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以下是一个测试代码，我在其中尝试通过 google api 发布一些查询，api 应该以 json 可解析字符串发送响应。当我打印字符串时，我清楚地看到了诸如 url 之类的实体，但是当我尝试从 JSON 对象中获取这些实体时，我得到了错误。

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.net.URL;
import java.net.URLConnection;

import org.json.JSONObject;



public class test {

    public static void main (String[] args) {

        try {
            // The request also includes the userip parameter which provides the end
            // user's IP address. Doing so will help distinguish this legitimate
            // server-side traffic from traffic which doesn't come from an end-user.
            URL url = new URL(
                    "https://ajax.googleapis.com/ajax/services/search/web?v=1.0&q=lahore");
            URLConnection connection = url.openConnection();


            String line;
            StringBuilder builder = new StringBuilder();
            BufferedReader reader = new BufferedReader(new InputStreamReader(connection.getInputStream()));
            while((line = reader.readLine()) != null) {
                builder.append(line);
                System.out.println(line);

            }

            JSONObject jsonObject = new JSONObject(builder.toString());
            System.out.println(jsonObject);

            String url1 = (String) jsonObject.get("url");
            System.out.println(url1);


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

    }

}

它抛出以下异常：

org.json.JSONException: JSONObject["url"] not found.
    at org.json.JSONObject.get(JSONObject.java:459)
    at org.json.JSONObject.getJSONArray(JSONObject.java:540)
    at test.main(test.java:37)

非常感谢任何帮助。

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1 回答 1

1

检查json字符串后：

您无法直接获取“url”，要获取“url”，您需要在将流读入 StringBuilder 后执行以下操作：

JSONObject jsonObject = new JSONObject(builder.toString());
JSONObject responseData = (JSONObject)jsonObject.get("responseData");
JSONArray results = (JSONArray)responseData.get("results");
for(int i = 0; i < results.length(); i++)
{
     JSONObject urlObject  = (JSONObject)results.get(i);
     System.out.println(urlObject.get("url"));
}
于 2013-11-12T01:29:19.850 回答