Android：在contentResolver.insert（）中返回null

Question

我有这段代码：

public static long addContactAddresses(Context context, String number) {
    long add_id = 0;
    final ContentResolver resolver = context.getContentResolver();
    final Cursor c = resolver.query(ConvAddresses.CONTENT_URI,
            new String[] { ConvAddressesColumns._ID,
                    ConvAddressesColumns.NUMBER },
            ConvAddressesColumns.NUMBER + "=?",
            new String[] { number }, null);
    try {
        if (c != null && c.moveToFirst()) {
            if ((c.getString(0)) != null) {
                add_id = Long.valueOf(c.getString(0));
            }
        } else {
            ContentValues values = new ContentValues();
            values.put(ConvAddressesColumns.NUMBER, number);
            Uri result = resolver.insert(ConvAddresses.CONTENT_URI, values);
            add_id = ContentUris.parseId(result);
        }
    } finally {
        if (c != null) {
            c.close();
        }
    }
    return add_id;
}

很少resolver.insert(ConvAddresses.CONTENT_URI, values);返回null，这对我来说是个问题......

查看插入实现：

public final Uri insert(Uri url, ContentValues values)
{
    IContentProvider provider = acquireProvider(url);
    if (provider == null) {
        throw new IllegalArgumentException("Unknown URL " + url);
    }
    try {
        long startTime = SystemClock.uptimeMillis();
        Uri createdRow = provider.insert(mPackageName, url, values);
        long durationMillis = SystemClock.uptimeMillis() - startTime;
        maybeLogUpdateToEventLog(durationMillis, url, "insert", null /* where */);
        return createdRow;
    } catch (RemoteException e) {
        // Arbitrary and not worth documenting, as Activity
        // Manager will kill this process shortly anyway.
        return null;
    } finally {
        releaseProvider(provider);
    }
}

RemoteException

时可以返回null

我想将我的代码编辑为：

Uri result = resolver.insert(ConvAddresses.CONTENT_URI, values);
if (result != null) {
    add_id = ContentUris.parseId(result);
} else {
    Log.w(TAG, "RemoteException in resolver.insert, let's try it again!");
    add_id = addContactAddresses(context, number);
}

这会导致无限循环吗？这不好吗？这个好吗？ 在这种情况下如何导致RemoteException？