2013-08-06 34 views
1

我試圖從一個應用程序發送一個字符串到另一個。我會返回一個字符串的響應。在這裏,myhost.com/views是我需要發送字符串值並從中獲得響應的第二個應用程序。但是,當我試圖發送它不執行此代碼。有人可以糾正我在哪裏我錯了嗎?錯誤:目標主機不能爲空,或在參數中設置

下面是我寫的代碼。

public static void sendData(String strval) throws IOException{ 
String doSend="https://myhost.com/views?strval="+strval; 
    HttpClient httpclient = new DefaultHttpClient(); 
    try { 
     System.out.println("inside try"); 
     URIBuilder builder = new URIBuilder(); 
     System.out.println("builder="+builder); 

     builder.setHost("myhost.com").setPath("/views"); 
     builder.addParameter("strval", strval); 
     System.out.println("add param,sethost,setpath complete"); 

     URI uri = builder.build(); 
     System.out.println("uri="+uri); 


     HttpGet httpget = new HttpGet(uri); 
     System.out.println("httpGet"+httpget); 

     HttpResponse response = httpclient.execute(httpget); 
     System.out.println(response.getStatusLine().toString()); 

     if (response.getStatusLine().getStatusCode() == 200) { 
      String responseText = EntityUtils.toString(response.getEntity()); 
      System.out.println("responseText="+responseText); 
      httpclient.getConnectionManager().shutdown(); 
     } else { 
      System.out.println("Server returned HTTP code " 
        + response.getStatusLine().getStatusCode()); 
     } 
     } catch (java.net.URISyntaxException bad) { 
     System.out.println("URI construction error: " + bad.toString()); 
     } 
     catch(Exception e){ System.out.println("e.getMessage=>"+e.getMessage()); } 

    } 

代碼運行,直到當我打印例外我看到excep.getMessage() - >

 java.lang.IllegalStateException: Target host must not be null, or set in parameters. 
    at org.apache.http.impl.client.DefaultRequestDirector.determineRoute(DefaultRequestDirector.java:789) 
    at org.apache.http.impl.client.DefaultRequestDirector.execute(DefaultRequestDirector.java:414) 
    at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:906) 
    at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:805) 
    at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:784) 
+1

你可以找到類路徑上的jar列表?您需要httpclient jar和httpcore jar。 – roby

+0

問題在於我使用的httpcore jar版本。 版本4.2.1是不兼容的,我使用4.2.3版本,它工作正常,直到 HttpGet httpget = new HttpGet(uri);更新了異常堆棧,因爲我在同一代碼中看到了其他錯誤。 – smiley

回答

7

此代碼不能將其識別爲一個有效的URI,因爲它缺少http。這是我添加的解決代碼:

builder.setScheme("http"); 
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