在HttpURLConnection类为什么不作为的JSONObject PARAMS的工作，但String作为PARAMS正在 [英] In HttpURLConnection Why don't JSONObject as Params work but String as Params are working

问题描述

的功能是：

private String register(String myurl) throws IOException {

        String resp = null;
        try {
            JSONObject parameters = new JSONObject();
           // parameters.put("jsonArray", ((makeJSON())));
            parameters.put("key", "key");//getencryptkey());
            URL url = new URL(myurl);
            HttpURLConnection conn = (HttpURLConnection) url.openConnection();
            //  conn.setReadTimeout(10000 /* milliseconds *///);
            //  conn.setConnectTimeout(15000 /* milliseconds */);
            conn.setRequestProperty("Content-Type", "application/json");
            conn.setDoOutput(true);
            conn.setDoInput(true);
            conn.setRequestMethod("POST");
            OutputStream out = new BufferedOutputStream(conn.getOutputStream());
            BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(out, "UTF-8"));
            writer.write(parameters.toString());
            writer.close();
            out.close();

            int responseCode = conn.getResponseCode();
            System.out.println("\nSending 'POST' request to URL : " + url);
            System.out.println("Response Code : " + responseCode);

            BufferedReader in = new BufferedReader(new InputStreamReader(conn.getInputStream()));
            String inputLine;
            StringBuffer response = new StringBuffer();


            while ((inputLine = in.readLine()) != null) {
                response.append(inputLine);
            }
            in.close();
            System.out.println("strngbuffr" + response.toString());
            resp = response.toString();

        } catch (Exception exception) {
            System.out.println("Exception: " + exception);
        }

        System.out.println("rsp"+ resp.toString());
        return resp.toString();
    }

我得到的回应code 200，这意味着连接是好的，但是我得到PHP端空变量，还有什么能错在这里？

I get the response code as 200, which means connection is okay however I get empty variables on PHP side, what can be wrong here?

早些时候，我发送一个JSON数组太多，但只是为了测试functonality我评论说出来，现在我只送一个变量键作为钥匙

Earlier I was sending a JSON array too but just to test functonality I commented that out now I am only sending one variable key as "key"

它令人惊讶地看到，这个样本code ++工程 - SAN的JSON阵列和键值对：

Its amazing to see, this sample code works - sans the JSON array and the key value pairs:

private String sendPost(String url) throws Exception {

        String USER_AGENT = "Mozilla/5.0";
        URL obj = new URL(url);

        HttpURLConnection con = (HttpURLConnection) obj.openConnection();

        //add reuqest header
        con.setRequestMethod("POST");
        con.setRequestProperty("User-Agent", USER_AGENT);
        con.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");

        String urlParameters ="sn=C02G8416DRJM&cn=&locale=&caller=&num=12345";
        // Send post request
        con.setDoOutput(true);
        DataOutputStream wr = new DataOutputStream(con.getOutputStream());
        wr.writeBytes(urlParameters);
        wr.flush();
        wr.close();

        int responseCode = con.getResponseCode();
        System.out.println("\nSending 'POST' request to URL : " + url);
        System.out.println("Post parameters : " + urlParameters);
        System.out.println("Response Code : " + responseCode);

        BufferedReader in = new BufferedReader(
                new InputStreamReader(con.getInputStream()));
        String inputLine;
        StringBuffer response = new StringBuffer();

        while ((inputLine = in.readLine()) != null) {
            response.append(inputLine);
        }
        in.close();

        //print result
        System.out.println("rvsp"+response.toString());

        return response.toString();
    }

所以它归结为更换这样的：

So it boils down to replacing this:

 JSONObject parameters = new JSONObject();
            parameters.put("jsonArray", new JSONArray(Arrays.asList(makeJSON())));
            parameters.put("key", getencryptkey()); 

本：

String urlParameters ="jArr="+makeJSON()+"Key="+getencryptkey();

和我仍然好奇。

推荐答案

我觉得这里的问题是不是在Java方面，如果这些参数是固定式的像JSON在你的情况下，JSON对象作为POST PARAMS方法如果收集这种方式在PHP端将工作：

I reckon the problem here is not at the Java side, If the parameters is of fixed type like in json in your case, the JSON Object as POST params method will work if collected this way on the php side:

<?php
    $json = file_get_contents('php://input');
    $obj = json_decode($json);
    print_r($obj);
    print_r("this is a test response");
?>

这篇关于在HttpURLConnection类为什么不作为的JSONObject PARAMS的工作，但String作为PARAMS正在的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！