通过GET方法发送值

Question

我一直试图通过GET方法发送数据到服务器，但我无法找到一种方法来做到这一点。我在异步任务中尝试过几个代码，但没有任何代码Web服务在CakePHP执行和格式是这样的：请

Base_URI/users/add.json?json={“email”: [email protected], “password”: “xxxxxxxxx”, “first_name”: “Xyz”, “last_name”: “Xyz”} 

的Android专家推测出路这个问题。由于

下面是代码：

List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(); 
      nameValuePairs.add(new BasicNameValuePair("email", UserDummy.email)); 
      nameValuePairs.add(new BasicNameValuePair("password", UserDummy.password)); 
      nameValuePairs.add(new BasicNameValuePair("first_name", UserDummy.fname)); 
      nameValuePairs.add(new BasicNameValuePair("last_name", UserDummy.lname)); 
      // Making HTTP request 
    try { 

     // check for request method 
     if (method == "POST") { 
      // request method is POST 
      // defaultHttpClient 
      DefaultHttpClient httpClient = new DefaultHttpClient(); 
      HttpPost httpPost = new HttpPost(url); 
      httpPost.setEntity(new UrlEncodedFormEntity(params)); 

      HttpResponse httpResponse = httpClient.execute(httpPost); 
      HttpEntity httpEntity = httpResponse.getEntity(); 
      is = httpEntity.getContent(); 

     } else if (method == "GET") { 
      // request method is GET 
      DefaultHttpClient httpClient = new DefaultHttpClient(); 
      String paramString = URLEncodedUtils.format(params, HTTP.UTF_8); 
      url += "?json={" + paramString+"}";                                      ; 
      HttpGet httpGet = new HttpGet(url); 

      HttpResponse httpResponse = httpClient.execute(httpGet); 
      HttpEntity httpEntity = httpResponse.getEntity(); 
      is = httpEntity.getContent(); 
     } 

    } catch (UnsupportedEncodingException e) { 
     e.printStackTrace(); 
     Log.v("XXX", e.getMessage()); 
    } catch (ClientProtocolException e) { 
     e.printStackTrace(); 
     Log.v("XXX", e.getMessage()); 
    } catch (IOException e) { 
     e.printStackTrace(); 
     Log.v("XXX", e.getMessage()); 
    } 

    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(); 
    } 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; 

HTTPGET不接受URL的格式，并给出错误，但是当我尝试在浏览器中正常工作。错误如下：

Illegal character in query at index 56 

Answer 1

调试后，最后与一整天时间不同的解决方案，我的问题解决了:)

我需要的部分参数编码而不是像这样的整个URL：

String url = "Base_URI/users/add.json?json="; 
    url =url + URLEncoder.encode("{\"email\":\""+email+"\",\"password\":\""+password+"\"}", "UTF-8"); 

感谢大家的支持！

Answer 2

在android中使用volley lib进行网络连接。

Answer 3

您的字符串比较不正确。

与此相比对象引用，并不会为你工作：

method == "POST" 

这一翻译，使用equals（）方法：

"POST".equals(method); 

你也可以这样做：

method.equals("POST"); 

但如果方法为空，可能会导致错误。

Answer 4

如果你想使用GET方法：

这种方法可以帮助你

/** 
* 
* @param url stands for API 
* @param 
*  params[i][0] stands for column's name 
*  params[i][1] stands for value which respective with column's name 
* @return InputStream which got from Server 
*/ 
public static int sendJSONObject(IGetUserData iUserId, String method, String url, String[]... params) { 
    InputStream mInputStream = null; 
    HttpClient mHttpClient = null; 
    HttpRequestWithEntity mHttpGet = null; 
    int status = Def.REQUEST_INVALID; 
    try { 
     mHttpClient = new DefaultHttpClient(); 

     mHttpGet = new HttpRequestWithEntity(url, method); 

     JSONObject mObject = new JSONObject(); 
     for (String[] pair : params) { 
      mObject.put(pair[0], pair[1]); 
     } 

     StringEntity mStringEntity = new StringEntity(mObject.toString()); 
     mStringEntity.setContentEncoding("UTF-8"); 
     mStringEntity.setContentType("application/json"); 

     mHttpGet.setEntity(mStringEntity);   
     HttpResponse mResponse = mHttpClient.execute(mHttpGet); 
     status = mResponse.getStatusLine().getStatusCode(); 

     Log.d(TAG, "status: " + status); 
     if (mResponse != null && 
       (status == Def.CREATED || status == Def.OK)) { 
      mInputStream = mResponse.getEntity().getContent(); 
      if(mInputStream != null){ 
       String json = StreamUtils.converStreamToString(mInputStream); 
       userId = JSONUtils.getUserId(json); 
       iUserId.sendUserId(userId); 
       Log.d("viet","userid = " + userId); 
      } 
     } 

    } catch (Exception e) { 
     Log.e(TAG, "Error during send"); 
     status = Def.NETWORK_ERROR; 
    } 
    return status; 
} 

---

就在这里为你所需要的。

mHttpClient = new DefaultHttpClient(); 

    mHttpGet = new HttpRequestWithEntity(url, method); 

    JSONObject mObject = new JSONObject(); 
    for (String[] pair : params) { 
     mObject.put(pair[0], pair[1]); 
    } 

    StringEntity mStringEntity = new StringEntity(mObject.toString()); 
    mStringEntity.setContentEncoding("UTF-8"); 
    mStringEntity.setContentType("application/json"); 

    mHttpGet.setEntity(mStringEntity);   
    HttpResponse mResponse = mHttpClient.execute(mHttpGet); 

这里是HttpRequestWithEntity.java

import java.net.URI; 
import java.net.URISyntaxException; 

import org.apache.http.client.methods.HttpEntityEnclosingRequestBase; 

public class HttpRequestWithEntity extends HttpEntityEnclosingRequestBase { 

    private String method; 

    public HttpRequestWithEntity(String url, String method) { 
     if (method == null || (method != null && method.isEmpty())) { 
      this.method = HttpMethod.GET; 
     } else { 
      this.method = method; 
     } 
     try { 
      setURI(new URI(url)); 
     } catch (URISyntaxException e) { 
      e.printStackTrace(); 
     } 
    } 

    @Override 
    public String getMethod() { 
     return this.method; 
    } 

}