Question

我通过JSON使用php包装器从服务器端返回值。但是当我将值返回给客户端时会发生以下错误。

这是我的客户端代码

        @Override
    protected Boolean doInBackground(String... arg0) {

        try {

            // Setup the parameters
            ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
            nameValuePairs.add(new BasicNameValuePair("FirstNameToSearch",
                    strNameToSearch));
            // Create the HTTP request
            HttpParams httpParameters = new BasicHttpParams();

            // Setup timeouts
            HttpConnectionParams
                    .setConnectionTimeout(httpParameters, 45000);
            HttpConnectionParams.setSoTimeout(httpParameters, 45000);

            HttpClient httpclient = new DefaultHttpClient(httpParameters);
            HttpPost httppost = new HttpPost(
                    "http://172.16.12.142/etsmobile/menuload.php");

            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

            HttpResponse response = httpclient.execute(httppost);
            HttpEntity entity = response.getEntity();

            String result = EntityUtils.toString(entity);       

            // Create a JSON object from the request response
            JSONObject jsonObject = new JSONObject(result);

            // Retrieve the data from the JSON object
            pasName = jsonObject.getString("Name");
            pasPost = jsonObject.getString("Post");
            pasStation = jsonObject.getString("Station");


        } catch (Exception ex) {
            ex.printStackTrace();
        }

        return true;
    }

这是我的服务器端代码

<?php

$firstname = $_POST["FirstNameToSearch"];

$con = mysql_connect("localhost", "root", "") or die("Unable to connect to MySQL");


if (mysqli_connect_errno()) {
    echo 'Database connection error: ' . mysqli_connect_error();
    exit();
}

$selected = mysql_select_db("ets", $con) or die("Could not select ets");

$userdetails = mysql_query("SELECT users.* FROM login, users WHERE username = '$firstname' and login.emp_no=users.emp_no");
$getUser_result = mysql_fetch_assoc($userdetails);


$name = $getUser_result['name'];
$post = $getUser_result['post'];
$station = $getUser_result['station'];

mysql_close($con);

$result_data = array('Name' => $name, 'Post' => $post, 'Station' => $station);
//print_r($result_data);
echo json_encode($result_data);
?>

这是我的JSON输出

{"Name":"Sameera Yatawara","Post":"Station Master","Station":"Dematagoda"}

Answer 1

尝试使用以下内容检索数据：

...
String result = EntityUtils.toString(entity);

JSONArray array = (JSONArray) new JSONTokener(result).nextValue(); 
JSONObject json = array.getJSONObject(0);
json.getString('name');
...

或

...
String result = EntityUtils.toString(entity);

JSONObject json = (JSONObject ) new JSONTokener(result).nextValue();
json.getString('name');
...

同时检查PHP脚本编码。我有类似的问题。我在文本编辑器（Notepad ++，..）中使用DOM将编码设置为UTF-8，然后开始工作。

org.json.JSONException：Value类型为java.lang.String的数组无法转换为JSONObject

1 个答案: