Question

我从android连接到restful web服务。我需要使用httpPost 添加信息。但这需要很长时间才能得到例外。 java.net.SocketException：操作超时

另一方面我可以制作httpGet，我不会采取任何异常

代码在这里。

ip等于我的计算机的本地IP是网络。 ip = 10.80.20.20

   HttpClient httpclient = new DefaultHttpClient();  
         String url="http://"+ip+"/projectt/source/applySurvey";
             HttpPost httppost = new HttpPost(url);  
           String strSurveyId=new Long(surveyId).toString();
             try {  
                 // Add your data  
                 List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);  
                 nameValuePairs.add(new BasicNameValuePair("surveyId",strSurveyId));  
                 nameValuePairs.add(new BasicNameValuePair("questionId", "4"));  

                 httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));  

                 // Execute HTTP Post Request  
                 HttpResponse response = httpclient.execute(httppost); 
                HttpEntity entity = response.getEntity();

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

我的清单文件在这里

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
      package="com.mobil.survey.project"
      android:versionCode="1"
      android:versionName="1.0">
      <uses-permission android:name="android.permission.INTERNET"/>
    <application android:icon="@drawable/icon" android:label="@string/app_name">
        <activity android:name=".categoryList"
                  android:label="@string/app_name">
            <intent-filter>
                <action android:name="android.intent.action.MAIN" />
                <category android:name="android.intent.category.LAUNCHER" />
            </intent-filter>
        </activity>
        <activity android:name=".firmsList" android:label="@string/app_name"></activity>
         <activity android:name=".surveiesList" android:label="@string/app_name"></activity>
          <activity android:name=".survey" android:label="@string/app_name"></activity>

    </application>


</manifest>

定义了哪个安静的Web服务web.xml就在这里

<?xml version="1.0" encoding="UTF-8"?>
 <web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
        <display-name>projectt</display-name>




       <servlet> 
             <servlet-name>FormValidator</servlet-name> 
             <servlet-class>com.project.servlets.FormValidator</servlet-class> 
       </servlet> 
       <servlet-mapping> 
                <servlet-name>FormValidator</servlet-name>      
                <url-pattern>/FormValidator</url-pattern>
       </servlet-mapping> 

        <servlet>
        <servlet-name>Jersey REST Service</servlet-name>
        <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
        <init-param>
            <param-name>com.sun.jersey.config.property.packages</param-name>
            <param-value>com.project.resources</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>Jersey REST Service</servlet-name>
        <url-pattern>/source/*</url-pattern>
    </servlet-mapping>

       <welcome-file-list>  
          <welcome-file>index.html</welcome-file>
          <welcome-file>index.htm</welcome-file>  
          <welcome-file>index.jsp</welcome-file> 
           <welcome-file>default.html</welcome-file>  
           <welcome-file>default.htm</welcome-file>
           <welcome-file>default.jsp</welcome-file>
       </welcome-file-list>        
</web-app>

和我连接的uri资源在这里

@Path("/applySurvey")
public class AppliedSurveyResource  {

    @GET
    @Produces( MediaType.APPLICATION_JSON)
    public List<WSCategory> getCategories() {
        List<WSCategory> categories = new ArrayList<WSCategory>();
        categories.addAll( CategoryProvider.instance.getModel() );
        return categories; 
    }   

@POST
    @Produces(MediaType.TEXT_HTML)
    @Consumes(MediaType.APPLICATION_FORM_URLENCODED)
    public void newTodo(
            @FormParam("surveyId") Long surveyId,
            @FormParam("questionId") Long questionId,
            @FormParam("cellId") String cellId,
            @Context HttpServletResponse servletResponse
    ) throws IOException {

        PRAppliedSurvey appliedSurvey=new PRAppliedSurvey();
        appliedSurvey.setSurveyId(surveyId);
        appliedSurvey.setQuestionId(questionId);
        appliedSurvey.setCellId(cellId);
        java.util.Date today = new java.util.Date();
        System.out.println(new java.sql.Timestamp(today.getTime()));
        appliedSurvey.setApplyDt(new java.sql.Timestamp(today.getTime()));
        try {
            appliedSurvey.store();
        } catch (Exception e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

        /*Todo todo = new Todo(id,summary);
        if (description!=null){
            todo.setDescription(description);
        }
        TodoDao.instance.getModel().put(id, todo);

        URI uri = uriInfo.getAbsolutePathBuilder().path(id).build();
        Response.created(uri).build();

        servletResponse.sendRedirect("../create_todo.html");
        */
    }




}

}

Answer 1

尝试使用AsyncTask在不同的线程中运行它。 UI线程在引发超时之前具有相当低的阈值。这是为了防止UI对用户造成冻结。

http://developer.android.com/reference/android/os/AsyncTask.html

httpPost需要很长时间并获得此异常：java.net.SocketException：操作超时

1 个答案: