如何使用PARSE REST API在JAVA Libgdx中使用约束进行查询?

时间:2017-01-19 17:06:22

标签: java rest parse-platform libgdx httprequest

我正在使用解析服务器后端并在JAVA(Libgdx)中开发我的前端。

我可以在没有如下参数的情况下检索简单查询的结果:

        HttpRequest httpGet = new HttpRequest(HttpMethods.GET);
        httpGet.setUrl("https://parseapi.back4app.com/classes/MESSAGE/");
        httpGet.setHeader("Content-Type", "application/json");
        httpGet.setHeader("X-Parse-Application-Id", app_id);
        httpGet.setHeader("X-Parse-REST-API-Key", app_key);
        Gdx.net.sendHttpRequest(httpGet,Parse.this);

但我没有设法使用限制条件。我尝试将参数放入查询中,但没有成功:

    HttpRequest httpGet = new HttpRequest(HttpMethods.GET);
    String params = "where={\"username\":\"unknown\"}";
    String encoded="";
    try {
        encoded=URLEncoder.encode(params,"utf-8");
    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    }
    httpGet.setUrl("https://parseapi.back4app.com/classes/MESSAGE/");
    httpGet.setHeader("Content-Type", "application/json");
    httpGet.setHeader("X-Parse-Application-Id", app_id);
    httpGet.setHeader("X-Parse-REST-API-Key", app_key);
    httpGet.setContent(encoded);
    Gdx.net.sendHttpRequest(httpGet,Parse.this); 

我还尝试将参数(UTF-8编码)直接放入url;它也没有工作。(如同这个问题https://www.parse.com/questions/how-to-interpret-rest-api-without-data-url-encode

(解析服务器REST API文档https://parseplatform.github.io/docs/rest/guide/#query-constraints

如果你们有任何想法,我会非常感激, 提前谢谢!

0 个答案:

没有答案