Get方法用頭參數不能正常工作的android

Question

我有以下的JSON：

{ 
    "category": [{ 
    "id": "90", 
    "user_id": "1", 
    "category_id": "27", 
    "name": "આણંદ કોમર્સિયલ લેયર", 
    "order": "0", 
    "created_at": "2014-05-03 17:09:54", 
    "updated_at": "2014-05-03 17:09:54", 
    "deleted": "0", 
    "subtopics": [{ 
     "id": "203", 
     "user_id": "1", 
     "category_id": "27", 
     "subcategory_id": "90", 
     "name": "આણંદ કોમર્સિયલ લેયર (સંકર જાત)", 
     "order": "0", 
     "details": "<p style=\"text-align:justify\"><img alt=\"\" src=\"/packages/wysiwyg/ckeditor/plugins/kcfinder/upload/images/1.png\" style=\"height:271px; width:237px\" /></p>\r\n\r\n<ul>\r\n\t<li style=\"text-align:justify\">પ્રથમ ઈંડું મુક્વાની સરેરાશ ઉંમર:૧૪૨ દિવસ</li>\r\n\t<li style=\"text-align:justify\">સરેરાશ વાર્ષિક ઈંડા ઉત્પાદન : ૩૦૦ ઈંડા</li>\r\n\t<li style=\"text-align:justify\">૪૦ અઠવાડીયાની ઉંમરે ઈંડાનું સરેરાશ વજન : ૫૨ ગ્રામ</li>\r\n\t<li style=\"text-align:justify\">૭૨ અઠવાડીયાની ઉંમરે ઈંડાનું સરેરાશ વજન : ૫૪ ગ્રામ</li>\r\n\t<li style=\"text-align:justify\">સારી જીવાદોરી</li>\r\n</ul>\r\n", 
     "mobile_detail": "<p style=\"text-align:justify\"><img alt=\"\" src=\"/packages/wysiwyg/ckeditor/plugins/kcfinder/upload/images/1.png\" style=\"height:271px; width:237px\" /></p>\r\n\r\n<ul>\r\n\t<li style=\"text-align:justify\">પ્રથમ ઈંડું મુક્વાની સરેરાશ ઉંમર:૧૪૨ દિવસ</li>\r\n\t<li style=\"text-align:justify\">સરેરાશ વાર્ષિક ઈંડા ઉત્પાદન : ૩૦૦ ઈંડા</li>\r\n\t<li style=\"text-align:justify\">૪૦ અઠવાડીયાની ઉંમરે ઈંડાનું સરેરાશ વજન : ૫૨ ગ્રામ</li>\r\n\t<li style=\"text-align:justify\">૭૨ અઠવાડીયાની ઉંમરે ઈંડાનું સરેરાશ વજન : ૫૪ ગ્રામ</li>\r\n\t<li style=\"text-align:justify\">સારી જીવાદોરી</li>\r\n</ul>\r\n", 
     "created_at": "2014-05-03 17:11:43", 
     "updated_at": "2014-05-11 13:41:31", 
     "deleted": "0", 
     "images": [], 
     "videos": [] 
    }] 
    }] 
} 

我有以下代碼：

public JSONObject getJSONFromUrl(String url) { 

    // Making HTTP request 
    InputStream is=null; 
    try { 
     // defaultHttpClient 
     DefaultHttpClient httpClient = new DefaultHttpClient(); 
     //HttpPost httpPost = new HttpPost(url); 
     HttpGet httpPost = new HttpGet(url); 
     httpPost.setHeader("Content-Type", "application/json"); 
     httpPost.setHeader("X-Requested-With", "XMLHttpRequest"); 
     httpPost.setHeader("Mobile-Tokon", "7c^4N:9Y*Tq;P^f"); 
     HttpResponse httpResponse = httpClient.execute(httpPost); 
     HttpEntity httpEntity = httpResponse.getEntity(); 
     is = httpEntity.getContent(); 

    } catch (UnsupportedEncodingException e) { 
     e.printStackTrace(); 
    } catch (ClientProtocolException e) { 
     e.printStackTrace(); 
    } catch (IOException e) { 
     e.printStackTrace(); 
    } 

    String json=""; 
    try { 
     BufferedReader reader = new BufferedReader(new InputStreamReader(is, "UTF-8"), 8000); 
     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 
    JSONObject jObj=null; 
    try { 
     jObj = new JSONObject(json); 
     Log.d("JSON",jObj.toString()); 
    } catch (JSONException e) { 
     Log.e("JSON Parser", "Error parsing data " + e.toString()); 
    } 

    // return JSON String 
    return jObj; 

} 

在我的Java類，我調用上面的函數是這樣的：

getJSONFromUrl("http://ikishan.192.168.1.87.xip.io/api/newapps/27?email=api.ikisan@aau.in&password=%~?7ON9Xjp;BcYu"); 

我得到一個錯誤：

java.lang.IllegalArgumentException: Invalid % sequence: %~? in query at index 83: http://ikishan.192.168.1.87.xip.io/api/newapps/27?email=api.ikisan@aau.in&password=%~?7ON9Xjp;BcYu

在這一行：

HttpGet httpPost = new HttpGet(url); 

任何想法如何，我可以解決這個問題？

編輯

Answer 1

嘗試像下面。如果你加密密碼，那麼你可以像下面這樣做。

// Encrypt Password to UTF-8 format 

private String encrypt(String pwd) { 
    String encryptedPwd = null; 
    try { 
     byte[] byteArray = pwd.getBytes("UTF-8"); 
     encryptedPwd = Base64.encodeToString(byteArray, Base64.DEFAULT).trim(); 
    } catch (UnsupportedEncodingException e) { 
     e.printStackTrace(); 
    } 
    return encryptedPwd; 
} 

你可以用下面的類

public class EncodingUtil{ 


/** 
* Decodes the passed UTF-8 String using an algorithm that's compatible with 
* JavaScript's <code>decodeURIComponent</code> function. Returns 
* <code>null</code> if the String is <code>null</code>. 
* 
* @param s The UTF-8 encoded String to be decoded 
* @return the decoded String 
*/ 
public static String decodeURIComponent(String s) 
{ 
    if (s == null) 
    { 
     return null; 
    } 

    String result = null; 

    try 
    { 
     result = URLDecoder.decode(s, "UTF-8"); 
    } 

    // This exception should never occur. 
    catch (UnsupportedEncodingException e) 
    { 
     result = s; 
    } 

    return result; 
} 

/** 
* Encodes the passed String as UTF-8 using an algorithm that's compatible 
* with JavaScript's <code>encodeURIComponent</code> function. Returns 
* <code>null</code> if the String is <code>null</code>. 
* 
* @param s The String to be encoded 
* @return the encoded String 
*/ 
public static String encodeURIComponent(String s) 
{ 
    String result = null; 

    try 
    { 
     result = URLEncoder.encode(s, "UTF-8") 
       .replaceAll("\\+", "%20") 
       .replaceAll("\\%21", "!") 
       .replaceAll("\\%27", "'") 
       .replaceAll("\\%28", "(") 
       .replaceAll("\\%29", ")") 
       .replaceAll("\\%7E", "~"); 
    } 

    // This exception should never occur. 
    catch (UnsupportedEncodingException e) 
    { 
     result = s; 
    } 

    return result; 
} 

/** 
* Private constructor to prevent this class from being instantiated. 
*/ 
private EncodingUtil() 
{ 
    super(); 
}} 

Answer 2

那是因爲你已經提到的網址並不像this指定SO回答一個有效的URL編碼URL。

a complete URL is always in its encoded form: you take the strings for the individual components (scheme, authority, path, etc.), encode each according to its own rules, and then combine them into the complete URL string. Trying to build a complete unencoded URL string and then encode it separately leads to subtle bugs, like spaces in the path being incorrectly changed to plus signs (which an RFC-compliant server will interpret as real plus signs, not encoded spaces).

了在IllegalArgumentException例外提到不應與GET請求直接發送，需要這些無效字符轉換成這種格式

Character From Windows-1252 From UTF-8 
    %    %25    %25 
    ~    %7E    %7E 
    ?    %3F    %3F 

看到this網址的完整列表

---

在Java中，構建URL的正確方法是使用URI類。

import java.net.URLEncoder; 

public class HelloWorld { 
    static String uri = "http://ikishan.192.168.1.87.xip.io/api/newapps/27"; 
    static String params = "?email=api.ikisan@aau.in&password=%~?7ON9Xjp;BcYu"; 

    public static void main(String []args){ 
     try { 
      String encodedParams = URLEncoder.encode(params, "ASCII").toString(); 
      System.out.println(uri + encodedParams); 
     } 
     catch(java.io.UnsupportedEncodingException uee) { 
      //do something 
     } 
    } 
} 

輸出

http://ikishan.192.168.1.87.xip.io/api/newapps/27%3Femail%3Dapi.ikisan%40aau.in%26password%3D%25%7E%3F7ON9Xjp%3BBcYu