Data.CONTACT_ID+ "=? AND " + Data.MIMETYPE+ "='"+ Phone.CONTENT_ITEM_TYPE+ "'" + " AND " + Phone.TYPE + "=?";
Log.i("type",""+contact_type);
if(contact_type==1)
{String[] phoneArgs = new String[] {String.valueOf(id), String.valueOf(Phone.TYPE_HOME)};
ops.add(ContentProviderOperation.newUpdate(Data.CONTENT_URI).withSelection(selectPhone,phoneArgs).withValue(Phone.NUMBER,getnum).build());}
else if(contact_type==2)
{String[] phoneArgs = new String[] {String.valueOf(id), String.valueOf(Phone.TYPE_MOBILE)};
ops.add(ContentProviderOperation.newUpdate(Data.CONTENT_URI).withSelection(selectPhone,phoneArgs).withValue(Phone.NUMBER,getnum).build());}
else if(contact_type==3)
{String[] phoneArgs = new String[] {String.valueOf(id), String.valueOf(Phone.TYPE_WORK)};
ops.add(ContentProviderOperation.newUpdate(Data.CONTENT_URI).withSelection(selectPhone,phoneArgs).withValue(Phone.NUMBER,getnum).build());}
else
{String[] phoneArgs = new String[] {String.valueOf(id), String.valueOf(Phone.TYPE_MOBILE)};
ops.add(ContentProviderOperation.newUpdate(Data.CONTENT_URI).withSelection(selectPhone,phoneArgs).withValue(Phone.NUMBER,getnum).build());}
您是否找到了解决方案?如果你发现你可以发布它?感谢 – nikmin