Android 从php文件到java数组的JSON

我知道关于这个主题有一些问题，但我读了它们，并尝试了解决方案，但没有成功：

PHP脚本给出这个json数组结果：data[x]=

         ["alon","62","1.82","22","0","70","0","1"]

这是数据[x]变量

我必须将这个结果转换成Java变量，如姓名、体重、身高等。。但是我不知道怎么

请帮帮我

我的职能：

        private class LongOperation  extends AsyncTask<String, Void, Void> {





        private final HttpClient Client = new DefaultHttpClient();
        private String Error = null;

        protected void onPreExecute() {


        }

        protected Void doInBackground(String... urls) {
            try {

                  HttpGet httpget = new HttpGet(urls[0]);
                ResponseHandler<String> responseHandler = new BasicResponseHandler();
                data[x] = Client.execute(httpget, responseHandler);

            } catch (ClientProtocolException e) {
                Error = e.getMessage();
                Toast.makeText(getApplicationContext(),"error2" , Toast.LENGTH_LONG).show();
                cancel(true);
            } catch (IOException e) {
                Error = e.getMessage();
                Toast.makeText(getApplicationContext(),"error34" , Toast.LENGTH_LONG).show();
                cancel(true);
            }

            return null;
        }

        public void onPostExecute(Void unused) {

            String name = null,weight = null;
            if (Error != null) {  
            } else {

                // here I have to do something with the arrays... 

            Toast.makeText(getApplicationContext(),"d:" + data[x] + "o:" + name + " : " + weight, Toast.LENGTH_LONG).show(); 
             }
            x++;
        }

    }

---

为此创建一个模态类

class myModal {
   private String name, weight, height, ...;
   public String getName() { return this.name; } 
   public void setName(String name) { this.name = name; }
   //and more getters and setters
}


JSONObject json = new JSONObject(data[x]); // in your sample its a JSONArray but its wrong formatted. make sure you encode it properply with php json_encode(array("data", yourdata))...);

myModal modal = new myModal();
modal.setName(json.getString("name"));

php应该类似于

<?php 
  $data = array("name" => "myname", "weight" => 20);
  print json_encode(  $data  ); 
?>

而json在本例中可以用

 JSONArray json = new JSONArray(data);
 for (int i = 0; i <= json.length();i++){
   JSONObject jsonObj = json.getJsonObject(i);
   myModal modal = new myModal();
   modal.setString(jsonObj.getString("name"));
  //and so on          
 }

---

确保阅读理解的基础知识

我试过你说的。仍然不起作用=[php显示：[数据，{name:alon，weight:62，height:1.82，age:22，weeks:0，target:70，ketzev:0，gender:1}].eclipse希望在代码中添加try和catch:JSONObject json=null；try{json=new JSONObjectdata[x]；}catch JSONException e{//TODO自动生成的catch块e.printStackTrace；}myModal modal=new myModal；尝试{modal.setNamejson.getStringname；}catch JSONException e{//TODO自动生成的catch块e.printStackTrace；}php文件：$data=arrayname=>$name，weight=>$weight，height=>$height，age=>$age，weeks=>$weeks，target=>$target，ketzev=>$ketzev，gender=>$gender；$length=count$array；echo json_encodearraydata，$data；