Question

我有点困惑这个问题，我即将解析JSON到我的Android应用程序。我用php创建了我自己的模型：

class Users extends CI_Model
{
function get_all()
{
$query = $this->db->get('usrs');

    foreach ($query->result() as $row)
    {
        //$name = $row->Username;
        $data = $row;
        echo json_encode($data);
    }
}
}

我的控制器：

 class Rest extends CI_Controller
 {
public function index()
{
    echo $this->load->model("Users");
        echo "<br />";
        echo $this->Users->get_all();
}

 }

但是当我跑这个时它只是给了我这个：

 {"UserAccessId":"1","Username":"Paul","Password":"Parreno","FirstName":"John Paul","MiddleName":"Pineda","LastName":"Parreno","Email":"johnpaul_sandwich_chicosci@yahoo.com"}

没有数组的名称。

这是我在Android应用程序中正在处理的代码：

   private static String url = "http://localhost/kwotted/index.php/rest";

   private static final String TAG_NAME = "Username";
    private static final String TAG_FNAME = "FirstName";
    private static final String TAG_LNAME = "LastName";
    private static final String TAG_DB = "";

    // contacts JSONArray
    JSONArray contacts = null;

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

        // Hashmap for ListView
        ArrayList<HashMap<String, String>> contactList = new ArrayList<HashMap<String, String>>();

        // Creating JSON Parser instance
        JSONParser jParser = new JSONParser();

        // getting JSON string from URL
        JSONObject json = jParser.getJSONFromUrl(url);

        try {
            // Getting Array of Contacts
            //contacts = json.getJSONArray(TAG_CONTACTS);
            contacts = json.getJSONArray(TAG_DB);

            // looping through All Contacts
            for(int i = 0; i < contacts.length(); i++){
                JSONObject c = contacts.getJSONObject(i);

                // Storing each json item in variable
                //String id = c.getString(TAG_ID);
                String name = c.getString(TAG_NAME);
                String fname = c.getString(TAG_FNAME);
                String lname = c.getString(TAG_LNAME);
                //String email = c.getString(TAG_EMAIL);
                //String address = c.getString(TAG_ADDRESS);
                //String gender = c.getString(TAG_GENDER);

                // Phone number is agin JSON Object
                /*JSONObject phone = c.getJSONObject(TAG_PHONE);
                String mobile = phone.getString(TAG_PHONE_MOBILE);
                String home = phone.getString(TAG_PHONE_HOME);
                String office = phone.getString(TAG_PHONE_OFFICE);
                */
                // creating new HashMap
                HashMap<String, String> map = new HashMap<String, String>();

                // adding each child node to HashMap key => value
                //map.put(TAG_ID, id);
                map.put(TAG_NAME, name);
                map.put(TAG_FNAME, fname);
                map.put(TAG_LNAME, lname);

                // adding HashList to ArrayList
                contactList.add(map);
            }
        } catch (JSONException e) {
            e.printStackTrace();
        }


        /**
         * Updating parsed JSON data into ListView
         * */
        ListAdapter adapter = new SimpleAdapter(this, contactList,
                R.layout.list_item,
                new String[] { TAG_NAME, TAG_FNAME, TAG_LNAME }, new int[] {
                //TAG_EMAIL, TAG_PHONE_MOBILE }, new int[] {
                        R.id.name, R.id.email, R.id.mobile });

        setListAdapter(adapter);

        // selecting single ListView item
        ListView lv = getListView();

        // Launching new screen on Selecting Single ListItem
        lv.setOnItemClickListener(new OnItemClickListener() {

            @Override
            public void onItemClick(AdapterView<?> parent, View view,
                    int position, long id) {
                // getting values from selected ListItem
                String name = ((TextView) view.findViewById(R.id.name)).getText().toString();
                String cost = ((TextView) view.findViewById(R.id.email)).getText().toString();
                String description = ((TextView) view.findViewById(R.id.mobile)).getText().toString();

                // Starting new intent
                Intent in = new Intent(getApplicationContext(), SingleMenuItemActivity.class);
                in.putExtra(TAG_NAME, name);
                in.putExtra(TAG_FNAME, cost);
                in.putExtra(TAG_LNAME, description);
                startActivity(in);

            }
        });



    }

但是android在行上抛出NullPointerException：

    contacts = json.getJSONArray(TAG_DB);

Answer 1

从这个JSON输出

{“UserAccessId”：“1”，“用户名”：“保罗”，“密码”：“Parreno”，“FirstName”：“John Paul”，“MiddleName”：“Pineda”，“LastName”：“Parreno”， “电子邮件”： “johnpaul_sandwich_chicosci@yahoo.com”}

它不是您正在处理的数组，而只是一个JSON对象。
也许您可以尝试以下方法：

JSONObject obj = (JSONObject) new JSONTokener(json_string_output).nextValue();
String name  = obj .getString(TAG_NAME);
String fname = obj .getString(TAG_FNAME);
String lname = obj .getString(TAG_LNAME);

您可以参考JSONTokener API

如何从php数组中获取Json数组名称json_encode？ Android的

1 个答案: