Question

在我的ContentProvider中导致错误的方法

public static Cursor getSuggestions(String query) {
    SQLiteDatabase db = dbHelper.getReadableDatabase();
    db.beginTransaction();
    try {
        String selection = Formula.FORMULA_NAME + " LIKE %?%";
        String[] selectionArgs = { query + "*" };
        Cursor cursor = dbHelper.getReadableDatabase().query(
                FORMULA_TABLE_NAME,
                new String[] { BaseColumns._ID,
                        SearchManager.SUGGEST_COLUMN_TEXT_1, 
                        BaseColumns._ID + " AS " + SearchManager.SUGGEST_COLUMN_INTENT_DATA_ID 
                        }, 
                        selection,
                selectionArgs, null, null, null);
        db.setTransactionSuccessful();
        return cursor;
    } catch (SQLiteException e) {
    } finally {
        db.endTransaction();
    }
    throw new SQLException("Failed to begin transaction");
}

数据库创建：

db.execSQL("CREATE TABLE " + FORMULA_TABLE_NAME + " (" +
                BaseColumns._ID + " INTEGER PRIMARY KEY AUTOINCREMENT," +
                SearchManager.SUGGEST_COLUMN_TEXT_1 + " TEXT," +
                Formula.CATEGORY + " TEXT" +
                ");");

使用的常数：

    public static final String FORMULA_NAME = SearchManager.SUGGEST_COLUMN_TEXT_1;
    public static final String CATEGORY = "category";

问题在于，在我的方法中，事务是不成功的，因为它会抛出错误：throw new SQLException("Failed to begin transaction");我要做的是搜索数据库作为搜索的一部分。当用户激活搜索框时，我将其设置为使此方法应该根据其名称返回包含可疑项目的游标。通过调试，我推断出问题出在我的Content Provider中的搜索方法。任何解决方案或想法？

Answer 1

我猜测行throw new SQLException("Failed to begin transaction");意味着在你的catch块中。

如果我们将所有内容简化为：

，该怎么办？

public static Cursor getSuggestions(String query) {
    SQLiteDatabase db = dbHelper.getReadableDatabase();
    String selection = Formula.FORMULA_NAME + " LIKE %?%";
    String[] selectionArgs = { query + "*" };
    Cursor cursor = db.query(
            FORMULA_TABLE_NAME,
            new String[] { BaseColumns._ID,
                    SearchManager.SUGGEST_COLUMN_TEXT_1, 
                    BaseColumns._ID + " AS " + SearchManager.SUGGEST_COLUMN_INTENT_DATA_ID 
                    }, 
            selection,
            selectionArgs, null, null, null);
    return cursor;
}

在Android搜索上搜索SQLite数据库

1 个答案: