如何讓用戶從聯系人中選擇一個電話號碼，然后獲取電話號碼，給定姓名和姓氏

Question

我到現在所做的是：

點擊某個按鈕時：

Intent intent = new Intent(Intent.ACTION_PICK);
intent.setType(ContactsContract.CommonDataKinds.Phone.CONTENT_TYPE);
startActivityForResult(intent, CONTACT_PICKER_RESULT); 

然后：

@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
    if (resultCode == RESULT_OK) {  
        switch (requestCode) {  
        case CONTACT_PICKER_RESULT:  
            if (data != null) {
                Uri uri = data.getData();
                if (uri != null) {
                    Cursor cursor = null;
                    Cursor cursorStructuredName = null;
                    try{    
                        cursor = getContentResolver().query(uri, null, null, null, null);      
                        boolean hasNext = cursor.moveToNext();
                        String contactId = cursor.getString(cursor.getColumnIndex(ContactsContract.Contacts._ID));
                        String hasPhoneNumber = cursor.getString(cursor.getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER));
                        String phoneNumber = null;      

                        Cursor phones = getContentResolver().query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null,ContactsContract.CommonDataKinds.Phone.CONTACT_ID +" = "+ contactId,null, null);

                        if (phones.moveToFirst()) {//does not get inside the if
                            phoneNumber = phones.getString(phones.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));
                        }
                        phones.close();


                        // projection
                        String[] projection = new String[] {ContactsContract.CommonDataKinds.StructuredName.DISPLAY_NAME, 
                                ContactsContract.CommonDataKinds.StructuredName.GIVEN_NAME,
                                ContactsContract.CommonDataKinds.StructuredName.MIDDLE_NAME,
                                ContactsContract.CommonDataKinds.StructuredName.FAMILY_NAME,
                                ContactsContract.CommonDataKinds.StructuredName.PREFIX,
                                ContactsContract.CommonDataKinds.StructuredName.SUFFIX};


                        String where = ContactsContract.Data.RAW_CONTACT_ID + " =?"; 
                        String[] whereParameters = new String[]{contactId};

                        //Request
                        cursorStructuredName = getContentResolver().query(ContactsContract.Data.CONTENT_URI, projection, where, whereParameters, null);

                        String displayName = null;
                        String givenName = null;
                        String middleName = null;                           
                        String familyName = null;
                        String prefix = null;
                        String suffix = null;
                        hasNext = cursorStructuredName.moveToFirst();
                        if (cursorStructuredName != null && hasNext) {//does not get here because hasNext equals false
                            displayName = cursorStructuredName.getString(cursorStructuredName.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.DISPLAY_NAME)); 
                            givenName = cursorStructuredName.getString(cursorStructuredName.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.GIVEN_NAME)); 
                            middleName = cursorStructuredName.getString(cursorStructuredName.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.MIDDLE_NAME)); 
                            familyName = cursorStructuredName.getString(cursorStructuredName.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.FAMILY_NAME)); 
                            prefix = cursorStructuredName.getString(cursorStructuredName.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.PREFIX)); 
                            suffix = cursorStructuredName.getString(cursorStructuredName.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.SUFFIX)); 
                        }

                    } finally {
                        if (cursor != null) {
                            cursor.close();
                        }
                        if (cursorStructuredName != null) {
                            cursorStructuredName.close();
                        }
                    }
                }
            }
            break;
        }
    }
}

我無法獲得姓氏，姓名等。我無法獲得所選的電話號碼。

正如我在標題中所說，我想要的是當用戶點擊某個按鈕時，將打開聯系人列表，當他選擇特定聯系人時，會顯示此聯系人電話號碼列表，然后當他選擇一個號碼時方法onAtivityResult將被調用，我將“解析”電話號碼，姓氏，姓名等。

Answer 1

好吧，經過更多的研究，我已經解決了我的問題。

下一個代碼允許用戶按下按鈕，然后打開聯系人應用程序，讓用戶選擇聯系人（僅顯示具有至少一個電話號碼的聯系人）。 在選擇聯系人之后，其所有號碼的列表都是打開的，讓用戶可以選擇一個（例如，在某些設備中，相同的聯系人出現的次數超過一次，每個電話號碼都有一個） ）。 之后調用onActivityResult（）方法，然后（如果一切正常），我將取消所選擇的電話號碼，給定名稱和所選聯系人的姓氏：

@Override
public void onClick(View v) {
    Intent intent = new Intent(Intent.ACTION_PICK);
    intent.setType(ContactsContract.CommonDataKinds.Phone.CONTENT_TYPE);
    startActivityForResult(intent, CONTACT_PICKER_RESULT);  
}

@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
    if (resultCode == RESULT_OK) {  
        switch (requestCode) {  
        case CONTACT_PICKER_RESULT:  
            handleContactSelection(data);
            break;
        }
    }
}

private void handleContactSelection(Intent data) {
    if (data != null) {
        Uri uri = data.getData();
        if (uri != null) {
            Cursor cursor = null;
            Cursor nameCursor = null;
            try {
                cursor = getContentResolver().query(uri, new String[]{  
                        ContactsContract.CommonDataKinds.Phone.NUMBER,
                        CommonDataKinds.Phone.CONTACT_ID} ,                                
                        null, null, null);

                String phoneNumber = null;                  
                String contactId = null;
                if (cursor != null && cursor.moveToFirst()) {
                    contactId = cursor.getString(cursor.getColumnIndex(CommonDataKinds.Phone.CONTACT_ID));
                    phoneNumber = cursor.getString(cursor.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));                               
                }

                String givenName = null;///first name.
                String familyName = null;//last name.

                String projection[] = new String[]{ContactsContract.CommonDataKinds.StructuredName.GIVEN_NAME,
                        ContactsContract.CommonDataKinds.StructuredName.FAMILY_NAME};
                String whereName = ContactsContract.Data.MIMETYPE + " = ? AND " + 
                        ContactsContract.CommonDataKinds.StructuredName.CONTACT_ID + " = ?";
                String[] whereNameParams = new String[] { ContactsContract.CommonDataKinds.StructuredName.CONTENT_ITEM_TYPE, contactId};

                nameCursor = getContentResolver().query(ContactsContract.Data.CONTENT_URI, 
                        projection, whereName, whereNameParams, ContactsContract.CommonDataKinds.StructuredName.GIVEN_NAME);

                if(nameCursor != null && nameCursor.moveToNext()) {
                    givenName = nameCursor.getString(nameCursor.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.GIVEN_NAME));
                    familyName = nameCursor.getString(nameCursor.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.FAMILY_NAME));
                }

                doSomething(phoneNumber,givenName,familyName);

            } finally {
                if (cursor != null) {
                    cursor.close();
                }

                if(nameCursor != null){
                    nameCursor.close();
                }
            }
        }
    }
}   

Answer 2

我相信這個會解決你的一些問題。 這基本上需要一個phoneno並查詢其聯系號。我使用它來將數據上傳到服務器。 至於電話號碼，我從收到的電話聽眾那里得到它。

String[] projection = new String[]{
                                ContactsContract.PhoneLookup.DISPLAY_NAME};
                        Uri contacturi= Uri.withAppendedPath(ContactsContract.PhoneLookup.CONTENT_FILTER_URI, Uri.encode(phoneno));
                    ContentResolver cr = getContentResolver();      
Cursor crc= cr.query(contacturi, projection, null, null, null);
                    if(crc.moveToNext()){
                            name=crc.getString(crc.getColumnIndex(ContactsContract.PhoneLookup.DISPLAY_NAME));

                        }
                        else{
                            name ="Unknown";
                        }

您需要在crc游標上添加所有必填字段。 即家庭，電子郵件，家庭。 希望這能給你一些見解： - /