Question

我是JSON和Android的新手我做简单的应用程序，我想从mysql中的数据库请求数据，我得到这个例外

这是我的代码

 final GetDataFromDB getdb = new GetDataFromDB();

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    tl = (TableLayout) findViewById(R.id.maintable);


    new Thread(new Runnable() {
        public void run() {
            data = getdb.getDataFromDB();
            System.out.println(data);

            runOnUiThread(new Runnable() {

                @Override
                public void run() {
                    ArrayList<Users> users = parseJSON(data);
                    addData(users);
                }
            });

        }
    }).start();
}

public ArrayList<Users> parseJSON(String result) {
    ArrayList<Users> users = new ArrayList<Users>();
    try {
        JSONObject jsonObj = new JSONObject(result);
        JSONArray jArray = jsonObj.getJSONArray("users");
        for (int i = 0; i < jArray.length(); i++) {
            JSONObject json_data = jArray.getJSONObject(i);
            Users user = new Users();
            user.setId(json_data.getInt("id"));
            user.setJenis_kendaraan(json_data.getString("jenis_kendaraan"));
            user.setBarang_sitaan(json_data.getString("barang_sitaan"));
            users.add(user);
        }
    } catch (JSONException e) {
        Log.e("log_tag", "Error parsing data " + e.toString());
    }
    return users;
}

这里是Users.java类

public class Users {

int id;
String jenis_kendaraan;
String barang_sitaan;

public int getId() {
    return id;
}
public void setId(int id) {
    this.id = id;
}
public String getJenis_kendaraan() {
    return jenis_kendaraan;
}
public void setJenis_kendaraan(String jenis_kendaraan) {
    this.jenis_kendaraan = jenis_kendaraan;
}
public String getBarang_sitaan() {
    return barang_sitaan;
}
public void setBarang_sitaan(String barang_sitaan) {
    this.barang_sitaan = barang_sitaan;
}

}

这是GetDataFromDB.java

public class GetDataFromDB {

public String getDataFromDB() {
    try {

        HttpPost httppost;
        HttpClient httpclient;
        httpclient = new DefaultHttpClient();
        httppost = new HttpPost(
                "http://192.168.0.102/koneksi/getdata.php"); // change this to your URL.....
        ResponseHandler<String> responseHandler = new BasicResponseHandler();
        final String response = httpclient.execute(httppost,
                responseHandler);

        return response.trim();

    } catch (Exception e) {
        System.out.println("ERROR : " + e.getMessage());
        return "error";
    }
}
}

这是我的PHP代码，（我正在使用AMPPS）

<?php

mysql_connect("localhost","root","mysql"); // host, username, password...
mysql_select_db("tilang"); // db name...

$q=mysql_query("SELECT id, jenis_kendaraan, barang_sitaan FROM pelanggaran");
while($row=mysql_fetch_assoc($q))
        $json_output[]=$row;

print(json_encode($json_output));

mysql_close();

?>

logcat的

06-03 10:01:22.795  30366-30366/com.example.indraaaeff.historyapp E/log_tag﹕ Error parsing data org.json.JSONException: Value error of type java.lang.String cannot be converted to JSONArray

这是我从php文件中获取的字符串。

[{"pid":"1","jenis_kendaraan":"Motor","barang_sitaan":"SIM"},   
{"pid":"2","jenis_kendaraan":"Mobil","barang_sitaan":"Lain-lain"},
{"pid":"3","jenis_kendaraan":"Motor","barang_sitaan":"SIM"}]

请帮我删除

Answer 1

 @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        tl = (TableLayout) findViewById(R.id.maintable);
        new loadUserTask().execute();
    }


 public class loadUserTask extends AsyncTask<Void,Void,ArrayList<User>> {

        @Override
        protected Void doInBackground(Void... params) {
            data = getdb.getDataFromDB();
            System.out.println(data);
            ArrayList<Users> users = parseJSON(data);
            return users;   
        }


        @Override
        protected void onPostExecute(ArrayList<User> users) {
            addData(users);
        }
    }


    public ArrayList<Users> parseJSON(String result) {
        ArrayList<Users> users = new ArrayList<Users>();
        try {
            JSONObject jsonObj = new JSONObject(result);
            JSONArray jArray = jsonObj.getJSONArray("users");
            for (int i = 0; i < jArray.length(); i++) {
                JSONObject json_data = jArray.getJSONObject(i);
                Users user = new Users();
                user.setId(json_data.getInt("id"));
                user.setJenis_kendaraan(json_data.getString("jenis_kendaraan"));
                user.setBarang_sitaan(json_data.getString("barang_sitaan"));
                users.add(user);
            }
        } catch (JSONException e) {
            Log.e("log_tag", "Error parsing data " + e.toString());
        }
        return users;
    }

java.lang.String类型的值错误无法转换为JSONArray

1 个答案: