JSON文件不工作

JSON文件不工作

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JSON文件不工作/不读jsonarray不能转换到的JSONObject

本文介绍了JSON文件不工作/不读jsonarray不能转换到的JSONObject的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

你好，我要访问我的Andr​​oid应用我的网页记录。通过这样做，我用JSON做到这一点，PHP。这是我的JSON文件的URL：

但问题是，它只是显示了PHP的code。我需要能够访问/读取该文件。 :(任何想法我这里我做什么？帮助是我pciated多少AP $ P $。感谢。

这是我到目前为止已经试过：

 ＆LT; PHP    包括（'connectdb.php'）;
    $的SQL =SELECT salesordercard_ code，location_from，location_to，业务员code从salesorderingcard
    $结果= mysql_query（$的SQL）;
    如果（$结果=== FALSE）{
     死亡（mysql_error（））; // TODO：更好的错误处理
    }
    $集=阵列（）;
    而（$ ROW1 = mysql_fetch_assoc（$结果））{
        $设置[] = $ ROW1;
    }    标题（内容类型：应用程序/ JSON'）;
    回声json_en code（$套）;
    ？＆GT;

MainActivity.class 

  @覆盖
保护无效doInBackground（虚空...... PARAMS）{
        //创建数组
        ArrayList的=新的ArrayList＆LT;＆HashMap的LT;字符串，字符串＆GT;＆GT;（）;
        // Retrive JSON从JSONfunctions.class指定网站的网址对象
        的JSONObject = JSONFunctions.getJSONfromURL（http://www.shoppersgroup.net/vanmanagement/results.php）;        尝试{
            //找到数组名称
            jsonarray = jsonobject.getJSONArray（上岗）;            的for（int i = 0; I＆LT; jsonarray.length（）;我++）{
                HashMap的＆LT;字符串，字符串＆GT;地图=新的HashMap＆LT;字符串，字符串＆GT;（）;
                的JSONObject = jsonarray.getJSONObject（ⅰ）;
                //Log.i（MainActivity.class.getName（），jsonobject.getString（MOVIE_NAME））;
                // Retrive JSON对象
                map.put（TAG_ code，jsonobject.getString（salesordercard_ code））;
                map.put（TAG_LOCATION_FROM，jsonobject.getString（location_from））;
                map.put（TAG_LOCATION_TO，jsonobject.getString（location_to））;
                //设置JSON对象到数组
                arraylist.add（地图）;
            }
        }赶上（JSONException E）{
            Log.e（错误，e.getMessage（））;
            e.printStackTrace（）;
        }
        返回null;
}

logcat的： 

 二月11日至18日：35：08.521：E / log_tag（1047）：错误分析数据org.json.JSONException：值[{\"salesman$c$c\":\"SLMAN001\",\"location_to\":\"MAIN\",\"location_from\":\"IN-TRANSIT\",\"salesordercard_$c$c\":\"SLESO0001\"},{\"salesman$c$c\":\"SLMAN001\",\"location_to\":\"MAIN\",\"location_from\":\"IN-TRANSIT\",\"salesordercard_$c$c\":\"SLESO0002\"}]型的org.json.JSONArray不能转换到的JSONObject


解决方案

首先作为JSON字符串

  [//它的一个阵列
    上岗，//它不是一个数组其指数为0，因此我们将不会使用索引0
    {//索引1
        salesordercard_ code：SLESO0001
        location_from：在途
        location_to：主
        推销员code：SLMAN001
    }，
    {//索引2
        salesordercard_ code：SLESO0002
        location_from：在途
        location_to：主
        推销员code：SLMAN001
    }
]

更改doInBackground（）方法像下面，

  @覆盖
保护无效doInBackground（虚空...... PARAMS）{
        //创建数组
        ArrayList的=新的ArrayList＆LT;＆HashMap的LT;字符串，字符串＆GT;＆GT;（）;
        // Retrive JSON从JSONfunctions.class指定网站的网址对象
        的JSONObject = JSONFunctions.getJSONfromURL（http://www.shoppersgroup.net/vanmanagement/results.php）;        尝试{
            //找到数组名称
            jsonarray =新JSONArray（jsonobject.toString（））;            的for（int i = 1; I＆LT; jsonarray.length（）;我++）{//从指数1 STRAT
                HashMap的＆LT;字符串，字符串＆GT;地图=新的HashMap＆LT;字符串，字符串＆GT;（）;
                JSONObject的jsonobj = jsonarray.getJSONObject（I）
                //Log.i（MainActivity.class.getName（），jsonobj.getString（MOVIE_NAME））;
                // Retrive JSON对象
                map.put（TAG_ code，jsonobj.getString（salesordercard_ code））;
                map.put（TAG_LOCATION_FROM，jsonobj.getString（location_from））;
                map.put（TAG_LOCATION_TO，jsonobj.getString（location_to））;
                //设置JSON对象到数组
                arraylist.add（地图）;
            }
        }赶上（JSONException E）{
            Log.e（错误，e.getMessage（））;
            e.printStackTrace（）;
        }
        返回null;
}

希望这将解决您的问题。

Hello I want to access my web records in my android app. By doing so, I use JSON to do that and PHP. This is the url of my json file: here

But the problem is it's just displaying the php code. I need to be able to access/read that file. :( Any ideas what I am doing in here? Help is much appreciated by me. thanks.

This is what I've tried so far:

    <?php

    include('connectdb.php');
    $sql = "SELECT salesordercard_code, location_from, location_to, salesmancode FROM salesorderingcard";
    $result = mysql_query($sql);
    if($result === FALSE) {
     die(mysql_error()); // TODO: better error handling
    }
    $set = array();
    while($row1 = mysql_fetch_assoc($result)) {
        $set[] = $row1;
    }

    header('Content-type: application/json');
    echo json_encode($set);


    ?>

MainActivity.class

@Override
protected Void doInBackground(Void... params) {
        // Create the array
        arraylist = new ArrayList<HashMap<String, String>>();
        // Retrive JSON Objects from the given website URL in JSONfunctions.class
        jsonobject = JSONFunctions.getJSONfromURL("http://www.shoppersgroup.net/vanmanagement/results.php");

        try {
            // Locate the array name
            jsonarray = jsonobject.getJSONArray("posts");

            for (int i = 0; i < jsonarray.length(); i++) {
                HashMap<String, String> map = new HashMap<String, String>();
                jsonobject = jsonarray.getJSONObject(i);
                //Log.i(MainActivity.class.getName(), jsonobject.getString("movie_name"));
                // Retrive JSON Objects
                map.put(TAG_CODE, jsonobject.getString("salesordercard_code"));
                map.put(TAG_LOCATION_FROM, jsonobject.getString("location_from"));
                map.put(TAG_LOCATION_TO, jsonobject.getString("location_to"));
                // Set the JSON Objects into the array
                arraylist.add(map);
            }
        } catch (JSONException e) {
            Log.e("Error", e.getMessage());
            e.printStackTrace();
        }
        return null;
}

Logcat:

            11-18 02:35:08.521: E/log_tag(1047): Error parsing data org.json.JSONException: Value [{"salesmancode":"SLMAN001","location_to":"MAIN","location_from":"IN-TRANSIT","salesordercard_code":"SLESO0001"},{"salesmancode":"SLMAN001","location_to":"MAIN","location_from":"IN-TRANSIT","salesordercard_code":"SLESO0002"}] of type org.json.JSONArray cannot be converted to JSONObject
解决方案

First as Json String

[ // Its an Array
    "posts",// Its not an Array and its index is 0 so we will not use index 0
    {//Index1
        "salesordercard_code": "SLESO0001",
        "location_from": "IN-TRANSIT",
        "location_to": "MAIN",
        "salesmancode": "SLMAN001"
    },
    {//index2
        "salesordercard_code": "SLESO0002",
        "location_from": "IN-TRANSIT",
        "location_to": "MAIN",
        "salesmancode": "SLMAN001"
    }
]

Change your doInBackground() method like below,

@Override
protected Void doInBackground(Void... params) {
        // Create the array
        arraylist = new ArrayList<HashMap<String, String>>();
        // Retrive JSON Objects from the given website URL in JSONfunctions.class
        jsonobject = JSONFunctions.getJSONfromURL("http://www.shoppersgroup.net/vanmanagement/results.php");

        try {
            // Locate the array name
            jsonarray = new JSONArray(jsonobject.toString());

            for (int i = 1; i < jsonarray.length(); i++) {// strat from index1
                HashMap<String, String> map = new HashMap<String, String>();
                JSONObject jsonobj = jsonarray.getJSONObject(i);
                //Log.i(MainActivity.class.getName(), jsonobj.getString("movie_name"));
                // Retrive JSON Objects
                map.put(TAG_CODE, jsonobj.getString("salesordercard_code"));
                map.put(TAG_LOCATION_FROM, jsonobj.getString("location_from"));
                map.put(TAG_LOCATION_TO, jsonobj.getString("location_to"));
                // Set the JSON Objects into the array
                arraylist.add(map);
            }
        } catch (JSONException e) {
            Log.e("Error", e.getMessage());
            e.printStackTrace();
        }
        return null;
}

Hope this will solve your problem.

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07-30 21:20
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