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php-JSONException错误:android客户端

WBOY
WBOYOriginal
2016-06-02 11:35:011105Durchsuche

phpandroid

java.文件,错误:int success = json.getInt(TAG_SUCCESS);

<code>   protected String doInBackground(String... args) {    // Building Parameters    List<namevaluepair> params = new ArrayList<namevaluepair>();    // getting JSON string from URL    JSONObject json = jParser.makeHttpRequest(url_all_products, "GET", params);    // Check your log cat for JSON reponse    Log.d("All Products: ", json.toString());    try {        // Checking for SUCCESS TAG        **int success = json.getInt(TAG_SUCCESS);**        if (success == 1) {            // products found            // Getting Array of Products            products = json.getJSONArray(TAG_PRODUCTS);            Log.d("level1: ", "@@@@@@@@@@@@@@@@@@@@@@$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$");            // looping through All Products            for (int i = 0; i  map = new HashMap<string string>();                // adding each child node to HashMap key => value                map.put(TAG_PID, id);                map.put(TAG_NAME, name);                // adding HashList to ArrayList                productsList.add(map);            }        } else {            // no products found            // Launch Add New product Activity            Log.d("level3: ", "jldksffffffffffffffffffffffffffffffffffffff");            Intent i = new Intent(getApplicationContext(),                    NewProductActivity.class);            // Closing all previous activities            i.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);            startActivity(i);        }    } catch (JSONException e) {        e.printStackTrace();    }    return null;}</string></namevaluepair></namevaluepair></code>

服务器的jsonArray如下,我已经通过jsonlint.con验证过了

<code>     {"tbl_user": {    "0": {        "id": "195",        "email": "aru@yahoo.com",        "password": "202cb962ac59075b964b07152d234b70",        "fname": "aru",        "lname": "sharma"    },    "1": {        "id": "196",        "email": "manu@yahoo.com",        "password": "202cb962ac59075b964b07152d234b70",        "fname": "manu",        "lname": "sharma"    },    "2": {        "id": "197",        "email": "rishi@yahoo.com",        "password": "202cb962ac59075b964b07152d234b70",        "fname": "rishi",        "lname": "sharma"    },    "success": 1}}</code>

然后PHP代码如下:

<code>    function getUsers() {$sql = "select * FROM tbl_user ORDER BY fname";try {    $db = getConnection();    $stmt = $db->query($sql);      $users = $stmt->fetchAll(PDO::FETCH_OBJ);    $users["success"] = 1;    $db = null;    echo '{"tbl_user": ' . json_encode($users) . '}';} catch(PDOException $e) {    echo '{"error":{"text":'. $e->getMessage() .'}}'; }}</code>

请帮我解决一下,谢谢。

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