我遇到与联系人有关的问题。我收到了手机联系人并将其存储在我的列表对象中。这是代码
Uri uri = ContactsContract.Data.CONTENT_URI;
String[] projection = {
ContactsContract.Data.CONTACT_ID,
ContactsContract.Data.DISPLAY_NAME,
ContactsContract.Data.PHOTO_ID,
ContactsContract.Data.DATA1
};
Cursor phones = getContentResolver().query(
uri, projection, ContactsContract.Data.MIMETYPE + "='" + ContactsContract.CommonDataKinds.Phone.CONTENT_ITEM_TYPE + "' AND " + ContactsContract.Data.DATA1 + "!=''", null, null);
if (phones.moveToFirst()) {
do {
long ID = phones.getLong(phones.getColumnIndex(projection[0]));
String DisplayName = phones.getString(phones.getColumnIndex(projection[1]));
String photoID = phones.getString(phones.getColumnIndex(projection[2]));
String Key = phones.getString(phones.getColumnIndex(projection[3]));
String photoURI = "null";
if(Key != null && Key.toString().trim().length() > 0 && (Key.startsWith("0") || Key.startsWith("+"))){
if (photoID != null) {
photoURI=String.valueOf(ID);;
//Console.WriteLine("*************************************> id="+ID+" uri="+photoURI.ToString());
}
ContactBean contactModel=new ContactBean(DisplayName,Key,photoID);
list.add(contactModel);
} else {
// No number!!
}
} while (phones.moveToNext());
}
我按照我的要求删除了所有联系人和电子邮件联系人。我的问题是我得到所有的联系人，包括重复的联系人。如果我有一个联系人保存了3次同名和号码，它将获得所有三个联系人。我不想要这个。有没有办法避免这种情况。 getContactResolver查询中的任何内容或我必须删除列表的重复项。任何解决方案或建议？
答案 0 :(得分：4)
你可以试试这个：
ContentResolver cr = getContentResolver();
Cursor cur = cr.query(ContactsContract.Contacts.CONTENT_URI, null, null, null, ContactsContract.Contacts.DISPLAY_NAME + " ASC ");
String lastnumber = "0";
if (cur.getCount() > 0)
{
while (cur.moveToNext())
{
String number = null;
String id = cur.getString(cur.getColumnIndex(ContactsContract.Contacts._ID));
String name = cur.getString(cur.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));
if (Integer.parseInt(cur.getString(cur.getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER))) > 0)
{
Cursor pCur = cr.query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null, ContactsContract.CommonDataKinds.Phone.CONTACT_ID + " = ?", new String[]
{ id }, null);
while (pCur.moveToNext())
{
number = pCur.getString(pCur.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));
Log.e("lastnumber ", lastnumber);
Log.e("number", number);
if (number.equals(lastnumber))
{
}
else
{
lastnumber = number;
Log.e("lastnumber ", lastnumber);
int type = pCur.getInt(pCur.getColumnIndex(Phone.TYPE));
switch (type)
{
case Phone.TYPE_HOME:
Log.e("Not Inserted", "Not inserted");
break;
case Phone.TYPE_MOBILE:
databaseHandler.insertContact(id, name, lastnumber, 0);
break;
case Phone.TYPE_WORK:
Log.e("Not Inserted", "Not inserted");
break;
}
}
}
pCur.close();
}
}
}
这里我首先在sqlite数据库中插入数据，然后按名称编写选择查询。
希望有所帮助
答案 1 :(得分：3)
使用PhoneNumberUtils.compare(a, b)过滤掉重复的数字
val contacts = ArrayList<MyContact>()
val uniqueMobilePhones = ArrayList<String>()
while (cursorPhones.moveToNext()) {
val displayName = cursorPhones.getString(cursorPhones.getColumnIndex(ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME))
val number = cursorPhones.getString(cursorPhones.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER))
val convertedNumber = convert(telman, number)
var duplicate = false
uniqueMobilePhones.forEach { addedNumber ->
if (PhoneNumberUtils.compare(addedNumber, number)) {
duplicate = true
}
}
if (!duplicate) {
uniqueMobilePhones.add(number)
contacts.add(MyContact(displayName, number, convertedNumber.replace(Regex("[ -+()]"), "")))
}
}
答案 2 :(得分：1)
String lastnumber = "0";
ContentResolver cr = getContentResolver();
Cursor cursor = cr.query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, Constants.PROJECTION, null, null, null);
if (cursor != null) {
try {
final int nameIndex = cursor.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME);
final int numberIndex = cursor.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER);
String name, number;
while (cursor.moveToNext()) {
name = cursor.getString(nameIndex);
number = cursor.getString(numberIndex).trim();
number = number.replaceAll("\\s", "");
if (number.equals(lastnumber)) {
} else {
lastnumber = number;
Contact contact = new Contact();
contact.name = name;
contact.phone = number;
mContactList.add(contact);
if (adapter != null)
adapter.notifyDataSetChanged();
System.out.println("ContactFragment.readContact ==>" + name);
}
}
} finally {
cursor.close();
}
}
答案 3 :(得分：1)
下面的代码是在联系人列表中查找重复的号码，以及在联系人列表中出现该号码的频率
public ArrayList<ContactToDelete> findDuplicates(ArrayList<Contact> listContainingDuplicates) {
ArrayList<Contact> duplicatesOrganised = new ArrayList();
ArrayList<ContactToDelete> setToReturn = new ArrayList();
// Collections.sort(listContainingDuplicates);
Collections.sort(listContainingDuplicates, new Comparator<Contact>() {
public int compare(Contact obj1, Contact obj2) {
// ## Ascending order
return obj1.getPhoneNumber().compareToIgnoreCase(obj2.getPhoneNumber()); // To compare string values
// return Integer.valueOf(obj1.empId).compareTo(Integer.valueOf(obj2.empId)); // To compare integer values
// ## Descending order
// return obj2.firstName.compareToIgnoreCase(obj1.firstName); // To compare string values
// return Integer.valueOf(obj2.empId).compareTo(Integer.valueOf(obj1.empId)); // To compare integer values
}
});
int ii, size = listContainingDuplicates.size();
//Orders all the duplicates together along with the unique(non-duplicate)
for (ii = 0; ii < size; ii++) {
if (ii + 1 == size) {
duplicatesOrganised.add(listContainingDuplicates.get(ii));
Log.i("DuplicateOrdered: ", listContainingDuplicates.get(ii).getPhoneNumber() + " " + listContainingDuplicates.get(ii).getName());
} else if (listContainingDuplicates.get(ii).getPhoneNumber().equals(listContainingDuplicates.get(ii + 1).getPhoneNumber())) {
duplicatesOrganised.add(listContainingDuplicates.get(ii));
Log.i("DuplicateOrdered: ", listContainingDuplicates.get(ii).getPhoneNumber() + " " + listContainingDuplicates.get(ii).getName());
} else {
duplicatesOrganised.add(listContainingDuplicates.get(ii));
Log.i("DuplicateOrdered: ", listContainingDuplicates.get(ii).getPhoneNumber() + " " + listContainingDuplicates.get(ii).getName());
}
}
int firstcome = 0;
int start = 0;
boolean present = false;
boolean duplicatefond = false;
int startsetToReturn = 0;
if (!duplicatesOrganised.isEmpty() &&
duplicatesOrganised.size() > 1 &&
!duplicatesOrganised.get(0).getPhoneNumber().equals(duplicatesOrganised.get(1).getPhoneNumber())) {
start = 1;
}
for (int i = 0; i < duplicatesOrganised.size(); i++) {
String currentNumber = duplicatesOrganised.get(i).getPhoneNumber();
if(setToReturn.size()>0){
for (int j = 0; j <setToReturn.size() ; j++) {
if(setToReturn.get(j).getNumber().equals(currentNumber)){
present = true;
}
}
}
int flag = 0;
if(!present) {
for (int j = i+1; j < duplicatesOrganised.size(); j++) {
if (duplicatesOrganised.get(j).getPhoneNumber().equals(currentNumber)) {
duplicatefond = true;
if (flag == 0) {
flag++;
setToReturn.add(new ContactToDelete(duplicatesOrganised.get(i).getPhoneNumber(), duplicatesOrganised.get(i).getName(), flag));
} else
{
flag++;
setToReturn.get(startsetToReturn).setRepeatValue(flag);
}
}
}
if(duplicatefond) {
startsetToReturn++;
duplicatefond = false;
}
}
firstcome = 0;
present = false;
}
startsetToReturn =0;
Log.e("setToReturn", setToReturn.toString());
return setToReturn;
}
我使用的两个模型类是获取完整的联系人列表，而另一个是获取双目编号
没有1个模型舱
public class ContactToDelete {
private String number;
private String name;
private int repeatValue;
private String contactID;
public ContactToDelete(String number, String name, int repeatValue, String contactID) {
this.number = number;
this.name = name;
this.repeatValue = repeatValue;
this.contactID = contactID;
}
public ContactToDelete() {
}
public ContactToDelete(String number, String name, int repeatValue) {
this.number = number;
this.name = name;
this.repeatValue = repeatValue;
}
public String getContactID() {
return contactID;
}
public void setContactID(String contactID) {
this.contactID = contactID;
}
public String getNumber() {
return number;
}
public void setNumber(String number) {
this.number = number;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public int getRepeatValue() {
return repeatValue;
}
public void setRepeatValue(int repeatValue) {
this.repeatValue = repeatValue;
}
}
没有2个模型舱
public class Contact {
public String name;
public String phoneNumber;
public Contact(String name, String phoneNumber, String contactID) {
this.name = name;
this.phoneNumber = phoneNumber;
this.contactID = contactID;
}
public String getContactID() {
return contactID;
}
public void setContactID(String contactID) {
this.contactID = contactID;
}
public String contactID;
public Contact() {
}
public Contact(String name, String phoneNumber) {
this.name = name;
this.phoneNumber = phoneNumber;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getPhoneNumber() {
return phoneNumber;
}
public void setPhoneNumber(String phoneNumber) {
this.phoneNumber = phoneNumber;
}
}
答案 4 :(得分：0)
由于我们正在查询原始联系人列表，因此使用内容提供程序/光标加载器具有多个联系人是显而易见的。我删除重复项的方法是重写哈希码和equals方法。下面是我的代码，可以避免将多个联系人添加到列表中。
import android.os.Parcel;
import android.os.Parcelable;
import android.text.TextUtils;
模型类包含以下字段。您可以根据需要进行修改。
private String name;
private String number;
private boolean isSelected;
现在在模型类中覆盖哈希码和equals方法。
@Override
public boolean equals(Object v) {
boolean retVal = false;
if (v instanceof SelectableContact){
SelectableContact ptr = (SelectableContact) v;
if(ptr != null) {
//We can add some regexpressions to ignore special characters or spaces but
// this consumes a lot of memory and slows down the contact loading.
if(!TextUtils.isEmpty(ptr.number) && !TextUtils.isEmpty(this.number) && ptr.number.equalsIgnoreCase(this.number)) {
retVal = true;
}//if
}
}
return retVal;
}
@Override
public int hashCode() {
int hash = 7;
hash = 17 * hash + (this.number != null ? this.number.hashCode() : 0);
return hash;
}
现在很好。如果列表项的内容相同，则会在添加到列表时声明拒绝。 看下面的示例。让我的模型类为Contact。
public class Contact implements implements Parcelable {
}
一旦您从contentProvider或ContactCursor加载程序获取了联系人， 执行此操作。
List<Contact> contactList = new ArraList<>;
Contact contact = new Contact();
if(!contactList.contains(contact)) {
//add contact to list.
}else {
//remove contact from list.
}
hashcode和equals方法将在添加之前比较列表项的内容。如果存在相同的内容，它将删除。
很好。
有关更多信息，请参阅 Why do I need to override the equals and hashCode methods in Java?