Question

我想从JSON获取f_index值到我的Android应用程序中。我收到错误＆＃34;没有f_index的值。

W/System.err: org.json.JSONException: No value for f_index

如果我删除下面的行，那么我可以正确获取lat，long和is_local值并进一步使用它们。

ill_id_data.add(ill_jo.getInt("f_index"));

我无法理解这里出了什么问题？

在应用程序中收到JSON，我将其传递给解析方法。

{
"server_response": [{
    "f_index": "3",
    "lat": "21.9159",
    "long": "78.1021",
    "is_local": "0"
}, {
    "f_index": "6",
    "lat": "25.0751",
    "long": "54.9476",
    "is_local": "0"
}, {
    "f_index": "7",
    "lat": "14.6773",
    "long": "75.542",
    "is_local": "1"
}, {
    "f_index": "8",
    "lat": "21.9159",
    "long": "78.1021",
    "is_local": "0"
}, {
    "f_index": "9",
    "lat": "25.8079",
    "long": "86.0449",
    "is_local": "1"
}]
}

在我的应用程序中解析JSON的代码。

ArrayList<Double> ill_lat_data = new ArrayList<Double>(3);
ArrayList<Double> ill_long_data = new ArrayList<Double>(3);
ArrayList<Interger> ill_id_data = new ArrayList<Integer>(3);
ArrayList<Integer> ill_is_local_data = new ArrayList<Integer>(3);

public void parseIndexLatLngJSON() {

    int count_ill = 0;

    try {
        ill_json_object = new JSONObject(json_index_latlng);
        ill_json_array = gm_json_object.getJSONArray("server_response");

        while (count_ill < ill_json_array.length()) {
            JSONObject ill_jo = ill_json_array.getJSONObject(count_ill);

            ill_id_data.add(ill_jo.getInt("f_index"));
            Log.d("tag_last_check", String.valueOf(ill_jo.getInt("f_index")));
            ill_lat_data.add(ill_jo.getDouble("lat"));
            ill_is_local_data.add(ill_jo.getInt("is_local"));
            ill_long_data.add(ill_jo.getDouble("long"));

            count_ill++;
        }
    } catch (JSONException e) {
        e.printStackTrace();
    }
}

这是我的php文件。

$sql="SELECT id,lat,`long` FROM newsphere_one WHERE lat != 0 OR `long` != 0;";

$con=mysqli_connect($host,$hostname,$password,$db);

$result = mysqli_query($con,$sql);

$response = array();

while($row=mysqli_fetch_array($result)){
array_push($response, array("f_index" => $row[0],"lat" => $row[1],"long" => $row[2],"is_local" => $row[3]));
}

echo json_encode(array("server_response"=>$response));

Answer 1

首先，

ill_id_data.add(ill_jo.getInt("f_index"));

尝试获取一个整数，而在json中它被定义为一个字符串。尝试获取字符串然后将其转换为int。尝试使用：

Integer.parseInt()

Answer 2

在php中

从

更改以下行

Property binding ngForOf not used by any directive on an embedded template

到

array_push($response, array("f_index" => $row[0],"lat" => $row[1],"long" => $row[2],"is_local" => $row[3]));

使用array_push($response, array("f_index" => intval($row[0]),"lat" => $row[1],"long" => $row[2],"is_local" => $row[3]));转换为整数

或

使用intval($row[0])转换为float

Answer 3

再次仔细观察以下两行

ill_json_object = new JSONObject(json_index_latlng);
ill_json_array = gm_json_object.getJSONArray("server_response");

在第1行中，您已将json_index_latlng指定给对象 ill_json_object

但在第2行中，您正在从 gm_json_object 访问服务器响应。

再次检查您的代码，祝你好运

错误：＆＃34; ColumnName＆＃34;没有值。在Android中获取JSON值

3 个答案: