Android 获取 phone 联系人并删除重复项

Question

我遇到了与联系人相关的问题。我得到了 phone 个联系人并将它们存储在我的列表对象中。这是它的代码

  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());
    }

我正在获取所有联系人，并且已根据我的要求删除了电子邮件联系人。我的问题是我正在获取所有联系人，包括重复联系人。如果我用相同的姓名和号码保存了 3 次联系人，它会获取所有三个联系人。我不想要这个。有什么办法可以避免这种情况。 getContactResolver 查询中的任何内容，或者我必须删除列表中的重复项。任何解决方案或建议？

Answer 1

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();
                }

            }
        }

这里我先在 sqlite 数据库中插入数据，然后写 select 按名称分组查询。

希望对您有所帮助

Answer 2

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("[ -+()]"), "")))
                    }
                }

Answer 3

 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();
        }
    }

Answer 4

使用内容 provider/cursor 加载程序有多个联系人是显而易见的，因为我们正在查询原始联系人列表。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，它将避免向列表中添加多个联系人。

import android.os.Parcel;
import android.os.Parcelable;
import android.text.TextUtils;

模型 class 包含以下字段。您可以根据需要修改。

private String name;
private String number;
private boolean isSelected;

现在覆盖模型中的 hashcode 和 equals 方法 class。

@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;
}

现在可以出发了。如果列表项的内容相同，则在添加到列表时会声明拒绝。看下面example.Let我的模特class联系方式

public class Contact implements implements Parcelable {

}

从 contentProvider 或 ContactCursor 加载程序获取联系人后，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 方法会在添加之前比较列表项的内容。如果存在相同的内容，它将被删除。

很高兴去。

更多信息请参考 Why do I need to override the equals and hashCode methods in Java?

Answer 5

这里的代码是查找联系人列表中的重复号码以及该号码在联系人列表中出现的频率

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class，一个是获取完整的联系人列表，另一个是获取重复号码

没有 1 个模型 Class

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;
    }
}

第 2 型号 Class

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;
    }
}

Android 获取 phone 联系人并删除重复项

Android get phone contacts and remove duplicates

java

android

android-contacts