如何在java中對url字符串進行編碼，以便不在空格中加上+符號？

Question

我有一個JSONParser類，它可以讓我創建HTTPRequests。所以這裏是類

package com.thesis.menubook; 

import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStream; 
import java.io.InputStreamReader; 
import java.io.UnsupportedEncodingException; 
import java.util.List; 

import org.apache.http.HttpEntity; 
import org.apache.http.HttpResponse; 
import org.apache.http.NameValuePair; 
import org.apache.http.client.ClientProtocolException; 
import org.apache.http.client.entity.UrlEncodedFormEntity; 
import org.apache.http.client.methods.HttpGet; 
import org.apache.http.client.methods.HttpPost; 
import org.apache.http.client.utils.URLEncodedUtils; 
import org.apache.http.impl.client.DefaultHttpClient; 
import org.json.JSONException; 
import org.json.JSONObject; 

import android.util.Log; 

public class JSONParser { 

    static InputStream is = null; 
    static JSONObject jObj = null; 
    static String json = ""; 

    // constructor 
    public JSONParser() { 

    } 

    // function get json from url 
    // by making HTTP POST or GET mehtod 
    public JSONObject makeHttpRequest(String url, String method, 
      List<NameValuePair> params) { 

     // 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, "utf-8"); 
       url += "?" + paramString; 
       HttpGet httpGet = new HttpGet(url); 
       Log.d("URL",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(); 
     } 

     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()); 
      Log.e("JSON Parser", "json string :" +json); 
     } 

     // return JSON String 
     return jObj; 

    } 
} 

我這樣訪問它。

params.add(new BasicNameValuePair("category", "MAIN DISH")); 
      JSONObject json = jsonParser.makeHttpRequest("http://"+ipaddress+"/MenuBook/selectMenu.php", "GET", params); 

我想保留參數的格式，如MAIN DISH。但是當我看着我的LogCat時，它會返回一個像這樣形成的url。

http://192.168.10.149:80/MenuBook/selectMenu.php?category=MAIN+DISH

然後使我的應用程序失敗並強制關閉，因爲我沒有像類主+ DISH

我想我的URL將要形成這樣的。

http://192.168.10.149:80/MenuBook/selectMenu.php?category='MAIN DISH'

然後將返回正確的結果。我在網上搜索，只發現解決方案，使空白+和%20這將不會返回正確的結果。

您可以提出任何解決方案？

Answer 1

您的問題體現了矛盾的條款。 URL編碼（實際上是表單編碼）已經被定義，並且它已經被定義爲用'+'替換空格，而不是引用有關的值元素。服務器端軟件需要理解並相應地運行。所有由Java提供的服務器端軟件，例如HttpServletRequest已經做到了。如果您的代碼不符合RFC，請修復它。

Answer 2

我用'MAIN+DISH'