類中的方法不能應用於給定的類型;必需的：HashMap中<字符串，整數>

Question

我在做這個程序，我得到這個錯誤：

錯誤：（93，34）誤差：類MainActivity.DownloadTask方法getPostDataString不能被應用到給定的類型; 要求：發現的HashMap ：字符串，整數 原因：實際的和正式的參數列表長度

protected String doInBackground(Object... params) { 
      URL url; 
      String response = ""; 
      try { 
       url = new URL("http://app.iseemobile.com/imenu/getDistrictRestaurants.php"); 
       HttpURLConnection conn = (HttpURLConnection) url.openConnection(); 
       conn.setReadTimeout(15000); 
       conn.setConnectTimeout(15000); 
       conn.setRequestMethod("POST"); 
       conn.setDoInput(true); 
       conn.setDoOutput(true); 
       OutputStream os = conn.getOutputStream(); 
       BufferedWriter writer = new BufferedWriter(
         new OutputStreamWriter(os, "UTF-8")); 
       writer.write(getPostDataString("district", 1)); 
       writer.flush(); 
       writer.close(); 
       os.close(); 
       int responseCode = conn.getResponseCode(); 
       if (responseCode == HttpsURLConnection.HTTP_OK) { 
        String line; 
        BufferedReader br = new BufferedReader(new InputStreamReader(conn.getInputStream())); 
        while ((line = br.readLine()) != null) { 
         response += line; 
        } 
       } else { 
        response = ""; 
       } 
      } catch (Exception e) { 
       e.printStackTrace(); 
      } 
      return response; 
    } 

不同用下面的方法：

private String getPostDataString(HashMap<String, Integer> params) throws UnsupportedEncodingException{ 
     StringBuilder result = new StringBuilder(); 
     boolean first = true; 
     for(Map.Entry<String, Integer> entry:params.entrySet()){ 
      if (first) 
       first = false; 
      else 
       result.append("&"); 

      result.append(URLEncoder.encode(entry.getKey(), "UTF-8")); 
      result.append("="); 
      result.append(URLEncoder.encode(String.valueOf(entry.getValue()), "UTF-8")); 
     } 
     return result.toString(); 
    } 

Answer 1

你應該通過地圖作爲getPostDataString方法

protected String doInBackground(Object... params) { 
     URL url; 
     String response = ""; 
     try { 
      url = new URL("http://app.iseemobile.com/imenu/getDistrictRestaurants.php"); 
      HttpURLConnection conn = (HttpURLConnection) url.openConnection(); 
      conn.setReadTimeout(15000); 
      conn.setConnectTimeout(15000); 
      conn.setRequestMethod("POST"); 
      conn.setDoInput(true); 
      conn.setDoOutput(true); 
      OutputStream os = conn.getOutputStream(); 
      BufferedWriter writer = new BufferedWriter(
        new OutputStreamWriter(os, "UTF-8")); 
      Map<String, Integer> inputMap = new HashMap<String, Integer>(); 
      map.put("district", 1); 
      writer.write(getPostDataString(map); 
      writer.flush(); 
      writer.close(); 
      os.close(); 
      int responseCode = conn.getResponseCode(); 
      if (responseCode == HttpsURLConnection.HTTP_OK) { 
       String line; 
       BufferedReader br = new BufferedReader(new InputStreamReader(conn.getInputStream())); 
       while ((line = br.readLine()) != null) { 
        response += line; 
       } 
      } else { 
       response = ""; 
      } 
     } catch (Exception e) { 
      e.printStackTrace(); 
     } 
     return response; 
} 

Answer 2

你傳遞一個字符串和一個整數，同時呼籲getPostDataString其中，因爲它期望HashMap<String, Integer>。 Java不會直接將字符串和整數轉換爲地圖。您需要在呼叫功能中創建地圖併發送。

替換下面的語句

writer.write(getPostDataString("district", 1)); 

與

Map<String, Integer> m = new HashMap<String, Integer>(); 
m.put("district",1); 
writer.write(getPostDataString(m));