Question

我正在解析JSON并尝试将JSON数据存储到SQLite数据库中，我的JSON中只有name属性并且有大约15条记录，但我无法将所有记录存储到SQLite中 - 仍然仅将最后解析的记录存储到JSON中。

以下是我的JSON的样子：

{
  "actors": [
    {
      "name": "Name 1"
    },
    {
      "name": "Name 2"
    },
    {
      "name": "Name 3"
    },
    .......
    {
      "name": "Name 15"
    }
  ]
}

我如何将所有记录存储到JSON中？正如您在我的JSON中看到的，我有15条记录，但在SQLite中，我只获取最后一条JSON记录的数据（即 - 名称15）

@Override
protected Boolean doInBackground(String... urls) {
    try {

        //------------------>>
        HttpGet httppost = new HttpGet(urls[0]);
        HttpClient httpclient = new DefaultHttpClient();
        HttpResponse response = httpclient.execute(httppost);

        // StatusLine stat = response.getStatusLine();
        int status = response.getStatusLine().getStatusCode();

        if (status == 200) {
            HttpEntity entity = response.getEntity();
            String data = EntityUtils.toString(entity);


            JSONObject jsono = new JSONObject(data);
            JSONArray jarray = jsono.getJSONArray("actors");

            for (int i = 0; i < jarray.length(); i++) {
                JSONObject object = jarray.getJSONObject(i);

                Actors actor = new Actors();

                actor.setName(object.getString("name"));                        

                actorsList.add(actor);                      

                sqliteDB.open();

                sqliteDB.deleteAll();                                           

                sqliteDB.insert(object.getString("name"));

                sqliteDB.close();
            }
            return true;
        }

        //------------------>>

    } catch (ParseException e1) {
        e1.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    } catch (JSONException e) {
        e.printStackTrace();
    }
    return false;
}

Answer 1

@Override
    protected Boolean doInBackground(String... urls) {
        try {

            //------------------>>
            HttpGet httppost = new HttpGet(urls[0]);
            HttpClient httpclient = new DefaultHttpClient();
            HttpResponse response = httpclient.execute(httppost);

            // StatusLine stat = response.getStatusLine();
            int status = response.getStatusLine().getStatusCode();

            if (status == 200) {
                HttpEntity entity = response.getEntity();
                String data = EntityUtils.toString(entity);


                JSONObject jsono = new JSONObject(data);
                JSONArray jarray = jsono.getJSONArray("actors");
                sqliteDB.open(); 
                sqliteDB.deleteAll();  
                for (int i = 0; i < jarray.length(); i++) {
                    JSONObject object = jarray.getJSONObject(i);

                    Actors actor = new Actors();

                    actor.setName(object.getString("name"));                        

                    actorsList.add(actor);                      

                    sqliteDB.insert(object.getString("name"));

                }
                 sqliteDB.close();
                return true;
            }

            //------------------>>

        } catch (ParseException e1) {
            e1.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        } catch (JSONException e) {
            e.printStackTrace();
        }
        return false;
    }

Answer 2

在插入记录之前的for循环中，您正在删除数据库中的所有数据，这就是为什么只有您获得的最后一条记录，请尝试下面的代码

sqliteDB.open();
        sqliteDB.deleteAll();  
        for (int i = 0; i < jarray.length(); i++) {
            JSONObject object = jarray.getJSONObject(i);

            Actors actor = new Actors();

            actor.setName(object.getString("name"));                        

            actorsList.add(actor);                      

            sqliteDB.insert(object.getString("name"));

        }
         sqliteDB.close();

如何将所有JSON记录存储到SQLite中

2 个答案: