[英]Sending Parameters to .net Restful WebService from Android
我正在嘗試發送一個采用交錯數組作為參數的調用webservice方法。 我構建了數組,但始終將null傳遞給Web服務。
這是我的java類:
package com.mitch.wcfwebserviceexample;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.StringEntity;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.message.BasicHeader;
import org.apache.http.protocol.HTTP;
import org.json.JSONArray;
import org.json.JSONObject;
import android.os.AsyncTask;
import android.os.Bundle;
import android.util.Log;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.Button;
import android.widget.TextView;
import android.app.Activity;
public class MainActivity extends Activity implements OnClickListener {
private String values ="";
Button btn;
TextView tv;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.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://10.0.2.2:1736";
private final static String SERVICE_URI = "http://10.0.2.2:65031/SampleService.svc";
private MainActivity host;
public AsyncTaskExample(MainActivity host)
{
this.host = host;
}
public String GetSEssion(String URL)
{
boolean isValid = true;
if(isValid)
{
String[][] val = {
new String[] {"Student.ID","123456"},
new String[] {"Student.username","user1"},
new String[] {"Student.password","123456"},
new String[] {"Student.location.id","12"}
};
HttpPost requestAuth = new HttpPost(URL +"/Login");
try
{
JSONObject json = new JSONObject();
// json.put("sessionId", sessionId);
JSONArray params = new JSONArray();
params.put(val);
json.put("authenParams", params);
StringEntity se = new StringEntity(json.toString());
se.setContentType(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));
requestAuth.setHeader("Accept","application/json");
requestAuth.setEntity(se);
DefaultHttpClient httpClientAuth = new DefaultHttpClient();
HttpResponse responseAuth = httpClientAuth.execute(requestAuth);
HttpEntity responseEntityAuth = responseAuth.getEntity();
char[] bufferAuth = new char[(int)responseEntityAuth.getContentLength()];
InputStream streamAuth = responseEntityAuth.getContent();
InputStreamReader readerAuth = new InputStreamReader(streamAuth);
readerAuth.read(bufferAuth);
streamAuth.close();
String rawAuthResult = new String(bufferAuth);
Result = rawAuthResult;
String d = null;
// }
} catch (ClientProtocolException e) {
Log.e("Client Protocol", e.getMessage());
} catch (IOException e) {
Log.e("Client Protocol", e.getMessage() );
} catch(Exception e)
{
Log.e("Client Protocol", 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();
}
}
}
下面是應該接收該參數的方法:我在下面的代碼中放置了一個斷點並進行了檢查。 該參數始終為空。
public string Login(string[][] value)
{
string[] tester = null;
string testerExample="";
foreach (string[] st in value)
{
tester = st;
}
foreach (string dt in tester)
{
testerExample = dt;
}
return testerExample;
}
這是IStudentService中的方法聲明:
[OperationContract]
[WebInvoke(
Method="POST", UriTemplate="Login", BodyStyle= WebMessageBodyStyle.WrappedRequest, ResponseFormat = WebMessageFormat.Json, RequestFormat = WebMessageFormat.Json)]
string Login(string[][] value);
我按照您的建議嘗試了,但沒有成功。 它返回“請求錯誤”。這是我粘貼的示例代碼。
HttpClient client = new DefaultHttpClient(); HttpPost post = new HttpPost("http://10.0.2.2:65031/SampleService.svc/login"); List<NameValuePair> pairs = new ArrayList<NameValuePair>(); pairs.add(new BasicNameValuePair("tester","abcd")); pairs.add(new BasicNameValuePair("sampletest","1234")); UrlEncodedFormEntity entity = new UrlEncodedFormEntity(pairs,HTTP.UTF_8); post.setEntity(entity); HttpResponse response = client.execute(post); HttpEntity responseEntity = response.getEntity(); char[] buffer = new char[(int)responseEntity.getContentLength()]; InputStream stream = responseEntity.getContent(); InputStreamReader reader = new InputStreamReader(stream); reader.read(buffer); stream.close(); String value = new String(buffer);
我終於讓它按我想要的方式工作。 問題是我以這種方式構建Array(請參見第1節),並將其傳遞給JSONObject或JSONArray。 我切換並使用JSONArray構建Array,並將其傳遞給JSONObject(請參閱第2節)。 它像一種魅力。
第1節:這樣做的方法錯誤-(如果您要遍歷數組並將其放入JSONArray中,則可能會以這種方式工作。如果可以直接完成,這將是太多的工作。)
String[][] Array = { new String[]{"Example", "Test"}, new String[]{"Example", "Test"}, }; JSONArray jar1 = new JSONArray(); jar1.put(0, Array); **// Did not work**
第2節:經過長時間的嘗試以及@vorrtex提供的一些非常有用的提示和提示,我的工作方式。
JSONArray jar1 = new JSONArray(); jar1.put(0, "ABC"); jar1.put(1, "Son"); jar1.put(2, "Niece"); JSONArray jarr = new JSONArray(); jarr.put(0, jar1); JSONArray j = new JSONArray(); j.put(0,"session"); JSONObject obj = new JSONObject(); obj.put("value", jarr); obj.put("test", j); obj.put("name","myName"); Log.d("Obj.ToString message: ",obj.toString()); StringEntity entity = new StringEntity(obj.toString());
查看Web服務,它正是我想要的。
謝謝您的幫助!!!
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