【发布时间】：2014-02-05 20:06:04
【问题描述】：
我是 Android 新手，我正在尝试从 android 访问 wcf 服务，它给出空指针异常， 这是我的 .net wcf 服务 IUserService
[ServiceContract]
public interface IUserService
{ [OperationContract]
[WebInvoke(Method="GET",ResponseFormat = WebMessageFormat.Json,UriTemplate = "GetName")]
string GetName();
}
这是我的用户服务
public class UserService : IUserService
{
public string GetName()
{
return "Hello ! ";
}
}
这是我的 xml
<service name="Lera.Template.Services.WCF.UserService">
<endpoint address="" binding="webHttpBinding" contract="Lera.Template.Services.WCF.IUserService"
behaviorConfiguration="httpBehavior">
<identity>
<dns value="localhost" />
</identity>
</endpoint>
<endpoint address="mex" binding="mexHttpBinding" contract="IMetadataExchange" />
<host>
<baseAddresses>
<add baseAddress="http://localhost:8888/Lera.Template.Services.WCF/UserService/" />
</baseAddresses>
</host>
</service>
我正在使用日食 这是我的主要活动
public class MainActivity extends Activity {
private String values ="";
Button btn;
TextView tv;
private static String url = "http://192.168.12.146:8888/Lera.Template.Services.WCF/UserService/GetName";
private static final String StringVal = "StringValue";
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
JSONParser jParser = new JSONParser();
// getting JSON string from URL
JSONObject json = jParser.getJSONFromUrl(url);
try {
String temp = json.getString(StringVal);
Toast toast = Toast.makeText(this ,temp , Toast.LENGTH_SHORT);
toast.show();
}
catch (JSONException e) {
e.printStackTrace();
}
catch (Exception ex) {
Log.e("final:", ex.toString());
}
}
}
我正在使用 json 解析器类 公共类 JSONParser {
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";
// constructor
public JSONParser() {
}
public JSONObject getJSONFromUrl(String url) {
// Making HTTP request
try {
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpGet httpget = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpget);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
Log.e("connection" , e.toString());
}
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(is, "iso-8859-1"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
json = sb.toString();//here json type is string
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// return JSON String
return jObj;
}
}
Wcf 服务在浏览器上运行良好，但是当我尝试从我的 android 应用程序访问时，它给出了空指针异常
给出空指针异常的行是
HttpResponse httpResponse = httpClient.execute(httpget);
请帮我解决这个问题，我已经尝试了网上所有可用的东西，但仍然无法克服这个问题。
作为 A.S.提到我更新了我的主要活动课程 这是我与异步调用的主要活动，但异常是 连接到 url 被拒绝
public class MainActivity extends Activity implements OnClickListener {
private String values ="";
Button btn;
TextView tv;
String uri = "http://192.168.0.144:8888/Lera.Template.Services.WCF/UserService/GetName";
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
btn = (Button)this.findViewById(R.id.btnAccess);
tv = (TextView)this.findViewById(R.id.tvAccess);
btn.setOnClickListener(this);
}
@Override
public void onClick(View arg0) {
try
{
AsyncTaskExample task = new AsyncTaskExample(this);
task.execute("");
String test = values;
tv.setText(values);
} catch(Exception e)
{
Log.e("Click Exception: ", e.getMessage());
}
}
public class AsyncTaskExample extends AsyncTask<String, Void,String>
{
private String Result="";
//private final static String SERVICE_URI = "http://10.0.2.2:8889";
private final static String SERVICE_URI = "http://192.168.12.146:8888/Lera.Template.Services.WCF/UserService";
private MainActivity host;
public AsyncTaskExample(MainActivity host)
{
this.host = host;
}
public String GetSEssion(String URL)
{
boolean isValid = true;
if(isValid)
{
HttpClient client = new DefaultHttpClient();
//http://192.168.0.144:8888/Lera.Template.Services.WCF/UserService/
// HttpPost post = new HttpPost(uri);
HttpGet httpget = new HttpGet(uri);
httpget.setHeader("Accept", "application/json");
httpget.setHeader("Content-type", "application/json; charset=utf-8");
try
{
HttpResponse response = client.execute(httpget) ;
BufferedReader rd = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));
String line ="";
while((line = rd.readLine()) != null)
{
System.out.println(line);
}
}catch(Exception e)
{
Log.e("Error:", e.getMessage());
}
}
return Result;
}
@Override
protected String doInBackground(String... arg0) {
android.os.Debug.waitForDebugger();
String t = GetSEssion(SERVICE_URI);
return t;
}
@Override
protected void onPostExecute(String result) {
// host.values = Result;
super.onPostExecute(result);
}
@Override
protected void onPreExecute() {
// TODO Auto-generated method stub
super.onPreExecute();
}
@Override
protected void onCancelled() {
// TODO Auto-generated method stub
super.onCancelled();
}
}
}
如有任何帮助，我们将不胜感激。 谢谢。
【问题讨论】：
-
是 NetworkOnMainThread 问题吗？
-
@A.S.：谢谢您的回复，这些是我得到的异常 02-05 06:30:26.695: E/connection(3229): org.apache.http.conn.HttpHostConnectException: Connection to 192.168.0.144:80 拒绝 02-05 06:30:30.135：E/Buffer 错误（3229）：转换结果时出错 java.lang.NullPointerException：lock == null 02-05 06:30:31.485：E/JSON 解析器（3229） : 解析数据 org.json.JSONException 时出错：在 02-05 06:30:34.745 的字符 0 处输入结束：E/final:(3229):java.lang.NullPointerException
-
是的，您不能在主线程上发出任何 http 请求，请注意 AsyncTask 中的一些 http 连接教程
-
@A.S.：我也试过了，但让我再试一次。
-
您真的应该尝试一下，因为 Android 会阻止 MainThread 上的任何 http 请求，因此您不会得到任何输出 => Nullpointer。