发布到服务器后获取响应 [英] Get Response after Posting to server

问题描述

要上传数据到服务器，我用下面的方法，成功地发送到服务器，在这里，我需要得到一个数字，它是注册ID。我有codeD，在服务器的PHP。我试图与邮差工作正常，但在这里我没有变。所以，请让我知道我有什么关系？

 静态字符串结果= NULL;
    私有静态无效后（字符串终点，地图＆LT;字符串，字符串＆GT; PARAMS）
            抛出IOException
        URL网址;
        尝试{
            URL =新的URL（端点）;
        }赶上（MalformedURLException的E）{
            抛出新抛出：IllegalArgumentException（无效URL：+终点）;
        }
        StringBuilder的健美=新的StringBuilder（）;
        迭代器＆LT;钥匙进入LT;字符串，字符串＆GT;＆GT; 。迭代= params.entrySet（）迭代（）;
        //构造使用的参数POST正文
        而（iterator.hasNext（））{
            进入＆LT;字符串，字符串＆GT;参数= iterator.next（）;
            bodyBuilder.append（param.getKey（））。追加（'='）
                    .append（param.getValue（））;
            如果（iterator.hasNext（））{
                bodyBuilder.append（'和;'）;
            }
        }
        串体= bodyBuilder.toString（）;
        Log.v（TWA报告客户端，发帖+体+'给+网址）;
        字节[]字节= body.getBytes（）;
        HttpURLConnection的康恩= NULL;
        尝试{
            康恩=（HttpURLConnection类）url.openConnection（）;
            conn.setDoOutput（真）;
            conn.setUseCaches（假）;
            conn.setFixedLengthStreamingMode（bytes.length）;
            conn.setRequestMethod（POST）;
            conn.setRequestProperty（内容类型，
            应用程序/ x-WWW的形式urlen codeD;字符集= UTF-8）;
            //张贴请求
            OutputStream的OUT = conn.getOutputStream（）;
            out.write（字节）;
            out.close（）;            //处理响应
            INT状态= conn.getResponse code（）;
            如果（状态！= 200）{
              抛出新IOException异常（+状态后，错误code失败）;
            }
        } {最后
            如果（参数conn！= NULL）{
                conn.disconnect（）;
            }
        }      }
 

解决方案

试试这个片段，我会努力给你，

  URL网址;
URL =新的URL（http://www.url.com/app.php）;
URLConnection的连接;
连接= url.openConnection（）;
HttpURLConnection的httppost =（HttpURLConnection类）连接;
httppost.setDoInput（真）;
httppost.setDoOutput（真）;
httppost.setRequestMethod（POST）;
httppost.setRequestProperty（用户代理，瓦特兰海克-版本-t1.914）;
httppost.setRequestProperty（Accept_Language，EN-US）;
httppost.setRequestProperty（内容类型，
        应用程序/ x-WWW的形式urlen codeD）;
DataOutputStream类DOS =新的DataOutputStream类（httppost.getOutputStream（））;
dos.write（二）; //字节[] B后运行数据字符串回复;
InputStream的时间= httppost.getInputStream（）;
StringBuffer的SB =新的StringBuffer（）;
尝试{
    INT CHR;
    而（（CHR = in.read（））！=  -  1）{
        sb.append（（char）的CHR）;
    }
    回复= sb.toString（）;
} {最后
    附寄（）;
}
 

To post data to the server, I use the following method, successfully posts to the server, here I need to get one number that is registration id. I have coded that in the server php. I tried with Postman works fine, but here I'm not getting. So please let me know what I have to do?

static String result=null;    
    private static void post(String endpoint, Map<String, String> params)
            throws IOException {
        URL url;
        try {
            url = new URL(endpoint);
        } catch (MalformedURLException e) {
            throw new IllegalArgumentException("invalid url: " + endpoint);
        }
        StringBuilder bodyBuilder = new StringBuilder();
        Iterator<Entry<String, String>> iterator = params.entrySet().iterator();
        // constructs the POST body using the parameters
        while (iterator.hasNext()) {
            Entry<String, String> param = iterator.next();
            bodyBuilder.append(param.getKey()).append('=')
                    .append(param.getValue());
            if (iterator.hasNext()) {
                bodyBuilder.append('&');
            }
        }
        String body = bodyBuilder.toString();
        Log.v("TWA CLIENT REPORT", "Posting '" + body + "' to " + url);
        byte[] bytes = body.getBytes();
        HttpURLConnection conn = null;
        try {
            conn = (HttpURLConnection) url.openConnection();
            conn.setDoOutput(true);
            conn.setUseCaches(false);
            conn.setFixedLengthStreamingMode(bytes.length);
            conn.setRequestMethod("POST");
            conn.setRequestProperty("Content-Type",
            "application/x-www-form-urlencoded;charset=UTF-8");
            // post the request
            OutputStream out = conn.getOutputStream();
            out.write(bytes);
            out.close();       

            // handle the response
            int status = conn.getResponseCode();
            if (status != 200) {
              throw new IOException("Post failed with error code " + status);
            }
        } finally {
            if (conn != null) {
                conn.disconnect();
            }
        }

      } 

解决方案

try this snippets, i will work to you,

URL url;
url = new URL("http://www.url.com/app.php");
URLConnection connection;
connection = url.openConnection();
HttpURLConnection httppost = (HttpURLConnection) connection;
httppost.setDoInput(true);
httppost.setDoOutput(true);
httppost.setRequestMethod("POST");
httppost.setRequestProperty("User-Agent", "Tranz-Version-t1.914");
httppost.setRequestProperty("Accept_Language", "en-US");
httppost.setRequestProperty("Content-Type",
        "application/x-www-form-urlencoded");
DataOutputStream dos = new DataOutputStream(httppost.getOutputStream());
dos.write(b); // bytes[] b of post data

String reply;
InputStream in = httppost.getInputStream();
StringBuffer sb = new StringBuffer();
try {
    int chr;
    while ((chr = in.read()) != -1) {
        sb.append((char) chr);
    }
    reply = sb.toString();
} finally {
    in.close();
}

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