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所以我从这里开始练习这段代码，我想知道结果返回到哪里？

例如，我故意更改 url 以使其无效，但我没有看到没有工作的响应！任何地方。我查看了console,logcat和 没有 任何 指示帖子 是否 成功 的迹象.

我怎样才能找到这个？谢谢！

public class MainActivity extends Activity implements OnClickListener {

TextView tvIsConnected;
EditText etName,etCountry,etTwitter;
Button btnPost;

Person person;
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    // get reference to the views
    tvIsConnected = (TextView) findViewById(R.id.tvIsConnected);
    etName = (EditText) findViewById(R.id.etName);
    etCountry = (EditText) findViewById(R.id.etCountry);
    etTwitter = (EditText) findViewById(R.id.etTwitter);
    btnPost = (Button) findViewById(R.id.btnPost);

    // check if you are connected or not
    if(isConnected()){
        tvIsConnected.setBackgroundColor(0xFF00CC00);
        tvIsConnected.setText("You are conncted");
    }
    else{
        tvIsConnected.setText("You are NOT conncted");
    }

    // add click listener to Button "POST"
    btnPost.setOnClickListener(this);

}

public static String POST(String url, Person person){
    InputStream inputStream = null;
    String result = "";
    try {

        // 1. create HttpClient
        HttpClient httpclient = new DefaultHttpClient();

        // 2. make POST request to the given URL
        HttpPost httpPost = new HttpPost(url);

        String json = "";

        // 3. build jsonObject
        JSONObject jsonObject = new JSONObject();
        jsonObject.accumulate("name", person.getName());
        jsonObject.accumulate("country", person.getCountry());
        jsonObject.accumulate("twitter", person.getTwitter());

        // 4. convert JSONObject to JSON to String
        json = jsonObject.toString();

        // ** Alternative way to convert Person object to JSON string usin Jackson Lib 
        // ObjectMapper mapper = new ObjectMapper();
        // json = mapper.writeValueAsString(person); 

        // 5. set json to StringEntity
        StringEntity se = new StringEntity(json);

        // 6. set httpPost Entity
        httpPost.setEntity(se);

        // 7. Set some headers to inform server about the type of the content   
        httpPost.setHeader("Accept", "application/json");
        httpPost.setHeader("Content-type", "application/json");

        // 8. Execute POST request to the given URL
        HttpResponse httpResponse = httpclient.execute(httpPost);

        // 9. receive response as inputStream
        inputStream = httpResponse.getEntity().getContent();

        // 10. convert inputstream to string
        if(inputStream != null)
            result = convertInputStreamToString(inputStream);
        else
            result = "Did not work!";

    } catch (Exception e) {
        Log.d("InputStream", e.getLocalizedMessage());
    }

    // 11. return result
    return result;
}

public boolean isConnected(){
    ConnectivityManager connMgr = (ConnectivityManager) getSystemService(Activity.CONNECTIVITY_SERVICE);
        NetworkInfo networkInfo = connMgr.getActiveNetworkInfo();
        if (networkInfo != null && networkInfo.isConnected()) 
            return true;
        else
            return false;    
}
@Override
public void onClick(View view) {

    switch(view.getId()){
        case R.id.btnPost:
            if(!validate())
                Toast.makeText(getBaseContext(), "Enter some data!", Toast.LENGTH_LONG).show();
            // call AsynTask to perform network operation on separate thread
            new HttpAsyncTask().execute("http://hmkcode.appspot.com/jsonservlet");
        break;
    }

}
private class HttpAsyncTask extends AsyncTask<String, Void, String> {
    @Override
    protected String doInBackground(String... urls) {

        person = new Person();
        person.setName(etName.getText().toString());
        person.setCountry(etCountry.getText().toString());
        person.setTwitter(etTwitter.getText().toString());

        return POST(urls[0],person);
    }
    // onPostExecute displays the results of the AsyncTask.
    @Override
    protected void onPostExecute(String result) {
        Toast.makeText(getBaseContext(), "Data Sent!", Toast.LENGTH_LONG).show();
   }
}

private boolean validate(){
    if(etName.getText().toString().trim().equals(""))
        return false;
    else if(etCountry.getText().toString().trim().equals(""))
        return false;
    else if(etTwitter.getText().toString().trim().equals(""))
        return false;
    else
        return true;    
}
private static String convertInputStreamToString(InputStream inputStream) throws IOException{
    BufferedReader bufferedReader = new BufferedReader( new InputStreamReader(inputStream));
    String line = "";
    String result = "";
    while((line = bufferedReader.readLine()) != null)
        result += line;

    inputStream.close();
    return result;

}   
}
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1 回答 1

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在 POST() 的 catch 块中添加一行

result="Exception: " + e.getMessage();

然后在 onPostExecute

Toast (....,result,...);
于 2014-06-14T07:52:03.417 回答