Android obj.getJSONArray未获取JSON响应

我正试图从

PHP

获取一个包含纬度和经度的响应，所以在我的应用程序收到这些坐标后，它应该根据地图上的坐标添加标记。据介绍，

JSON

响应是有效的，但应用程序似乎不想添加标记

.java代码，用于检索

JSON

响应：

@Override
        protected void onPostExecute(String s) {
            super.onPostExecute(s);
            //hiding the progressbar after completion

            try {
                //converting response to json object
                JSONObject obj = new JSONObject(s);

                //if no error in response
                if (!obj.getBoolean("error")) {
                    Toast.makeText(getApplicationContext(), obj.getString("message"), Toast.LENGTH_SHORT).show();

                    JSONArray locatieArray = obj.getJSONArray("locatie");
                    for (int i = 0; i < locatieArray.length(); i++) {
                        JSONObject locatie = locatieArray.getJSONObject(i);
                        // check latitudine and longitudine is not null and if not null then cast these values and call the addMarker() method.
                        if(!locatie.isNull("latitudine") && !locatie.isNull("longitudine")) {
                            latitudine_sql =Double.valueOf(locatie.getString("latitudine"));
                            longitudine_sql = Double.valueOf(locatie.getString("longitudine"));
                            addMarker(latitudine_sql, longitudine_sql); // this method is implemented below
                        }
                        tip_problema_sql = locatie.getString("tip_problema");
                    }

                } else {
                    Toast.makeText(getApplicationContext(), "Some error occurred", Toast.LENGTH_SHORT).show();
                }
            } catch (JSONException e) {
                e.printStackTrace();
            }
        }
    }

JSON

应用程序检索到的响应：

{
"error": false,
"message": "Alerta raportata",
"locatie": {
    "1": {
        "latitudine": 37.4254,
        "longitudine": -122.08,
        "tip_problema": "Semafor nefunctional"
    },
    "2": {
        "latitudine": 37.4259,
        "longitudine": -122.088,
        "tip_problema": "Semafor nefunctional"
    },
    "3": {
        "latitudine": 37.4259,
        "longitudine": -122.088,
        "tip_problema": "Semafor nefunctional"
    },
    "4": {
        "latitudine": 37.4207,
        "longitudine": -122.085,
        "tip_problema": "Semafor nefunctional"
    }
}

---

}地点不是JSONArray。根据您在问题中提到的回答，这是一个JSONObject

您正在尝试访问JSONArray，但在您的响应中，JSONObject中的“locatie”不是JSONArray

{
} //This indicates JSONObject.

[
] //This indicates JSONArray.

JSONArray locatiarray=obj.getJSONArray（“locatie”）

您需要在服务器端修改php响应。您的回答应该如下所示：

 {
    "error": false,
    "message": "Alerta raportata",
    "locatie": [
       {
            "latitudine": 37.4254,
            "longitudine": -122.08,
            "tip_problema": "Semafor nefunctional"
        },
        {
            "latitudine": 37.4259,
            "longitudine": -122.088,
            "tip_problema": "Semafor nefunctional"
        },
        {
            "latitudine": 37.4259,
            "longitudine": -122.088,
            "tip_problema": "Semafor nefunctional"
        },
         {
            "latitudine": 37.4207,
            "longitudine": -122.085,
            "tip_problema": "Semafor nefunctional"
        }
    ]

    }

若如图所示更改服务器端的响应，则无需更改应用程序端代码。它会很好地工作。

类似于演示：

<?php 

// Declare two dimensional associative 
// array and initilize it 
$arr['locatie'] [] = array ( 
    1=>array( 
        "latitudine"=>1, 
        "longitudine"=>"Doorbell", 
        "tip_problema"=>199 
    ), 
    2=>array( 
        "latitudine"=>2, 
        "longitudine"=>"Bottle", 
        "tip_problema"=>99 
    ), 
    3=>array( 
        "latitudine"=>3, 
        "longitudine"=>"Washing Machine", 
        "tip_problema"=>7999 
    )
); 

// Function to convert array into JSON 
echo json_encode($arr); 

?> 

请根据您对数组值的要求进行设置，如：latitudine、longitudine、tip_problem a use

foreach

loop来生成它

---

我希望它能帮助你

“locatie”：[“1”：{“latitudine”：37.4254，“longitudine”：-122.08，“tip_problema”：“Semafor nefunctional”}]类似于此格式。您还没有使用JSONArray，它是一个JSONObject。@ViralPatel关于如何更改.php以获得该响应的任何提示？我是

PHP

问题的初学者，无法获得返回

locatie

between[]的响应需要帮助，请询问。有关如何更改.PHP以获得该响应的提示？我是

PHP

事务的初学者，在[]之间无法得到返回

locatie

的响应@jarvis@Costin我对PHPYes一点也不了解它实际上帮助了我！我用这种方法在

中实现了数组，现在它开始工作了！谢谢
<?php 

// Declare two dimensional associative 
// array and initilize it 
$arr['locatie'] [] = array ( 
    1=>array( 
        "latitudine"=>1, 
        "longitudine"=>"Doorbell", 
        "tip_problema"=>199 
    ), 
    2=>array( 
        "latitudine"=>2, 
        "longitudine"=>"Bottle", 
        "tip_problema"=>99 
    ), 
    3=>array( 
        "latitudine"=>3, 
        "longitudine"=>"Washing Machine", 
        "tip_problema"=>7999 
    )
); 

// Function to convert array into JSON 
echo json_encode($arr); 

?> 

{"locatie":[{"1":{"latitudine":1,"longitudine":"Doorbell","tip_problema":199},"2":{"latitudine":2,"longitudine":"Bottle","tip_problema":99},"3":{"latitudine":3,"longitudine":"Washing Machine","tip_problema":7999}}]}