在 post 请求中将 JSON 传输到服务器

服务器有两个参数：String and JSON。 提示，正确我转移JSON和 POST 请求中的字符串？

try {
    HttpClient httpClient = new DefaultHttpClient();
    HttpPost httpPost = new HttpPost("my_url");
    List parameters = new ArrayList(2);
    JSONObject jsonObject = new JSONObject();
    jsonObject.put("par_1", "1");
    jsonObject.put("par_2", "2");
    jsonObject.put("par_3", "3");
    parameters.add(new BasicNameValuePair("action", "par_action"));
    parameters.add(new BasicNameValuePair("data", jsonObject.toString()));
    httpPost.setEntity(new UrlEncodedFormEntity(parameters));
    HttpResponse httpResponse = httpClient.execute(httpPost);
    Log.v("Server Application", EntityUtils.toString(httpResponse.getEntity())+" "+jsonObject.toString());

} catch (UnsupportedEncodingException e) {
    Log.e("Server Application", "Error: " + e);
} catch (ClientProtocolException e) {
    Log.e("Server Application", "Error: " + e);
} catch (IOException e) {
    Log.e("Server Application", "Error: " + e);
} catch (JSONException e) {
    e.printStackTrace();
}

我不太确定您的问题是什么，但这是我发送 JSON 的方式（使用您的数据示例）。

Android / JSON 构建：

JSONObject jo = new JSONObject();
jo.put("action", "par_action");
jo.put("par_1", "1");
jo.put("par_2", "2");
jo.put("par_3", "3");

Android / 发送 JSON：

URL url = new URL("http://domaintoreceive.com/pagetoreceive.php");

HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url.toURI());

// Prepare JSON to send by setting the entity
httpPost.setEntity(new StringEntity(jo.toString(), "UTF-8"));

// Set up the header types needed to properly transfer JSON
httpPost.setHeader("Content-Type", "application/json");
httpPost.setHeader("Accept-Encoding", "application/json");
httpPost.setHeader("Accept-Language", "en-US");

// Execute POST
response = httpClient.execute(httpPost);

PHP/服务器端：

<?php
if (file_get_contents('php://input')) {
    // Get the JSON Array
    $json = file_get_contents('php://input');
    // Lets parse through the JSON Array and get our individual values
    // in the form of an array
    $parsedJSON = json_decode($json, true);

    // Check to verify keys are set then define local variable, 
    // or handle however you would normally in PHP.
    // If it isn't set we can either define a default value
    // ('' in this case) or do something else
    $action = (isset($parsedJSON['action'])) ? $parsedJSON['action'] : '';
    $par_1 = (isset($parsedJSON['par_1'])) ? $parsedJSON['par_1'] : '';
    $par_2 = (isset($parsedJSON['par_2'])) ? $parsedJSON['par_2'] : '';
    $par_3 = (isset($parsedJSON['par_3'])) ? $parsedJSON['par_3'] : '';

    // Or we could just use the array we have as is
    $sql = "UPDATE `table` SET 
                `par_1` = '" . $parsedJSON['par_1'] . "',
                `par_2` = '" . $parsedJSON['par_2'] . "',
                `par_3` = '" . $parsedJSON['par_3'] . "'
            WHERE `action` = '" . $parsedJSON['action'] . "'";
}