如何去在我的Andr​​oid Aplication code JSON值？ [英] How to decode JSON values in my Android Aplication?

问题描述

好家伙我试图去codeA JSON值进入我的Andr​​oid应用程序，但我真的不知道我在做什么错在我的code ...

这是我的PHP文件：

 ＆LT; PHP

    ...
    回声json_en code（$ mostrar_player）;
    ...

？＆GT;
 

这将返回我是这样的： [1，管理员，123，海军上将，邮件，0]

在我的Andr​​oid应用程序，我需要把该值转换成数组那里我可以选择separeted价值，但如何？

这是我的Andr​​oid code code：

 的EditText campoLogin =（EditText上）findViewById（R.id.campo_de_login）;
        EditText上campoSenha =（EditText上）findViewById（R.id.campo_de_senha）;

        吐司（Conectando ......）;
        Conectar康恩=新Conectar（？的http：//website/chat/login.php登录=+ campoLogin.getText（）+＆安培; senha =+ campoSenha.getText（））;
        Log.i（ -  RESPOSTA  - ，Conn.response）;
        //这里Conn.response我有我的反应[1，管理员，管理员，管理员，AFFS，0]

        字符串json_str = Conn.response;
        JSONObject的JSON =新的JSONObject（json_str）;
        JSONArray jArray = json.getJSONArray（味精）;

        //Log.i("JSON",json_data.getString("msg））;
 

解决方案

尝试

 ＆LT;？PHP的回声json_en code（$ mostra_player，JSON_FORCE_OBJECT）; ？＆GT;
 

这迫使是一个对象。

然后你可以访问它在Java / Android版

  JSONObject的OBJ =（JSONObject的）新JSONTokener（json_string_output）.nextValue（）;
串yourdata = obj.getString（yourdata）;
 

如果你想要一个阵列的尝试

 ＆LT;？PHP的回声json_en code（阵列（数据=＆GT; $ mostra_player））; ？＆GT;

        JSONObject的JSON =新的JSONObject（responsedata）;
        JSONArray jArray = json.getJSONArray（上岗）;
        的for（int i = 0; I＆LT; jArray.length（）;我++）{
            JSONObject的E = jArray.getJSONObject（我）;
            JSONObject的jObject =新的JSONObject（e.getString（数据））;

            Log.d（yourdata，jObject.getString（yourdata））;
            Log.d（MOREDATA，jObject.getString（MOREDATA））;
        }
 

Hello guys I'm trying to decode a JSON value into my Android Application, But I'm not really sure What Am I doing wrong in my code...

This is my PHP file:

<?php 

    ... 
    echo json_encode($mostrar_player);
    ...

?>

Which Will return me something like this: ["1","Admin","123","Adm","messages","0"]

in my Android Application I need to put that value into an Array there I can select values separeted, but how?

this is my Android Code code:

        EditText campoLogin = (EditText)findViewById(R.id.campo_de_login);
        EditText campoSenha = (EditText)findViewById(R.id.campo_de_senha);

        toast("Conectando...");
        Conectar Conn = new Conectar("http://website/chat/login.php?login=" + campoLogin.getText() + "&senha=" + campoSenha.getText() );
        Log.i("-RESPOSTA-",Conn.response); 
        // HERE IN Conn.response I have my response ["1","Admin","Admin","Admin","affs","0"]

        String json_str = Conn.response;
        JSONObject json = new JSONObject(json_str);
        JSONArray jArray = json.getJSONArray("msg");

        //Log.i("JSON",json_data.getString("msg"));

解决方案

try

 <?php   echo json_encode($mostra_player, JSON_FORCE_OBJECT); ?> 

which forces to be an object.

Then you can access it in java/android with

JSONObject obj = (JSONObject) new JSONTokener(json_string_output).nextValue();
String yourdata= obj.getString("yourdata");

If you want an Array try

<?php echo json_encode(array("data" => $mostra_player)); ?>

        JSONObject json = new JSONObject(responsedata);
        JSONArray jArray = json.getJSONArray("posts");
        for (int i = 0; i < jArray.length(); i++) {
            JSONObject e = jArray.getJSONObject(i);
            JSONObject jObject = new JSONObject(e.getString("data"));

            Log.d("yourdata", jObject.getString("yourdata"));
            Log.d("moredata", jObject.getString("moredata"));
        }

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