JSONException error: Rest php server-Android Client

Question

.爪哇文件。这就是错误所在--->int success = json.getInt（TAG_SUCCESS）;

       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 < products.length(); i++) {
                    JSONObject c = products.getJSONObject(i);
                    // Storing each json item in variable
                    String id = c.getString(TAG_PID);
                    String name = c.getString(TAG_NAME);
                    Log.d("level2: ", "lksdjflsdjf0wrewrwje************************");
                    // creating new HashMap
                    HashMap<String, String> 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;
    }

来自服务器的 Json 数组如下所示。我已经从 jsonlint.com 验证了它

     {
"tbl_user": {
    "0": {
        "id": "195",
        "email": "[e-mail]",
        "password": "202cb962ac59075b964b07152d234b70",
        "fname": "aru",
        "lname": "sharma"
    },
    "1": {
        "id": "196",
        "email": "[e-mail]",
        "password": "202cb962ac59075b964b07152d234b70",
        "fname": "manu",
        "lname": "sharma"
    },
    "2": {
        "id": "197",
        "email": "[e-mail]",
        "password": "202cb962ac59075b964b07152d234b70",
        "fname": "rishi",
        "lname": "sharma"
    },
    "success": 1
}

}

我想我无法从这里读出成功的价值。我无法在此 Json 字符串中看到错误。请帮忙。

服务器上创建 json 的 Rest Php 代码是

    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() .'}}'; 
}

}

Answer 1

试试这个：

.PHP：

try {
$db = getConnection();
$stmt = $db->query($sql);  
$users["tbl_user"] = $stmt->fetchAll(PDO::FETCH_OBJ);
$users["success"] = 1;
$db = null;
echo json_encode($users) 
} catch(PDOException $e) {
echo '{"error":{"text":'. $e->getMessage() .'}}'; 
}

在 Java 文件中：

JSONObject json = jParser.makeHttpRequest(url_all_products, "GET", params);
int success = json.getInt(TAG_SUCCESS);

Answer 2

尝试这样做，我们必须首先获取JSON对象，然后再获取其中的内部值。

JSONObject json = new JSONObject(result);
int success =   json.getInt(TAG_SUCCESS);