0

我正在通过 http url 连接将字符串从一个 servlet 发送到另一个:

final HttpURLConnection http = (HttpURLConnection) url.openConnection();     // here url is the url of the second servlet
http.setRequestMethod("POST");
http.setDoOutput(true);
http.setDoInput(true);
http.setUseCaches(false);
final OutputStream outstr = http.getOutputStream();
outstr.write(sb.toString().getBytes());
outstr.flush();
outstr.close();

我面临的问题是将其作为来自另一个 servlet 的请求来阅读。我尝试在第二个 servlet 的 getPost 方法中编写以下代码,但这不起作用:

try {
        int len = req.getContentLength();
        byte[] input = new byte[len];
        ServletInputStream sin = req.getInputStream();
        int c, count = 0;
        while ((c = sin.read(input, count, input.length - count)) != -1) {
            count += c;
        }
        sin.close();
        String inString = new String(input);
        String decodedString = URLDecoder.decode(inString, "UTF-8");
        log.info("Response received - ");
        log.info(decodedString);
        resp.getWriter().write(decodedString);
        resp.getWriter().flush();
        resp.getWriter().close();
    } catch (IOException e) {

    }

有人可以帮助获取第一个 servlet 发送的字符串并显示它的正确方法是什么?

4

2 回答 2

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您可以将您的字符串作为参数写入该帖子的 servlet 中,并在作为参数接收的 servlet 中读取它。

那个帖子的Servlet:

outstr.write(("your_param=" + sb.toString()).getBytes());

接收:

String yourParamValue = request.getParameter("your_param");
于 2013-10-25T07:34:36.823 回答
0

试试这个代码,它应该可以解决这个问题

try {

        BufferedReader reader = request.getReader();
        StringBuilder inputStringBuilder = new StringBuilder();
        String line = null;
        while ((line = reader.readLine()) != null) {
            inputStringBuilder.append(line).append("\n");

        }

        String requestData = inputStringBuilder.toString();
        reader.close();

        resp.getWriter().write(requestData);
        resp.getWriter().flush();
        resp.getWriter().close();

    } catch (IOException e) {

        e.printStacktrace();
    }
于 2013-10-25T07:47:29.863 回答