智能搜索android中的联系人

Question

在Android开发者网站上检索联系人教程列表之后,我设法实现了联系人搜索功能.到目前为止,这是我的代码

private void retrieveContactRecord(String phoneNo) {
        try {
            Log.e("Info", "Input: " + phoneNo);
            Uri uri = Uri.withAppendedPath(ContactsContract.PhoneLookup.CONTENT_FILTER_URI,
                    Uri.encode(phoneNo));
            String[] projection = new String[]{ContactsContract.PhoneLookup._ID, ContactsContract.PhoneLookup.DISPLAY_NAME};


            String sortOrder = ContactsContract.PhoneLookup.DISPLAY_NAME + " COLLATE LOCALIZED ASC";
            ContentResolver cr = getContentResolver();
            if (cr != null) {
                Cursor resultCur = cr.query(uri, projection, null, null, sortOrder);
                if (resultCur != null) {
                    while (resultCur.moveToNext()) {
                        String contactId = resultCur.getString(resultCur.getColumnIndex(ContactsContract.PhoneLookup._ID));
                        String contactName = resultCur.getString(resultCur.getColumnIndexOrThrow(ContactsContract.PhoneLookup.DISPLAY_NAME));
                        Log.e("Info", "Contact Id : " + contactId);
                        Log.e("Info", "Contact Display Name : " + contactName);
                        break;
                    }
                    resultCur.close();
                }
            }
        } catch (Exception sfg) {
            Log.e("Error", "Error in loadContactRecord : " + sfg.toString());
        }
    }

这是捕获,这段代码非常好,但我需要在这里实现智能搜索.我想26268匹配Amanu以及094 526 2684.我相信它被称为T9字典.

我试着寻找其他项目的线索,但我找不到任何东西.任何指针将不胜感激!

Answer 1

for ContentProvidercontact 不支持它。所以我所做的是将所有联系人转储到 a 中List，然后使用 aRegEx来匹配名称。

public static String[] values = new String[]{" 0", "1", "ABC2", "DEF3", "GHI4", "JKL5", "MNO6", "PQRS7", "TUV8", "WXYZ9"};

/**
 * Get the possible pattern
 * You'll get something like ["2ABC","4GHI"] for input "14"
 */
public static List<String> possibleValues(String in) {

    if (in.length() >= 1) {
        List<String> p = possibleValues(in.substring(1));
        String s = "" + in.charAt(0);
        if (s.matches("[0-9]")) {
            int n = Integer.parseInt(s);

            p.add(0, values[n]);
        } else {
            // It is a character, use it as it is
            p.add(s);
        }

        return p;
    }
    return new ArrayList<>();
}

....然后编译模式。我曾经(?i)使它不区分大小写

List<String> values = Utils.possibleValues(query);
StringBuilder sb = new StringBuilder();
for (String value : values) {
    sb.append("[");
    sb.append(value);
    sb.append("]");
    if (values.get(values.size() - 1) != value) {
    sb.append("\\s*");
    }
}

Log.e("Utils", "Pattern = " + sb.toString());

Pattern queryPattern = Pattern.compile("(?i)(" + sb.toString() + ")");

在此之后你就会知道该怎么做。