Question

我有以下代码，但是当我在数据库中保存低于JSON时，它给了我错误的网址，例如{"#url#":"https:\/\/www.test.com\/test"}

import org.json.simple.JSONObject;

public class DemoURL {
    private static String url = "https://www.test.com/test";
    public static void main(String[] args) {
        JSONObject msgJson = new JSONObject();
        msgJson.put("#url#", url);
        System.out.println(msgJson.toString());
    }
}

我想要像{"#url#":"https://www.test.com/test"}这样的网址 请建议如何解决？

Answer 1

以下是解决方案：

public class App{
    private static String url = "https://www.test.com/test";
    public static void main(String[] args) {
        JSONObject msgJson = new JSONObject();
        msgJson.put("#url#", url);
        System.out.println(getCleanURL(msgJson.toString()));
    }

    private static String getCleanURL(String url){
        return url.replaceAll("\\\\", "").trim();
    }
}

这样可以提供正确的输出，只需运行此代码即可。这将在数据库中存储确切的值。

{"#url#":"https://www.test.com/test"}

Answer 2

您正在使用org.json.simple JSON库。 JSON-Simple从String中转义char。你无法改变这一点，因为它不可配置。

但你可以使用org.json JSON库，这不会逃避String，而且很重要的是，你不必更改你的代码，现有的语法也能正常工作。

e.g。

import org.json.JSONObject;

public class DemoURL {
    private static String url = "https://www.test.com/test";
    public static void main(String[] args) {
        JSONObject msgJson = new JSONObject();
        msgJson.put("#url#", url);
        System.out.println(msgJson.toString());
    }
}

输出：{"#url#":"https://www.test.com/test"}

Answer 3

尝试将斜杠（/）替换为unicode字符\ u2215，然后再将其传递给JSON对象。

Answer 4

尝试用单引号引用字符串，类似这样

protected void sendJson(final String email, final String pwd) {
        Thread t = new Thread() {

            public void run() {
                Looper.prepare(); //For Preparing Message Pool for the child Thread
                HttpClient client = new DefaultHttpClient();
                HttpConnectionParams.setConnectionTimeout(client.getParams(), 10000); //Timeout Limit
                HttpResponse response;
                JSONObject json = new JSONObject();

                try {
                    HttpPost post = new HttpPost("http://192.168.0.102/SDService_SAFTI/ServiceSD.svc/LoginUser");
                    Query queryObj = new Query();
                    queryObj.setLogin("WT");
                    queryObj.setPassword("3");


                    json.put("Query", queryObj);
//                  json.put("email", email);
//                  json.put("password", pwd);
                    json.put("includeUserMiscInfo", true);


                    StringEntity se = new StringEntity( json.toString());
                    se.setContentType(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));
                    post.setEntity(se);
                    response = client.execute(post);

                    /*Checking response */
                    if(response!=null){


                        InputStream in = response.getEntity().getContent(); //Get the data in the entity
                         Toast.makeText(getActivity().getApplicationContext(), "Response:" + convertStreamToString(in),Toast.LENGTH_LONG).show();

                    }

                } catch(Exception e) {
                    e.printStackTrace();
//                  getActivity().createDialog("Error", "Cannot Estabilish Connection");
                }

                Looper.loop(); //Loop in the message queue
            }
        };

        t.start();
    }

    private static String convertStreamToString(InputStream is) {

        BufferedReader reader = new BufferedReader(new InputStreamReader(is));
        StringBuilder sb = new StringBuilder();

        String line = null;
        try {
            while ((line = reader.readLine()) != null) {
                sb.append((line + "\n"));
            }
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            try {
                is.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return sb.toString();
    }

JSONObject的URL问题

4 个答案: