Question

我正在向Web服务发布一些数据，我收到了一个json回复，我希望将其传递给jsonobject。

网络服务的回复是：

{
    "ValidateLoginResult": [
        {
            "ErrorMessage": "Wrong username pass",
            "PropertyName": null
        }
    ]
}

我希望将错误消息和属性名称传递给变量。我尝试过使用JSONobject和JSONarray，但没有运气

                HttpClient httpclient = new DefaultHttpClient();

                // 2. make POST request to the given URL
                HttpPost httpPost = new HttpPost(serverURL);
                StringEntity se = new StringEntity(data);
                httpPost.setEntity(se);

                // 7. Set some headers to inform server about the type of the content
                httpPost.setHeader("Accept", "application/json");
                httpPost.setHeader("Content-type", "application/json");

                // 8. Execute POST request to the given URL
                HttpResponse httpResponse = httpclient.execute(httpPost);
                // 9. Getting Reply
                inputStream = httpResponse.getEntity().getContent();
                ...
                ...
                 JSONArray json = new JSONArray(convertInputStreamToString(inputStream));



                JSONObject json_LL = json.getJSONObject(0);

                String str_value=json_LL.getString("ErrorMessage");

Answer 1

您正在JSONObject收到回复，并且您正试图在JSONArray中收到回复..这就是您收到错误的原因..

试试这种方式......

try {
    JSONObject result = new JSONObject(response);

    if(data.has("ValidateLoginResult"){
        JSONArray array = result.getJSONArray("ValidateLoginResult");

        for (int i = 0; i < array.length(); i++) {
        JSONObject obj = array.getJSONObject(i);
            String ErrorMessage= ""+obj.getString("ErrorMessage");
            String PropertyName= ""+obj.getString("PropertyName");
        }
    }

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

或

如果你想要一行答案..试试这个..

// going directly to array object..
JSONObject result = new JSONObject(response).getJSONArray("ValidateLoginResult").getJSONObject(0);

String ErrorMessage= ""+result.getString("ErrorMessage");
String PropertyName= ""+result.getString("PropertyName");

将json的回复从webservice传递给变量

1 个答案: