Android错误将字符串解析为json


Android Error Parsing string to json

在android上需要帮助=(已经坚持了很长时间!我的代码如图所示。

public class AllUsersActivity extends ListActivity {
// Progress Dialog
private ProgressDialog pDialog;
// Creating JSON Parser object
JSONParser jParser = new JSONParser();
ArrayList<HashMap<String, String>> usersList;
// url to get all users list
private static String url_all_users = "http://10.0.2.2/android_connect/get_all_users.php";
// JSON Node names
private static final String TAG_SUCCESS = "success";
private static final String TAG_USERS = "users";
private static final String TAG_UID = "UserID";
private static final String TAG_FIRSTNAME = "FirstName";
// users JSONArray
JSONArray users = null;
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.all_users);
    // Hashmap for ListView
    usersList = new ArrayList<HashMap<String, String>>();
    // Loading users in Background Thread
    new LoadAllusers().execute();
    // Get listview
    ListView lv = getListView();
    // on seleting single product
    // launching Edit Product Screen
    lv.setOnItemClickListener(new OnItemClickListener() {
        public void onItemClick(AdapterView<?> parent, View view,
                int position, long id) {
            // getting values from selected ListItem
            String uid = ((TextView) view.findViewById(R.id.uid)).getText()
                    .toString();
            // Starting new intent
            Intent in = new Intent(getApplicationContext(),
                    UserDetailsActivity.class);
            // sending uid to next activity
            in.putExtra(TAG_UID, uid);
            // starting new activity and expecting some response back
            startActivityForResult(in, 100);
        }
    });
}
// Response from Edit Product Activity
@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
    super.onActivityResult(requestCode, resultCode, data);
    // if result code 100
    if (resultCode == 100) {
        // if result code 100 is received
        // means user edited/deleted product
        // reload this screen again
        Intent intent = getIntent();
        finish();
        startActivity(intent);
    }
}
/**
 * Background Async Task to Load all product by making HTTP Request
 * */
class LoadAllusers extends AsyncTask<String, String, String> {
    /**
     * Before starting background thread Show Progress Dialog
     * */
    @Override
    protected void onPreExecute() {
        super.onPreExecute();
        pDialog = new ProgressDialog(AllUsersActivity.this);
        pDialog.setMessage("Loading users. Please wait...");
        pDialog.setIndeterminate(false);
        pDialog.setCancelable(false);
        pDialog.show();
    }
    /**
     * getting All users from url
     * */
    protected String doInBackground(String... args) {
        // Building Parameters
        List<NameValuePair> params = new ArrayList<NameValuePair>();
        // getting JSON string from URL
        JSONObject json = jParser.makeHttpRequest(url_all_users, "GET", params);
        // Check your log cat for JSON reponse
        Log.d("All users: ", json.toString());
        try {
            // Checking for SUCCESS TAG
            int success = json.getInt(TAG_SUCCESS);
            if (success == 1) {
                // users found
                // Getting Array of users
                users = json.getJSONArray(TAG_USERS);
                // looping through All users
                for (int i = 0; i < users.length(); i++) {
                    JSONObject c = users.getJSONObject(i);
                    // Storing each json item in variable
                    String id = c.getString(TAG_UID);
                    String name = c.getString(TAG_FIRSTNAME);
                    // creating new HashMap
                    HashMap<String, String> map = new HashMap<String, String>();
                    // adding each child node to HashMap key => value
                    map.put(TAG_UID, id);
                    map.put(TAG_FIRSTNAME, name);
                    // adding HashList to ArrayList
                    usersList.add(map);
                }
            } 
        } catch (JSONException e) {
            e.printStackTrace();
        }
        return null;
    }
    /**
     * After completing background task Dismiss the progress dialog
     * **/
    protected void onPostExecute(String file_url) {
        // dismiss the dialog after getting all users
        pDialog.dismiss();
        // updating UI from Background Thread
        runOnUiThread(new Runnable() {
            public void run() {
                /**
                 * Updating parsed JSON data into ListView
                 * */
                ListAdapter adapter = new SimpleAdapter(
                        AllUsersActivity.this, usersList,
                        R.layout.list_item, new String[] { TAG_UID,
                                TAG_FIRSTNAME},
                        new int[] { R.id.uid, R.id.name });
                // updating listview
                setListAdapter(adapter);
            }
        });
    }
}

有人能帮帮我吗!我得到的错误是分析数据组织时出错。JSONException:值无法转换为JSONObject

这是JSON字符串

Array{"Users":[{"UserID":"1","FirstName":"lalawee","Email":"12345","Password":null},{"UserID":"2","FirstName":"shadowblade721","Email":"12345","Password":null},{"UserID":"3","FirstName":"dingdang","Email":"12345","Password":null},{"UserID":"4","FirstName":"solidsnake0328","Email":"12345","Password":null}],"success":1}

错误是否存在于JSON解析器类中?

EDIT:这里是输出上面数组的php脚本的代码。有人能告诉我在json字符串之前输出单词数组的php脚本出了什么问题吗?很抱歉给你添麻烦。我是编码新手。一直在关注在线教程,但已经坚持了几天了。

<?php
/*
* Following code will list all the Users
*/
// array for JSON response
$response = array();
// include db connect class
require_once __DIR__ . '/db_connect.php';
// connecting to db
$db = new DB_CONNECT();
// get all Users from Users table
$result = mysql_query("SELECT * FROM Users") or die(mysql_error());
// check for empty result
if (mysql_num_rows($result) > 0) {
// looping through all results
// Users node
$response["Users"] = array();
while ($row = mysql_fetch_array($result)) {
    // temp user array
    $user[] = array();
    $user["UserID"] = $row["UserID"];
    $user["FirstName"] = $row["FirstName"];
    $user["Email"] = $row["Email"];
    $user["Password"] = $row["Password"];
    // push single User into final response array
    array_push($response["Users"], $user);
}
// success
$response["success"] = 1;
echo $response;
// echoing JSON response
echo json_encode($response);
}
else {
// no Users found
$response["success"] = 0;
$response["message"] = "No Users found";
// echo no users JSON
echo json_encode($response);
}
?>

使用GSON库,它是谷歌的官方库,工作起来很有魅力。没有HashMaps等,直接对象。

看看这里。

如错误所示,JSON字符串不代表有效的JSON对象。

这样尝试(一开始没有Array):

{"Users":[{"UserID":"1","FirstName":"lalawee","Email":"12345","Password":null},{"UserID":"2","FirstName":"shadowblade721","Email":"12345","Password":null},{"UserID":"3","FirstName":"dingdang","Email":"12345","Password":null},{"UserID":"4","FirstName":"solidsnake0328","Email":"12345","Password":null}],"success":1}Array{"Users":[{"UserID":"1","FirstName":"lalawee","Email":"12345","Password":null},{"UserID":"2","FirstName":"shadowblade721","Email":"12345","Password":null},{"UserID":"3","FirstName":"dingdang","Email":"12345","Password":null},{"UserID":"4","FirstName":"solidsnake0328","Email":"12345","Password":null}],"success":1}

您可以在此处检查您的JSON String的有效性:http://jsonlint.com/