3 
public void doView(RenderRequest renderRequest, RenderResponse renderResponse) throws IOException, PortletException {

    Map countryList = new HashMap();

    String str = "http://10.10.10.25/TEPortalIntegration/CustomerPortalAppIntegrationService.svc/PaymentSchedule/PEPL/Unit336";

    try {
        URL url = new URL(str);

        URLConnection urlc = url.openConnection();

        BufferedReader bfr = new BufferedReader(new InputStreamReader(urlc.getInputStream()));

        String line, title, des;

        while ((line = bfr.readLine()) != null) {

            JSONArray jsa = new JSONArray(line);

            for (int i = 0; i < jsa.length(); i++) {
                JSONObject jo = (JSONObject) jsa.get(i);

                title = jo.getString("Amount"); 

                countryList.put(i, title);
            }

            renderRequest.setAttribute("out-string", countryList);

            super.doView(renderRequest, renderResponse);
        }
    } catch (Exception e) {

    }
}

我正在尝试json从 liferay portlet 类访问对象，并且我想将任何字段的值数组传递json给 jsp 页面。

4

1 回答 1

3

在将其转换为 JSON 数组之前，您需要阅读完整的响应。这是因为响应中的每一行都将是一个（无效的）JSON 片段，无法单独解析。稍作修改，您的代码应该可以工作，突出显示如下：

// fully read response
final String line;
final StringBuilder builder = new StringBuilder(2048);

while ((line = bfr.readLine()) != null) {
    builder.append(line);
}

// convert response to JSON array
final JSONArray jsa = new JSONArray(builder.toString());

// extract out data of interest
for (int i = 0; i < jsa.length(); i++) {
    final JSONObject jo = (JSONObject) jsa.get(i);
    final String title = jo.getString("Amount"); 

    countryList.put(i, title);
}
于 2012-12-18T07:09:31.810 回答