在Java Servlet中上传文件

更新时间：2023-12-03 22:43:46

不提供文件路径浏览器的安全功能。

您的代码中有可用的文件内容（ InputStream filecontent ），所以您可以使用它或使用 FileItem 的简便方法，例如

  item.write（new File （ /path/to/myfile.txt））;

I have a Java Dynamic Web Project, and I'm using TomCat v7.0.

I am new to web projects and I didn't quite understand how I can upload a file in one of my jsp pages. Since my project is intended to be only local, I thought I could use a multipart form in which the person would choose the file (and this part goes fine) and later retreive the file path from my Servlet. I can't complete this part though, it appears to only give me the name of the file, not its entire path.

Can anyone point me to the right direction? I've read several posts about Apache File Upload and retreiving information from the multipart form but nothing seems to help me.

How can I get the file path from a form or alternatively how can I get the uploaded file to use in my Java classes?

Thanks in advance.

.jsp:

<form method="post" action="upload" enctype="multipart/form-data">
<input type="file" name="filePath" accept="application/vnd.ms-excel, application/vnd.openxmlformats-officedocument.spreadsheetml.sheet"></input>
<input type="submit" value="Enviar"></input>
</form>

Java Servlet:

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException
{
    PrintWriter out = response.getWriter();
    out.println("<html><body>");

    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();

                out.println("<h1>"+fieldname+" / "+fieldvalue+"</h1>");
            }
            else
            {
                // Process form file field (input type="file").
                String fieldname = item.getFieldName();
                String filename = item.getName();
                InputStream filecontent = item.getInputStream();
                String s = filecontent.toString();
                out.println("<h1>"+s+" / "+filename+"</h1>");
                item.write(null);
            }
        }
    }
    catch (FileUploadException e)
    {
        throw new ServletException("Cannot parse multipart request.", e);
    }
    catch (Exception e)
    {
        e.printStackTrace();
    }

    out.println("</body></html>");
}

Not providing the file path is a security feature of the browser.

You have the file contents available in your code (InputStream filecontent) so you could use that or use one of the convenience methods on FileItem, e.g.

item.write(new File("/path/to/myfile.txt"));

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在Java Servlet中上传文件

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