java - 在 java 中读取请求标头为“Accept-Encoding: gzip, deflate”的 HttpServeltRequest 正文

Question

我正在尝试从中读取正文，HttpServletRequest但正文未显示。

包含以下HttpServletRequest信息。

 body: id=8652976&event=test&payload[chargify]=testing
headers: 
  Content-Type: application/x-www-form-urlencoded
  X-Chargify-Webhook-Signature: ed57683a9d8a3f25869dbf138ce5c66f
  Accept: "*/*; q=0.5, application/xml"
  X-Chargify-Webhook-Id: "8652976"
  Accept-Encoding: gzip, deflate
  Content-Length: "47"

我正在尝试使用以下代码读取请求正文。但是在inStream.readLine()给出空值时

if ("gzip".equalsIgnoreCase(request.getHeader("Accept-Encoding"))) {

  GZIPInputStream gzipInputStream = new GZIPInputStream(request.getInputStream());
  Reader decoder = new InputStreamReader(gzipInputStream, "UTF-8");
  BufferedReader br = new BufferedReader(decoder);
  String inputLine;
  while ((inputLine = br.readLine()) != null) {
    body.append(inputLine).append(System.getProperty("line.separator"));
  }
   gzipInputStream.close();
 } else {
    InputStreamReader input = new InputStreamReader(request.getInputStream());
    BufferedReader inStream = new BufferedReader(input);
    String inputLine;
    while ((inputLine = inStream.readLine()) != null) {
        body.append(inputLine).append(System.getProperty("line.separator"));
    }
    inStream.close();
}

Answer 1

有更简单的方法可以处理所有这些样板代码并将您从这些样板代码中解救出来。

检查Apache commons-fileupload 以获取文档

下面是一个示例代码，展示了使用这个库是多么容易。

// Create a new file upload handler
ServletFileUpload upload = new ServletFileUpload();

// Parse the request
FileItemIterator iter = upload.getItemIterator(request);
while (iter.hasNext()) {
    FileItemStream item = iter.next();
    String name = item.getFieldName();
    InputStream stream = item.openStream();
    if (item.isFormField()) {
        System.out.println("Form field " + name + " with value "
            + Streams.asString(stream) + " detected.");
    } else {
        System.out.println("File field " + name + " with file name "
            + item.getName() + " detected.");
        // Process the input stream
        ...
    }
}