从手机中获取所有联系人有时会返回 NULL答案

【问题标题】：Getting all the contacts from the phone sometimes returns NULL从手机中获取所有联系人有时会返回 NULL
【发布时间】：2017-05-25 15:08:41
【问题描述】：

我制作了一个应用程序，其中我需要电话簿中可用的所有联系人。我在列表中显示这些数字。应用程序工作正常，但有时应用程序强制关闭，因为光标返回 null。这并不总是发生，但有时会发生。现在我该如何处理？？？？

代码

public static JSONArray getAllContactList(Context context) {
        Cursor c = context.getContentResolver().query(ContactsContract.Contacts.CONTENT_URI, null, null, null, null);
        alAllContacts = new ArrayList<ContactModel>();
        while (!(c == null) && c.moveToNext()) {
            String id = c.getString(c.getColumnIndexOrThrow(ContactsContract.Contacts._ID));
            String number = c.getString(c.getColumnIndexOrThrow(ContactsContract.Contacts.HAS_PHONE_NUMBER));

            if (number.equalsIgnoreCase("1")) {
//                    Cursor phones = getActivity().getContentResolver().query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null,
//                            ContactsContract.CommonDataKinds.Phone.CONTACT_ID + " = '" + id + "'", null, null);

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


                while (phones.moveToNext()) {

                    String contactName = phones.getString(phones.getColumnIndex(ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME));
                    String contactNumber = phones.getString(phones.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));

                    contactNumber = contactNumber.replace("+", "");

                    if (contactNumber.length() > 10) {
                        contactNumber = contactNumber.substring(2);
                    }
//                    contactNumber.replace("+91", "");
                    alAllContacts.add(new ContactModel(contactName, contactNumber));

//
                }

            }


        }
        c.close();


        JSONArray jsonArray = new JSONArray();
        for (int i = 0; i < alAllContacts.size(); i++) {
            jsonArray.put(alAllContacts.get(i).getJSONObject());
        }


        return jsonArray;

    }

logcat 说我在这一行得到空指针

 while (phones.moveToNext()) {

有时我会因为对话框正在运行而强制关闭，所以我显示进度条的代码也是正确的

public static void showProgress(Context context, String msg, boolean isVisible) {


        if (progressDialog == null) {
            progressDialog = new ProgressDialog(context);
            progressDialog.setMessage(msg);
            progressDialog.setCancelable(false);
        }

        if (isVisible) {
            progressDialog.show();
        } else if (isVisible == false) {
            if (progressDialog.isShowing()) {
                progressDialog.dismiss();
            }
        }

    }

【问题讨论】：

请参考我的此链接以获取联系人stackoverflow.com/questions/19972085/…
@BhanuSharma 先生，我正在正确获取联系人，但有时我得到空指针，那么我该如何处理该空指针
如何处理空指针异常？ while (phones != null && phones.moveToNext())

标签： android android-contacts android-progressbar

【解决方案1】：

我使用此代码获取所有联系人，它完成了这项工作。

private static void backupContacts(Context context) {

    String[] pCurProjection = new String[]{
            ContactsContract.CommonDataKinds.Phone.NUMBER,
            ContactsContract.CommonDataKinds.Phone.TYPE,
            ContactsContract.CommonDataKinds.Phone.LABEL,
            ContactsContract.CommonDataKinds.Phone.IS_PRIMARY
    };

    String contactSelection = ContactsContract.Contacts.IN_VISIBLE_GROUP + " = '1'";

    String[] contactProjection = new String[]{
            ContactsContract.Contacts._ID,
            ContactsContract.Contacts.DISPLAY_NAME
    };

    Cursor contactCursor = context.getContentResolver().query(
            ContactsContract.Contacts.CONTENT_URI,
            contactProjection,
            contactSelection,
            null,
            ContactsContract.Contacts.DISPLAY_NAME + " COLLATE LOCALIZED ASC");

    int contactId;
    String contactName;
    if (null != contactCursor) {
        while (contactCursor.moveToNext()) {

            contactId = contactCursor.getInt(
                    contactCursor.getColumnIndex(ContactsContract.Contacts._ID));
            contactName = contactCursor.getString(
                    contactCursor.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));

            System.out.println(contactId + ": " + contactName);

            Cursor pCur = context.getContentResolver().query(
                    ContactsContract.CommonDataKinds.Phone.CONTENT_URI,
                    pCurProjection,
                    ContactsContract.CommonDataKinds.Phone.CONTACT_ID + " = ?",
                    new String[]{String.valueOf(contactId)},
                    null);

            if (null != pCur) {
                while (pCur.moveToNext()) {

                    int id = pCur.getColumnIndex(
                            ContactsContract.CommonDataKinds.Phone._ID);

                    String number = (pCur.getString(
                            pCur.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER)));
                    String type = pCur.getString(
                            pCur.getColumnIndex(ContactsContract.CommonDataKinds.Phone.TYPE));//this line returns Null but it does not effect on the code(Does not Throws NullPointerException). if you found fix for it please let me know
                    int isPrimary = pCur.getInt(
                            pCur.getColumnIndex(ContactsContract.CommonDataKinds.Phone.IS_PRIMARY));

                    System.out.println(id + ": " + number + "\n" + type + "\n" + isPrimary);
                }
                pCur.close();
            }
        }
        contactCursor.close();
    }
}

如果您有任何问题，请告诉我:)

【讨论】：