Android Json不能与HttpPost一起使用，可能与setEntity有关

我正在使用此代码将其发送到我的php文件。 文件看起来是这样的

file_put_contents('dump.txt', "POST: \n" . print_r($_POST, true) . "\n\n\n GET: \n" . print_r($_GET, true));

我发送的json如下所示：

public void postData(String url,JSONObject json) {
    HttpClient httpclient = new DefaultHttpClient();
    String object = "?data=" +json.toString();
    System.out.println("object:" + object);
    try {
        HttpPost httppost = new HttpPost(url.toString());
        httppost.setHeader("Content-type", "application/json");
        System.out.println("url:" + url.toString());
        StringEntity se = new StringEntity(object); 
        System.out.println("json:" + json.toString());

        se.setContentEncoding(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));
        httppost.setEntity(se); 

        HttpResponse response = httpclient.execute(httppost);
        String temp = EntityUtils.toString(response.getEntity());
        System.out.println("response:" + response);

        int statusCode = response.getStatusLine().getStatusCode();
        System.out.println("Statuscode " + statusCode);

        if(statusCode==200) {
            Toast.makeText(LbsGeocodingActivity.this, "Verstuurd",  Toast.LENGTH_LONG).show();
        }
        else if(statusCode!=200) {
            Toast.makeText(this, statusCode, Toast.LENGTH_LONG).show();
        }
        /* try {
                String responseBody = EntityUtils.toString(response.getEntity());
        Toast.makeText(this, responseBody, Toast.LENGTH_LONG).show();
        } catch (ParseException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
        } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
        }*/
    } 
    catch (ClientProtocolException e) {

    } 
    catch (IOException e) {
    }
}

日志文件如下所示：

01-08 09:36:00.278: I/System.out(1124): object:?data={"id":"69403","timestamp":"08-01-2013 09:36:00","longitude":"-122.084095","latitude":"37.422005"}
01-08 09:36:00.298: I/System.out(1124): json:{"id":"69403","timestamp":"08-01-2013 09:36:00","longitude":"-122.084095","latitude":"37.422005"}
01-08 09:36:01.038: I/System.out(1124): response:org.apache.http.message.BasicHttpResponse@412d1b38
01-08 09:36:01.038: I/System.out(1124): Statuscode 200

出于安全原因，我省略了url

但是我在dump.txt中得到一个空文件

POST: 
Array
(
)



 GET: 
Array
(
)

编辑

setEntity（或附近的某个地方）有问题，因为当我已经在url中粘贴？data=时，我从PHP文件中得到了以下信息：

POST: 
Array
(
)



 GET: 
Array
(
    [data] => 
)

---

我在应用程序中使用我的

DefaultHttpClient

发送JSON的方式非常有效：

public static HttpUriRequest createPostForJSONObject(
        JSONObject params, String url) {
    HttpPost post = new HttpPost(url);
    post.setEntity(createStringEntity(params));
    return post;
}

private static HttpEntity createStringEntity(JSONObject params) {
    StringEntity se = null;
    try {
        se = new StringEntity(params.toString(), "UTF-8");
        se.setContentType("application/json; charset=UTF-8");
    } catch (UnsupportedEncodingException e) {
        Log.e(TAG, "Failed to create StringEntity", e);
        exception = e;
    }
    return se;
}

要在代码中运行此请求，您只需执行以下操作：

public void postData(String url,JSONObject json) {
    HttpClient httpclient = new DefaultHttpClient();
    try {
          HttpPost post = createPostForJSONObject(json, url);
          HttpResponse response = httpClient.execute(post);
          // DO YOUR THING
    } catch Exception {
         // HANDLE EXCEPTION
    }
}

---

谢谢你的快速回复，我马上就去测试。我面临一个小问题。如何从代码中调用此函数requestNewVideoMessageWithThreadId？这里没有发送HttpPost的方法发送HttpPost您需要做的就是调用

defaultHttpClent.execute（request）

Wehre我需要发送它吗？在一个新函数中或在“return se