Question

您好我正在尝试将首选选项添加到我的应用程序中我开始使用sqlite来保存图像的URL但是我在使用此代码将图片网址添加到收藏夹时遇到问题：

public void AddtoFavorite(Post pj) {
        SQLiteDatabase db = getWritableDatabase();
        ContentValues values = new ContentValues();
        values.put(KEY_IMAGE_CATNAME, pj.getName());
        values.put(KEY_IMAGE_URL, pj.getImgUrl());
        db.insert(TABLE_NAME, null, values);
        db.close();
    }

它已成功添加，但是当我想检查图像URL是否存在于数据库中时

  public List<Post> getFavRow(String url) {
        List<Post> dataList = new ArrayList();
        @SuppressLint("Recycle") Cursor cursor = getWritableDatabase().rawQuery("SELECT * FROM Favorite WHERE imageurl = '" + url + "'", null);
        if (cursor.moveToFirst()) {
            do {
                Post contact = new Post();
                contact.setName(cursor.getString(0));
                contact.setImgUrl(cursor.getString(1));
                dataList.add(contact);
            } while (cursor.moveToNext());
        }
        return dataList;
    }

总是我得到空列表。

Answer 1

你应该在你的getFavRow（）函数中使用getReadableDatabase（）而不是getWritableDatabase（）

SQLiteDatabase db = this.getReadableDatabase();

Answer 2

尝试

WHERE imageurl LIKE'％“+ url +”％'“

Answer 3

试试这个

public List<Post> getFavRow(String url) {
List<Post> dataList = new ArrayList();    

 String[] selectQuery = {"YOUR", "TABLE", "COLUMNS"};
 String selection = "IMAGEURL = ?";
 String[] selectionArgs = {url};
 Cursor cursor = database.query(yourTable, selectQuery, selection, selectionArgs, null, null, null);
 while(cursor.moveToNext()){
    Post contact = new Post();
    contact.setName(cursor.getString(cursor.getColumnIndex("NAME")));
    contact.setImgUrl(cursor.getString(cursor.getColumnIndex("IMAGEURL")));
    dataList.add(contact);
  }
 cursor.close();
 return dataList;
}

从sqlite中搜索一个值总是返回空结果

3 个答案: