java - 如何让用户从联系人中选择电话号码,然后获取电话号码、名字和姓氏

标签 java android android-contacts

到目前为止我所做的是:

点击某个按钮时:

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;
        }
    }
}

我无法获取姓氏、名字等信息。 我无法获取所选的电话号码。

正如我在标题中所说,我想要的是当用户点击某个按钮时,联系人列表将被打开,当他选择一个特定的联系人时,就会出现一个联系人电话号码的列表,然后当他选择将调用 onAtivityResult 方法的数字 我将“解析”电话号码、姓氏、名字等。

最佳答案

好吧,经过更多的研究,我已经解决了我的问题。

下一个代码让用户按下一个按钮,然后打开联系人应用程序,让用户选择一个联系人(只显示至少有一个电话号码的联系人)。 选择一个联系人后,会打开一个包含所有号码的列表,让用户可以选择一个(并非所有设备都有此选项,例如,在某些设备中,同一个联系人会多次出现,每个电话号码对应一个). 之后调用 onActivityResult() 方法,然后(如果一切正常)我提取所选的电话号码、名字和姓氏(所选联系人的):

@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();
                }
            }
        }
    }
}   

关于java - 如何让用户从联系人中选择电话号码,然后获取电话号码、名字和姓氏,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/21074910/

相关文章:

android - 如何在一次 SQLite 查询中获取联系人 ID、电子邮件、电话号码?联系人 Android 优化

java - 播放响亮声音的自动音量控制

android - 如何模拟一个非常慢的安卓手机?

android - 从联系人选择器中过滤掉 Facebook 联系人

android - 前 Lollipop 设备上的奇怪自定义工具栏

android - : Lorg/apache/http/client/methods/HttpPost 解析失败

android - 亚行 shell : Add contacts with extras

java - 使用嵌入式 Web 浏览器(例如 Chrome)作为 Java 桌面应用程序的 GUI 工具包?

java - Android 找不到类 java.awt.datatransfer.StringSelection?尝试将字符串放在剪贴板上

java - Java中的多个客户端到服务器通信程序