Question

我需要调用“HTTPS”api并将JSON params传递给服务器现在我正在尝试这样：

HttpClient httpclient = new DefaultHttpClient();

            HttpPost httppost = new HttpPost("https://myapi call goes here...");
            httppost.setHeader("Content-Type", "application/json");

            try
            {

                JSONObject j = new JSONObject();
                j.put("email", "my email value goes here");

                StringEntity s = new StringEntity(j.toString());

                httppost.setEntity(s);
                HttpResponse response = httpclient.execute(httppost);

                /* Checking response */
                if(response != null)
                {
                    responseBody = EntityUtils.toString(response.getEntity());


                }
                if(responseBody.equals("ok"))
                {



                }
            }
            catch (final Exception e)
            {
                e.printStackTrace();


            }

但它给了我例外： SSL Peer未经验证：没有同行证书。
请注意，我只能使用 HTTPS 来拨打api。

任何人都有任何想法？

Answer 1

看看这个answer。它已得到很好的解释，可能是您纠正代码所需要的。

Answer 2

使用以下代码，还需要将json放在url中。检查代码。

public void login(Bundle bndl) {

    try {
        // /////////////////////////
        final int TIMEOUT_MILLISEC = 8000;
        JSONObject json = new JSONObject();
        json.put("email", bndl.getString("email"));
        json.put("psw", bndl.getString("psw"));
        HttpParams httpParams = new BasicHttpParams();
        HttpConnectionParams.setConnectionTimeout(httpParams,
                TIMEOUT_MILLISEC);
        HttpConnectionParams.setSoTimeout(httpParams, TIMEOUT_MILLISEC);
        HttpClient client = new DefaultHttpClient(httpParams);

        String url = "http://www.abc.com/login.php/?email="
                + bndl.getString("email") + "&psw=" + bndl.getString("psw");

        HttpPost request = new HttpPost(url);
        request.setEntity(new ByteArrayEntity(json.toString().getBytes(
                "UTF8")));
        request.setHeader("json", json.toString());
        HttpResponse response = client.execute(request);
        HttpEntity entity = response.getEntity();
        // If the response does not enclose an entity, there is no need
        if (entity != null) {
            InputStream instream = entity.getContent();

            String result = RestClient.convertStreamToString(instream);

        }
    } catch (Exception ex) {

    }
}

以下是支持类，只需复制和pasete。它的名字是 RestClient

import java.io.BufferedReader;
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.HttpClient;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.DefaultHttpClient;
import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

import android.util.Log;
public class RestClient {

    public static String convertStreamToString(InputStream is) {
    /*
     * To convert the InputStream to String we use the BufferedReader.readLine()
     * method. We iterate until the BufferedReader return null which means
     * there's no more data to read. Each line will appended to a StringBuilder
     * and returned as String.
     */
    BufferedReader reader = new BufferedReader(new InputStreamReader(is));
    StringBuilder sb = new StringBuilder();

    String line = null;
    try {
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");
        }
    } catch (IOException e) {
        e.printStackTrace();
    } finally {
        try {
            is.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    return sb.toString();
}

/* This is a test function which will connects to a given
 * rest service and prints it's response to Android Log with
 * labels "Praeda".
 */
public static void connect(String url)
{

    HttpClient httpclient = new DefaultHttpClient();

    // Prepare a request object
    HttpGet httpget = new HttpGet(url); 

    // Execute the request
    HttpResponse response;
    try {
        response = httpclient.execute(httpget);
        // Examine the response status
        Log.i("Praeda",response.getStatusLine().toString());

        // Get hold of the response entity
        HttpEntity entity = response.getEntity();
        // If the response does not enclose an entity, there is no need
        // to worry about connection release

        if (entity != null) {

            // A Simple JSON Response Read
            InputStream instream = entity.getContent();
            String result= convertStreamToString(instream);
            Log.i("Praeda",result);

            // A Simple JSONObject Creation
            JSONObject json=new JSONObject(result);
            Log.i("Praeda","<jsonobject>\n"+json.toString()+"\n</jsonobject>");

            // A Simple JSONObject Parsing
            JSONArray nameArray=json.names();
            JSONArray valArray=json.toJSONArray(nameArray);
            for(int i=0;i<valArray.length();i++)
            {
                Log.i("Praeda","<jsonname"+i+">\n"+nameArray.getString(i)+"\n</jsonname"+i+">\n"
                        +"<jsonvalue"+i+">\n"+valArray.getString(i)+"\n</jsonvalue"+i+">");
            }

            // A Simple JSONObject Value Pushing
            json.put("sample key", "sample value");
            Log.i("Praeda","<jsonobject>\n"+json.toString()+"\n</jsonobject>");

            // Closing the input stream will trigger connection release
            instream.close();
        }


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

使用HTTPS请求发送JSON参数

2 个答案: