来自 Http 客户端的 HttpURLConnection 更新答案

【问题标题】：HttpURLConnection update from Http Client来自 Http 客户端的 HttpURLConnection 更新
【发布时间】：2015-12-25 03:44:58
【问题描述】：

您好，我想知道是否有人可以帮助我解决以下问题，我有一个当前已填充的数据库。我曾经使用 http 客户端调用它，它运行良好，但现在我正在尝试更新代码，因为它已被弃用以使用 httpurlconnection，但我没有成功。我查阅了一些教程并尝试了一些东西，但似乎没有用。数据库通过 php 文件调用并以 json 格式返回。如果我要从浏览器调用 php 文件，则响应如下： [{"id":"15","logo":"logo url ","title":"title"}]

我在控制台上得到的错误如下：java.lang.NullPointerException: Attempt to invoke virtual method 'void java.io.InputStream.close()' on a null object reference

这对我来说没有多大意义，因为脚本会提取信息

我有以下代码，我留下了注释部分以防万一我需要它，它还包括我用来调用数据库的旧方式谢谢！：

公共无效负载新闻（）{ InputStream is = null;

    String result = "";
 ArrayList<NameValuePair>();

    try {
        URL url = new URL("http://databasecall.php");
        //HttpClient httpclient = new DefaultHttpClient();
        HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
        //urlConnection.setRequestMethod("GET");
        //urlConnection.setRequestProperty("Content-length", "0");
        //urlConnection.setUseCaches(false);
        //urlConnection.setAllowUserInteraction(false);
        //urlConnection.setConnectTimeout(15000);
        //urlConnection.setReadTimeout(15000);
        //urlConnection.connect();
        int responseCode = urlConnection.getResponseCode();
        Log.i("Tag:", Integer.toString(responseCode)); //tag 200
        //HttpPost httppost = new HttpPost("http://databasecall.php");
        //httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
        //HttpResponse response = httpclient.execute(httppost);
        //HttpEntity entity = response.getEntity();
        //is = entity.getContent();

    /*}catch(Exception e){
        Log.e("log_tag", "Error in http connection "+e.toString());
    }*/

        //convert response to string
        //try{
        //BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
        if (responseCode == HttpURLConnection.HTTP_OK) {
            InputStream in = new BufferedInputStream(urlConnection.getInputStream());

            BufferedReader reader = new BufferedReader(new InputStreamReader(in));
            //BufferedReader reader = new BufferedReader(new InputStreamReader(urlConnection.getInputStream(), "iso-8859-1"), 8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null) {
                sb.append(line + "\n");
            }
            is.close();
            result = sb.toString();
            Log.i("Tag:", result);
        }
        }catch(Exception e){
            Log.e("log_tag", "Error converting result " + e.toString());
        }

【问题讨论】：

您正在打开 InputStream in .. 但您正在关闭 is.close();。哎呀:)
做到了！哈哈流行是对的，总是那些小东西我应该把它放在哪里？

标签： java php android httpurlconnection

【解决方案1】：

更新的 API

try {
        String urlParameters =  "name=toni&class=one&param3=ok"; 
        byte[] postData = urlParameters.getBytes(Charset.forName("UTF-8"));
        int postDataLength = postData.length;
        String request = "http://rocks.php";
        URL url = new URL(request);
        HttpURLConnection cox = (HttpURLConnection) url.openConnection();
        cox.setDoOutput(true);
        cox.setDoInput(true);
        cox.setInstanceFollowRedirects(false);
        cox.setRequestMethod("POST");
        cox.setRequestProperty("Content-Type",
                "application/x-www-form-urlencoded");
        cox.setRequestProperty("charset", "utf-8");
        cox.setRequestProperty("Content-Length",
                Integer.toString(postDataLength));
        cox.setUseCaches(false);
        OutputStreamWriter writer = new OutputStreamWriter(
                cox.getOutputStream());

        writer.write(urlParameters);
        writer.flush();
        String line;
        BufferedReader reader = new BufferedReader(new InputStreamReader(
                cox.getInputStream()));
        while ((line = reader.readLine()) != null) {
            System.out.println(line);
        }
        writer.close();
        reader.close();
    } catch (Exception e) {

        result = e.toString();
        Sucess = false;
        e.printStackTrace();
 }

【讨论】：