Question

我希望我的地图的搜索视图基于该地点的名称和Latlang。

我有一个Hashmap（）;

但这一行： map.put（BaseColumns._ID，＆＃34; rowid AS＆＃34; + BaseColumns._ID）;

给出错误：错误的第二个参数类型，必需的字符串，找到Latlang。

我尝试将BaseColumns._ID更改为Latlang，但我无法做到。

我该如何解决这个问题？

private static final String TAG = "DictionaryDatabase";

//The columns we'll include in the dictionary table
public static final String KEY_WORD = SearchManager.SUGGEST_COLUMN_TEXT_1;
public static final Latlang KEY_DEFINITION = SearchManager.SUGGEST_COLUMN_TEXT_2;

private static final String DATABASE_NAME = "dictionary";
private static final String FTS_VIRTUAL_TABLE = "FTSdictionary";
private static final int DATABASE_VERSION = 2;

private final DictionaryOpenHelper mDatabaseOpenHelper;
private static final HashMap<String,Latlang> mColumnMap = buildColumnMap();

/**
 * Constructor
 * @param context The Context within which to work, used to create the DB
 */
public DictionaryDatabase(Context context) {
    mDatabaseOpenHelper = new DictionaryOpenHelper(context);
}

private static HashMap<String,Latlang> buildColumnMap() {
    HashMap<String,Latlang> map = new HashMap<String,Latlang>();
    map.put(KEY_WORD, KEY_DEFINITION);
    map.put(KEY_WORD, KEY_DEFINITION);
    map.put(BaseColumns._ID, "rowid AS " +
            BaseColumns._ID);
    map.put(SearchManager.SUGGEST_COLUMN_INTENT_DATA_ID, "rowid AS " +
            SearchManager.SUGGEST_COLUMN_INTENT_DATA_ID);
    map.put(SearchManager.SUGGEST_COLUMN_SHORTCUT_ID, "rowid AS " +
            SearchManager.SUGGEST_COLUMN_SHORTCUT_ID);
    return map;
}

Answer 1

您将地图声明为：

HashMap<String,Latlang> map = new HashMap<String,Latlang>();

但是，您的数据是字符串：

map.put(SearchManager.SUGGEST_COLUMN_SHORTCUT_ID, "rowid AS " +
            SearchManager.SUGGEST_COLUMN_SHORTCUT_ID);

我建议您像这样声明地图：

HashMap<String, String> map = new HashMap<String, String>();

或确保您的值实际上是LatLng：

map.put(BaseColumns._ID, new LatLng(latitude, longitude));

android-如何使用String和Lat lang对map的searchview进行拼接？

1 个答案: