從PHP變量不會顯示在Android上（使用JSON）

Question

我想使用模糊（尚未使整個模糊計算）的Android應用程序，所以計算將在PHP中完成。對於初學者，我只是首先在php中進行簡單的計算，然後嘗試發送給android。但它不顯示任何內容..也沒有錯誤。

這裏是我的PHP代碼calc.php

 <?php 

$calcresult = 56 * 100 * 2051/49; 

echo json_encode($calcresult); 
?> 

的簡單的例子，這是我的java代碼JSONActivity.class

package com.example.ta2; 
import java.io.BufferedReader; 
import java.io.InputStream;import java.io.InputStreamReader; 
import org.apache.http.HttpResponse; 
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.JSONObject; 

import android.app.Activity; 
import android.os.Bundle; 
import android.widget.TextView; 
import android.widget.Toast; 

public class AturanKonsumsi extends Activity { 
    private JSONObject jObject; 

     private String xResult =""; 
    private String url = "http://10.0.2.2/calc.php"; 
     @Override 
     public void onCreate(Bundle savedInstanceState) { 
     super.onCreate(savedInstanceState); 
     setContentView(R.layout.daftarmakanan); 
      TextView txtResult = (TextView)findViewById(R.id.TextViewResult); 
      xResult = getRequest(url); 
      try { 
       parse(txtResult); 
      } catch (Exception e) { 
       e.printStackTrace(); 
      } 

     } 
     private void parse(TextView txtResult) throws Exception { 
      jObject = new JSONObject(xResult); 

      JSONArray menuitemArray = jObject.getJSONArray("calcresult"); 
      String sret=""; 
     for (int i = 0; i < menuitemArray.length(); i++) { 
       System.out.println(menuitemArray.getJSONObject(i) 
         .getString("calcresult").toString()); 
       sret +=menuitemArray.getJSONObject(i).getString(
         "calcresult").toString()+"\n"; 
     } 
     txtResult.setText(sret); 
    } 

     public String getRequest(String Url){ 

      String sret=""; 
      HttpClient client = new DefaultHttpClient(); 
      HttpGet request = new HttpGet(Url); 
      try{ 
      HttpResponse response = client.execute(request); 
       sret =request(response); 

      }catch(Exception ex){ 
       Toast.makeText(this,"Gagal "+sret, Toast.LENGTH_SHORT).show(); 
     } 
      return sret; 

    } 

     public static String request(HttpResponse response){ 
        String result = ""; 
        try{ 
         InputStream in = response.getEntity().getContent(); 
         BufferedReader reader = new BufferedReader(new InputStreamReader(in)); 
        StringBuilder str = new StringBuilder(); 
        String line = null; 
        while((line = reader.readLine()) != null){ 
         str.append(line + "\n"); 
        } 
         in.close(); 
         result = str.toString(); 
        }catch(Exception ex){ 
         result = "Error"; 
        } 
        return result; 
       } 
       } 

當我運行Android應用程序不顯示$ calcresult中的值，也沒有錯誤。感謝您的幫助

Answer 1

calcresult沒有引用您的PHP腳本正在生產的json。檢查輸出。爲了使你的Java代碼工作，你需要像這樣創建JSON：

<?php 
    $calcresult = 56 * 100 * 2051/49; 
    $json = array('calcresult' => array($calcresult)); 
    echo json_encode($json); 
?> 

或者，你可以簡化你的java。