如何从SQlite数据库中的drawable存储ID并再次检索它?

时间:2017-05-04 15:20:20

标签: java android sqlite android-sqlite

我想在我的SQlite数据库中存储一个drawable(如R.drawable.image)的ID,然后检索该ID以放入listView。我不想将图像保存为BLOB我只想存储ID。

现在我创建一个这样的表,我存储了一个drawable的ID:

 db.execSQL("CREATE TABLE cards \n" +
            "(\n" +
            "card_id INT PRIMARY KEY, \n" +
            "cpr VARCHAR(11) NOT NULL,\n" +
            "card_type VARCHAR(30), \n" +
            "front_photo VARCHAR(30), back_photo BLOB,\n" +
            "FOREIGN KEY (cpr) REFERENCES users(cpr)\n" +
            ")\n" +
            ";\n");

    db.execSQL("INSERT INTO cards (card_id , cpr, card_type, front_photo) VALUES(1, '170492-1802','Rejsekort', 'R.drawable.rejsekort_f');"
    );
    db.execSQL("INSERT INTO cards (card_id , cpr, card_type, front_photo) VALUES(2, '170492-1802','Bank Card', 'R.drawable.bankcard_f');"
    );
    db.execSQL("INSERT INTO cards (card_id , cpr, card_type, front_photo) VALUES(3, '170492-1802','Drivers Licence', 'R.drawable.bankcard_f');"
    );
    db.execSQL("INSERT INTO cards (card_id , cpr, card_type, front_photo) VALUES(4, '170492-1802','Sundhedskort', 'R.drawable.bankcard_f');"
    );
    db.execSQL("INSERT INTO cards (card_id , cpr, card_type, front_photo) VALUES(5, '170492-1802','Ungdomskort', 'R.drawable.bankcard_f');"
    );

然后我尝试获取该表的所有ID并将它们添加到Integer ArrayList中,以便我可以在ListView中显示它们:

ArrayList<Integer> images=new ArrayList<Integer>();
public void addImagesToListview() {
    DBHelper dbHelper = new DBHelper(MainActivity.this);
    db = dbHelper.getReadableDatabase();
    Cursor getImages = db.rawQuery("Select front_photo from cards", null);
    getImages.moveToFirst();
    while(!getImages.isAfterLast()) {
        String image = getImages.getString(0);
        int image2 = Integer.parseInt(image);
        images.add(image2);
        getImages.moveToNext(); }

但由于某种原因,应用程序崩溃了?我可以在我的ArrayList中添加ID的drawables,如下所示:

images.add(R.drawable.image);

那么为什么当我将该ID存储在数据库中并再次检索它时,它是否有效,所以它看起来像这样:

images.add("ID from database");

1 个答案:

答案 0 :(得分:0)

将您的数据库更改为

front_photo INT

当您插入行时,插入评估的ID,而不是表示它的字符串。

int photoID = R.drawable.rejsekort_f;
db.execSQL("INSERT INTO cards (card_id , cpr, card_type, front_photo) VALUES(1, '170492-1802','Rejsekort', " + photoID + ");";
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