SQLite Android rawQuery 仅返回准确值

Question

我有一个 SQLite 数据库，用于存储一些食品和饮料以及多语言应用程序的卡路里值。我想在 android 应用程序中显示它们，但我的查询不会返回值，除非我写出确切的值。

我搜索牛奶时的数据库示例部分：

public class SearchFragment extends Fragment {

Button searchbutton;
EditText editText;
View v;
Cursor cursor;
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
}

@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
                         Bundle savedInstanceState) {
    v = inflater.inflate(R.layout.fragment_search, container, false);
    addListenerOnButton(v);
    return v;
}

private void addListenerOnButton(View v) {
    searchbutton = v.findViewById(R.id.searchfragment_button);
    editText = v.findViewById(R.id.searchfragment_edittext);
    searchbutton.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            String keystring = editText.getText().toString();
            new FoodQuery().execute(keystring);
        }
    });
}

private class FoodQuery extends AsyncTask<String, Void, Void>{

    @Override
    protected Void doInBackground(String [] keys) {
        String name_col;
        DietAppDatabaseHelper daDBHelper;
        SQLiteDatabase daDB;
        daDBHelper = new DietAppDatabaseHelper(getContext());
        try {
            daDB = daDBHelper.getReadableDatabase();
            if (Locale.getDefault().getLanguage() == "tr")
                name_col = "name_tr";
            else
                name_col = "name_eng";
            String query = "SELECT * FROM food_data WHERE "
                    + name_col + " = ?" ;
            cursor = daDB.rawQuery(query, new String[]{"%" + keys[0] + "%"});
            int i = cursor.getCount();
            daDB.close();
        } catch (SQLiteException e){
            Log.d("SearchFragment", "doInBackground: " + e);
        }
        return null;
    }
}
}

当我调试“int i = cursor.getCount();”时行，上面的代码不会向光标返回任何值。因此我总是出错。但不知何故，当我将 rawQuery 的 selectionArgs 部分从 "%" + keys[0] + "%" 更改为仅 keys[0] 并输入任何行的直接值（例如：Milk (Whole)）大小写时，我确实得到了1 导致光标。 这是数据库助手类

public class DietAppDatabaseHelper extends SQLiteOpenHelper {
private static String DB_NAME = "food_data.db";
private static String DB_PATH = "";
private static final int DB_VERSION = 1;

private SQLiteDatabase mDataBase;
private final Context mContext;
private boolean mNeedUpdate = false;

public DietAppDatabaseHelper(Context context) {
    super(context, DB_NAME, null, DB_VERSION);
    if (android.os.Build.VERSION.SDK_INT >= 17)
        DB_PATH = context.getApplicationInfo().dataDir + "/databases/";
    else
        DB_PATH = "/data/data/" + context.getPackageName() + "/databases/";
    Log.d("Dietpath", DB_PATH);
    this.mContext = context;

    copyDataBase();

    this.getReadableDatabase();
}

public void updateDataBase() throws IOException {
    if (mNeedUpdate) {
        File dbFile = new File(DB_PATH + DB_NAME);
        if (dbFile.exists())
            dbFile.delete();

        copyDataBase();

        mNeedUpdate = false;
    }
}

private boolean checkDataBase() {
    File dbFile = new File(DB_PATH + DB_NAME);
    return dbFile.exists();
}

private void copyDataBase() {
    if (!checkDataBase()) {
        this.getReadableDatabase();
        this.close();
        try {
            copyDBFile();
        } catch (IOException mIOException) {
            throw new Error("ErrorCopyingDataBase");
        }
    }
}

private void copyDBFile() throws IOException {
    InputStream mInput = mContext.getAssets().open(DB_NAME);
    //InputStream mInput = mContext.getResources().openRawResource(R.raw.info);
    OutputStream mOutput = new FileOutputStream(DB_PATH + DB_NAME);
    byte[] mBuffer = new byte[1024];
    int mLength;
    while ((mLength = mInput.read(mBuffer)) > 0)
        mOutput.write(mBuffer, 0, mLength);
    mOutput.flush();
    mOutput.close();
    mInput.close();
}

public boolean openDataBase() throws SQLException {
    mDataBase = SQLiteDatabase.openDatabase(DB_PATH + DB_NAME, null, SQLiteDatabase.CREATE_IF_NECESSARY);
    return mDataBase != null;
}

@Override
public synchronized void close() {
    if (mDataBase != null)
        mDataBase.close();
    super.close();
}

@Override
public void onCreate(SQLiteDatabase db) {

}

@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
    if (newVersion > oldVersion)
        mNeedUpdate = true;
}
}

Answer 1

= 运算符进行完全匹配。如果要使用% 等通配符，请使用LIKE 运算符，如

String query = "SELECT * FROM food_data WHERE "
                + name_col + " LIKE ?" ;