【问题标题】:how to use JSON responce in a url如何在 url 中使用 JSON 响应
【发布时间】:2014-07-15 14:38:13
【问题描述】:

我正在使用 RESTful WebServices 开发一个 android 应用程序

假设, 我正在发送一个 url http 请求为

一些网络服务/数据/访问

并将数据发送为 {"serviceMessageCode":1,"serviceMessageText":"aaaaaa","items":null}

我想用获得的密钥发送另一个请求

somewebService/rest/services/secure/getcategories?apikey=aaaaaa

    int sMC = jsonObj.getInt("serviceMessageCode");

            if (sMC == 1) {
                smt = jsonObj.getString("serviceMessageText");

我可以用吗

somewebService/rest/services/secure/getcategories?apikey=smt

我想我不应该这样做,有人告诉我如何做到这一点..!!

请帮忙……

【问题讨论】:

    标签: android json web-services


    【解决方案1】:

    没有理由不能通过 GET 参数传递某些数据。它实际上取决于后端服务器上的 Rest API。您是否使用任何 REST 客户端或基础 apache http 包类向服务器发出请求?

    已编辑:

    BufferedReader in = null;
    
    try {
        HttpClient httpclient = new DefaultHttpClient();
    
        HttpGet request = new HttpGet();
        String uri = String.format("http://somewebService/rest/services/secure/getcategories?apikey=%s", Config.API_KEY); // API_KEY is constant value written somewhere or could you pass it as method argument
        URI website = new URI(uri);
        request.setURI(website);
        HttpResponse response = httpclient.execute(request);
        in = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));
    
        String line = null;
        StringBuilder builder = new StringBuilder();
        while(null != (line = in.readLine())) {
            builder.append(line);
        }
    
        in.close();
    } catch(Exception e) {
        Log.e("log_tag", "Error in http connection "+e.toString());
    }
    

    【讨论】:

    • 我正在使用 apache http 包
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