Question

我正在编写一个servlet，它从客户端接收xml文件并使用它。

我的问题是，在servletinputstream（我得到的：request.getInputStream（））中有一些开头和结尾的上传信息：

-----------------------------186292285129788
Content-Disposition: form-data; name="myFile"; filename="TASKDATA - Kopie.XML"
Content-Type: text/xml

<XML-Content>

-----------------------------186292285129788--

是否有一种智能解决方案可以将这些线从servletinputstream中删除？

问候

Answer 1

这是multipart/form-data标头（在RFC2388中指定）。获取一个完整的multipart/form-data解析器，而不是重新发明自己的解析器。 Apache Commons FileUpload是该作业的事实标准API。删除/WEB-INF/lib中所需的JAR文件，然后就像以下一样简单：

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    try {
        List<FileItem> items = new ServletFileUpload(new DiskFileItemFactory()).parseRequest(request);
        for (FileItem item : items) {
            if (item.isFormField()) {
                // Process regular form field (input type="text|radio|checkbox|etc", select, etc).
                String fieldname = item.getFieldName();
                String fieldvalue = item.getString();
                // ... (do your job here)
            } else {
                // Process form file field (input type="file").
                String fieldname = item.getFieldName();
                String filename = FilenameUtils.getName(item.getName());
                InputStream filecontent = item.getInputStream();
                // ... (do your job here)
            }
        }
    } catch (FileUploadException e) {
        throw new ServletException("Cannot parse multipart request.", e);
    }

    // ...
}

再一次，不要重塑自己。你真的不希望得到维护后果。

Answer 2

如果您的问题是要以流式传输方式读取文件（性能方面），请查看此链接 http://commons.apache.org/proper/commons-fileupload/streaming.html

（来自链接）：

// Check that we have a file upload request
boolean isMultipart = ServletFileUpload.isMultipartContent(request);

// 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.getInputStream();
    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
        ...
    }
}

Servlet：从servletinputstream剪切上传头

2 个答案: