Question

所以我一般都是json解析的新手。在社区的帮助下，我能够让json返回。但是，对于我的生活无法弄清楚如何访问/获取某些值。我想要/访问＆＃34; halfjpeg＆＃34; ＆＃34;内容中的价值＆＃34;部分。

[
  {
    "publicId": "1337",
    "name": "mypainting",
    "orientation": "vertical",
    "provider": "user",
    "providerFullName": "unknown",
    "location": {
      "coordinates": [
        -71,
        42
      ],
      "userCityCode": "BOS",
      "userCityName": "Boston",
      "userStateCode": "MA",
      "userStateName": "Massachusetts",
      "userCountryCode": "US",
      "userCountryName": "United States",
      "userZipCode": "02108",
      "latitude": 42
      "longitude": -71
    },
    "status": {
      "isLive": false,
      "isCertified": false,
      "hasVirusWarning": true
    },
    "content": {
      "halfJpeg": "userhalfpaintlocation.com",
      "fullJpeg": "userfullpaintlocation.com",
      "hugeJpeg": "userhugepaintlocation.com"
    },
    "createdAt": xxxxxxxx,
    "updatedAt": xxxxxxxx,
    "policy": {
      "isAdmin": false,
      "isUser": true
    }
  }
]

我的获取JSON的代码如下：

URL url = null;
        try {
            url = new URL("getthepaintingurl.com");
        } catch (MalformedURLException e) {
            e.printStackTrace();
        }
        HttpURLConnection conn = null;
        try {
            conn = (HttpURLConnection) url.openConnection();
        } catch (IOException e) {
            e.printStackTrace();
        }
        try {
            conn.setRequestMethod("GET");
        } catch (ProtocolException e) {
            e.printStackTrace();
        }

        try {
            System.out.println("Response Code: " + conn.getResponseCode());
        } catch (IOException e) {
            e.printStackTrace();
        }
        InputStream in = null;
        try {
            in = new BufferedInputStream(conn.getInputStream());
        } catch (IOException e) {
            e.printStackTrace();
        }

        BufferedReader reader = new BufferedReader(new InputStreamReader(in));
        StringBuilder out = new StringBuilder();
        String line;
        try {
            while ((line = reader.readLine()) != null) {
                out.append(line + "\n");

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

        System.out.println(out.toString());

        try {
            JSONObject jObject = new JSONObject(out.toString());
        } catch (JSONException e) {
            e.printStackTrace();
        }

Answer 1

build and install

只是你知道一些语法，阅读上面的内容（它应该有助于澄清！）。

您期望的回答是JSONArray的形式。所以它看起来像这样：

//create json array from the buffer string, and get the inner json object
JSONObject response = new JSONArray(out.toString()).getJSONObject(0);

//get the json object at key "content"
JSONObject content = response.getJSONObject("content");

//access the value of "halfJpg"
String halfJpgSrc = content.getString("halfJpeg");

注释应该自己解释，但实际上你从请求的输出中创建了一个JSONArray，并访问包含所有信息的内部json对象（索引0）。然后，您只需导航到您发送的数据。在这种情况下，内部数据存储在键/值对中，因此如何访问JSON响应中的任何其他内容应该是非常自我解释的。

以下是关于这两个对象的更多文档（JSONArray / JSONObject）

http://www.w3schools.com/json/json_syntax.asp JSONObject

从返回的json

1 个答案: