Question

这是json代码，httpurlconnection从json

获取数据

protected Void doInBackground(Void...params){
        InputStream is=null;

        String result="";

        String urlDate="http://pangkortourism.ga/fyp/player.php";

        BufferedReader reader=null;

        try{

            URL urlp=new URL(urlDate);

            try {
                HttpURLConnection c = (HttpURLConnection) urlp.openConnection();
                c.setRequestMethod("GET");
                c.setReadTimeout(10000);
                c.connect();
                reader = new BufferedReader(new InputStreamReader(c.getInputStream(), "UTF-8"));

                StringBuilder buf = new StringBuilder();
                String line = null;
                while ((line = reader.readLine()) != null) {
                    buf.append(line);
                }

                result = buf.toString();
            }
            finally {
                if (reader != null) {
                    reader.close();
                }
            }

        } catch (IOException e) {


            Log.e("MalformedURLException", "MalformedURLException " + e.toString());
        }



        //parse json data

        try{

            // Remove unexpected characters that might be added to beginning of the string

                    result=result.substring(result.indexOf("["));

            JSONArray jArray =new JSONArray(result);

            for(int i=0;i<jArray.length();i++){

                JSONObject json_data = jArray.getJSONObject(i);

                Product p=new Product();

                p.setid(json_data.getString("id"));

                p.setitem(json_data.getInt("item"));
                records.add(p);
            }

        }

        catch(Exception e){

            Log.e("ERROR", "Error pasting data "+e.toString());
        }
        return null;

    }

mysql数据库数据

json代码

[
    {"0":"1","id":"1","1":"pizza","item":"pizza"},
    {"0":"2","id":"2","1":"burger","item":"burger"}
]

android studio中的错误

04-07 11：58：12.527 23232-23303 / com.khor.newtry6 E /错误：错误粘贴 data org.json.JSONException：类型为item的值pizza java.lang.String无法转换为int 04-07 11：58：12.542 23232-23232 / com.khor.newtry6 E / size：0

Answer 1

“item”字段是String不是整数因此

以下更改：

p.setitem(json_data.getString("item"));

Answer 2

从代码中删除以下行，因为它会签名以指示json arrya

结果= result.substring（result.indexOf（＆＃34; [＆＃34））;

然后尝试后可能会有效。并使用p.setitem(json_data.getInt("item"));

更改项p.setitem(json_data.getString("item"));

Answer 3

尝试使用p.setitem（json_data.getString（＆＃34; item＆＃34;））; INSTEAD OF

p.setitem（json_data.getInt（＆＃34;项目＆＃34））;

并在方法p.setitem（）中更改参数数据类型; to String ..

即。签名为p.setitem（）;方法应该是：

public void setitem（String item）{

this.item = item;

}

粘贴数据org.json.JSONException时出错

3 个答案: