Question

我试图从HttpServletRequest读取身体，但身体没有显示。

HttpServletRequest包含以下信息。

 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

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并保存所有这些锅炉铭牌代码。

检查Apache commons-fileupload 对于Documentation

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
        ...
    }
}

读取HttpServeltRequest主体，其中请求标头是java中的“Accept-Encoding：gzip，deflate”

1 个答案: