Question

我想知道为什么这段代码不能执行？我正在尝试通过POST方法从我的设备发送数据，但没有错误。该应用程序通过沟通“我的应用程序已停止。”

在我的设备上完成

这是执行：

KlientNameValue kn = new KlientNameValue(getApplicationContext());
kn.new  MyAsyncTask().execute(zam.klient.getNazwa(),zam.klient.getNip(),zam.klient.getAdres());

这是代码：

public class KlientNameValue {

List<NameValuePair> KlientNameValuePairs = new ArrayList<NameValuePair>();
Context context;
public KlientNameValue(Context context) {
    super();
    this.context=context;
}



 public class MyAsyncTask extends AsyncTask<String, Void, Void> {

      @Override protected Void doInBackground(String... params) { 
          // TODO     Auto-generated method stub 
          postData(params[0], params[1], params[2]);
      return null;

      }

      @Override
      protected void onPostExecute(Void result) {

      Toast.makeText(context , "Zlecenie zostało wysłane",
     Toast.LENGTH_LONG).show();
          }

      void postData(String nazwa, String nip, String adres) {
          HttpClient httpclient = new DefaultHttpClient();
          HttpPost httppost = new HttpPost("here is my default link :)");

      try { // Add your data

      KlientNameValuePairs = new ArrayList<NameValuePair>();
      KlientNameValuePairs.add(new BasicNameValuePair("Kli_imie", nazwa));
      KlientNameValuePairs.add(new BasicNameValuePair("Kli_adres", adres));
      KlientNameValuePairs.add(new BasicNameValuePair( "Kli_nr_telefonu",
      nip)); 



      httppost.setEntity(new UrlEncodedFormEntity( KlientNameValuePairs));
      HttpResponse response = httpclient.execute(httppost);
      //httppost.setEntity(new UrlEncodedFormEntity( 
      //      ZamowienieNameValuePairs)); // HttpResponse response1 =



      } catch (IOException e) { // TODO Auto-generated catch block
      e.printStackTrace(); }

      }

      }

}

错误：

02-15 17:45:24.695: E/AndroidRuntime(21890): at android.widget.Toast.<init>(Toast.java:94) 
02-15 17:47:19.343: W/SingleClientConnManager(22288): Invalid use of SingleClientConnManager: connection still allocated.
 02-15 17:47:19.343: W/SingleClientConnManager(22288): Make sure to release the connection before allocating another one.

Answer 1

无效使用SingleClientConnManager：仍然分配了连接。

您正在执行两次完全错误的http请求，然后再使用它。因此，请删除第二个httpclient.execute(httppost);，因为您已经执行了此http请求。

并调用此

httpResponse.getEntity().consumeContent();

调用上述方法以指示不再需要此实体的内容。由于此方法调用，所有实体实现都应该释放所有已分配的资源

Answer 2

public static DefaultHttpClient getThreadSafeClient() {

    DefaultHttpClient client = new DefaultHttpClient();
    ClientConnectionManager mgr = client.getConnectionManager();
    HttpParams params = client.getParams();

    client = new DefaultHttpClient(
        new ThreadSafeClientConnManager(params,
            mgr.getSchemeRegistry()), params);

    return client;
}

使用此代码，以便您的免费资源和分配或使用的例外不会出现当两个或多个线程与单个org.apache.http.impl.client.DefaultHttpClient交互时，可能会发生此异常或者只是为每个请求提供新对象，无论何时何地，您都可以获取或发布http客户端请求与服务器交互以获取数据或下载大文件像

String post_url="http://www.google.com";
    DefaultHttpClient client = new DefaultHttpClient();
    HttpPost httpost = new HttpPost(Post_url);     
    httpost.setHeader("Accept", "application/json");
    httpost.setHeader("Content-type", "application/json");
    HttpResponse httpResponse = client.execute(httpost);
    String response= EntityUtils.toString(httpResponse.getEntity());

如您所愿在代码中检索响应

SingleClientConnManager的使用无效：仍然分配了连接

2 个答案: