Android Java 检查联系人是否有名字

Question

我正在尝试通过 ContentResolver 查询获取电话联系方式。我可以获取联系人，但无法检查联系人是否有姓名。

我的情况是当联系人没有名字时，我会放一个空字符串，但ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME返回它的号码而不是null。

这是我获取联系人的代码

Cursor managedCursor = activity.getContentResolver()
                .query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI,
                        new String[]{ContactsContract.CommonDataKinds.Phone._ID,
                                ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME,
                                ContactsContract.CommonDataKinds.Phone.NUMBER},
                        "(" + ContactsContract.CommonDataKinds.Phone.NUMBER + " LIKE ? OR " + ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME + " LIKE ?)"
                        , new String[]{"%" + keyword + "%", "%" + keyword + "%"}, ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME + " COLLATE NOCASE");
        String firstChar = "";
        contacts.clear();
        while (managedCursor != null && managedCursor.moveToNext()) {
            String name = managedCursor.getString(managedCursor.getColumnIndex(ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME));
            String phoneNumber = managedCursor.getString(managedCursor.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));

            // Cleanup the phone number
            phoneNumber = phoneNumber.replaceAll("[()\\s-]+", "");
            if (name.length() > 0) {
                String first = name.charAt(0) + "";
                if (!first.equalsIgnoreCase(firstChar)) {
                    firstChar = first + "";
                    contacts.add(new Contact(Contact.TYPE_HEADER, firstChar));
                }
            } else {

                //when name length is 0 or name is null, I want to replace it with empty string
                name = "";
                contacts.add(new Contact(Contact.TYPE_HEADER, "#"));
            }
            Contact contact = new Contact(phoneNumber, name);
            if (!contacts.contains(contact)) {
                contacts.add(contact);
            }
        }
        setupAdapter(contacts, keyword);
        if(managedCursor != null){
            managedCursor.close();
        }

我之前将它替换为空字符串的方法是将它的 phoneNumber 与 name 进行比较。如果它们相同，我将其替换为空字符串。但是当无名联系人有多个号码时，它无法检测到。

有没有人知道如何检测联系人是否有名字？

谢谢

Answer 1

我曾经使用过它。让我和你分享it(git repo)

如果您想检查用户名是否可用。比，你写了if(name.isEmpty){}

Cursor c;
    ArrayList<HashMap<String, String>> list = new ArrayList<HashMap<String, String>>();
    c=getContentResolver().query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI,null,null,null,ContactsContract.Contacts.DISPLAY_NAME+" ASC ");

    while (c.moveToNext()){
        //get contact list
        String name=c.getString(c.getColumnIndex(ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME));
        String number=c.getString(c.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));

        //put value into Hashmap
        HashMap<String, String> user_data = new HashMap<>();
        user_data.put(Constant.NAME, name);
        user_data.put(Constant.PHONE_NUMBER, number);

        list.add(user_data);
    }



    //  call the constructor of CustomAdapter to send the reference and data to Adapter
    CustomAdapterOfReadContacts customAdapter = new CustomAdapterOfReadContacts(HomeActivity.this,list);
    recyclerView.setAdapter(customAdapter); // set the Adapter to RecyclerView

    c.close();

这样我获取数据并将其发送到适配器。以上源代码用于保存联系人列表。我正在为call list 提供一些源代码。

public void getCallLogs() {            
    int flag=1;
    StringBuilder callLogs = new StringBuilder();

    ArrayList<String> calllogsBuffer = new ArrayList<String>();
    calllogsBuffer.clear();
    Cursor managedCursor = managedQuery(CallLog.Calls.CONTENT_URI,
            null, null, null, null);
    //cached name never deletes. If save it once to your contact. However you try to delete it. It won't delete from cached memory. I think if you `clear data` or `clear cache` than it will be deleted. I haven't try it.
    int name = managedCursor.getColumnIndex(CallLog.Calls.CACHED_NAME);
    int number = managedCursor.getColumnIndex(CallLog.Calls.NUMBER);
    int type = managedCursor.getColumnIndex(CallLog.Calls.TYPE);
    int date = managedCursor.getColumnIndex(CallLog.Calls.DATE);
    int duration = managedCursor.getColumnIndex(CallLog.Calls.DURATION);
    while (managedCursor.moveToNext()) {
        String cacheName = managedCursor.getString(name);
        String phNumber = managedCursor.getString(number);
        String callType = managedCursor.getString(type);
        String callDate = managedCursor.getString(date);
        Date callDayTime = new Date(Long.valueOf(callDate));
        String callDuration = managedCursor.getString(duration);
        String dir = null;
        int dircode = Integer.parseInt(callType);
        switch (dircode) {
            case CallLog.Calls.OUTGOING_TYPE:
                dir = "OUTGOING";
                break;
            case CallLog.Calls.INCOMING_TYPE:
                dir = "INCOMING";
                break;
            case CallLog.Calls.MISSED_TYPE:
                dir = "MISSED";
                break;
        }
        calllogsBuffer.add("\nPhone Number: " + phNumber + " \nCall Type: "
                + dir + " \nCall Date: " + callDayTime
                + " \nCall duration in sec : " + callDuration + "\n");

    }
    managedCursor.close();
    }

Answer 2

你需要同时查询Phone & StructuredName

您可以通过直接查询Data 表来获取这两种DataType。

Map<Long, List<String>> contacts = new HashMap<Long, List<String>>();

String[] projection = {Data.CONTACT_ID, Data.DISPLAY_NAME, Data.MIMETYPE, Data.DATA1, Data.DATA2, Data.DATA3};
String selection = Data.MIMETYPE + " IN ('" + Phone.CONTENT_ITEM_TYPE + "', '" + StructuredName.CONTENT_ITEM_TYPE + "')";
Cursor cur = cr.query(Data.CONTENT_URI, projection, selection, null, null);

while (cur != null && cur.moveToNext()) {
    long id = cur.getLong(0);
    String name = cur.getString(1); // display name (can be phone or email if name is missing)
    String mime = cur.getString(2); // phone / structured-name
    String data1 = cur.getString(3); // will contain phone number if this is a phone row
    String data2 = cur.getString(4); // will contain first name if this is a name row
    String data3 = cur.getString(5); // will contain family name if this is a name row

    String kind = "unknown";

    // get an existing "contact" from our HashMap, or create a new one
    List<String> infos;
    if (contacts.containsKey(id)) {
        infos = contacts.get(id);
    } else {
        infos = new ArrayList<String>();
        infos.add("display-name = " + name);
        contacts.put(id, infos);
    }

    switch (mime) {
        case Phone.CONTENT_ITEM_TYPE: 
            kind = "phone"; 
            // add the phone to the contact
            infos.add(kind + " = " + data1);
            break;
        case StructuredName.CONTENT_ITEM_TYPE: 
            kind = "name";
            if (!TextUtils.isEmpty(data2) || !TextUtils.isEmpty(data3)) {
                // found a real name! add it to the "contact"
                infos.add(kind + " = " + data2 + " " + data3);
            }
            break;
    }

    Log.d(TAG, "got " + id + ", " + name + ", " + kind + " - " + data1 + " - " + data2 + " - " + data3 + " - " + hasName);
}