javax.servlet.http.Part到java.io.File

Question

javax.servlet.http.Part到java.io.File

我需要帮助代码,用于将javax.servlet.http.Part转换为java.io.File

我发现这个有用的代码,但我需要帮助正确实现代码.

private void processFilePart(Part part, String filename) throws IOException
    {
        filename = filename.substring(filename.lastIndexOf('/') + 1).substring(filename.lastIndexOf('\\') + 1);
        String prefix = filename;
        String suffix = "";
        if (filename.contains("."))
        {
            prefix = filename.substring(0, filename.lastIndexOf('.'));
            suffix = filename.substring(filename.lastIndexOf('.'));
        }
        File file = File.createTempFile(prefix + "_", suffix, new File(location));
        if (multipartConfigured)
        {
            part.write(file.getName());
        }
        else
        {
            InputStream input = null;
            OutputStream output = null;
            try
            {
                input = new BufferedInputStream(part.getInputStream(), DEFAULT_BUFFER_SIZE);
                output = new BufferedOutputStream(new FileOutputStream(file), DEFAULT_BUFFER_SIZE);
                byte[] buffer = new byte[DEFAULT_BUFFER_SIZE];
                for (int length = 0; ((length = input.read(buffer)) > 0);)
                {
                    output.write(buffer, 0, length);
                }
            }
            finally
            {
                if (output != null)
                    try
                    {
                        output.close();
                    }
                    catch (IOException logOrIgnore)
                    {
                    }
                if (input != null)
                    try
                    {
                        input.close();
                    }
                    catch (IOException logOrIgnore)
                    {
                    }
            }
        }
        put(part.getName(), file);
        part.delete();
    }

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我试图编辑代码,以便创建java.io.File作为结果,但我总是有问题.

private void processFilePart(Part part, String filename) throws IOException
    {
        int DEFAULT_BUFFER_SIZE = 2048;

        filename = filename.substring(filename.lastIndexOf('/') + 1).substring(filename.lastIndexOf('\\') + 1);
        String prefix = filename;
        String suffix = "";
        if (filename.contains("."))
        {
            prefix = filename.substring(0, filename.lastIndexOf('.'));
            suffix = filename.substring(filename.lastIndexOf('.'));
        }
        File file = new File(filename);
        InputStream input = null;
        OutputStream output = null;
        try
        {
            input = new BufferedInputStream(part.getInputStream(), DEFAULT_BUFFER_SIZE);
            output = new BufferedOutputStream(new FileOutputStream(file), DEFAULT_BUFFER_SIZE);
            byte[] buffer = new byte[DEFAULT_BUFFER_SIZE];
            for (int length = 0; ((length = input.read(buffer)) > 0);)
            {
                output.write(buffer, 0, length);
            }
        }
        finally
        {
            if (output != null)
                try
                {
                    output.close();
                }
                catch (IOException logOrIgnore)
                {
                }
            if (input != null)
                try
                {
                    input.close();
                }
                catch (IOException logOrIgnore)
                {
                }
        }

        // how to get the result

        part.delete();
    }

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转换对象的正确方法是什么？

Answer 1

Mau*_*rez 6

fileupload示例应用程序 - 取自http://docs.oracle.com/javaee/6/tutorial/doc/glraq.html

fileupload示例说明了如何实现和使用文件上载功能.

fileupload示例应用程序由一个servlet和一个HTML表单组成,该表单向servlet发出文件上载请求.

此示例包含一个非常简单的HTML表单,其中包含两个字段:文件和目标.输入类型文件使用户能够浏览本地文件系统以选择文件.选择文件后,它将作为POST请求的一部分发送到服务器.在此过程中,对带有输入类型文件的表单应用了两个强制性限制:

必须将enctype属性设置为multipart/form-data的值.

它的方法必须是POST.

以这种方式指定表单时,整个请求将以编码形式发送到服务器.然后,servlet处理请求以处理传入的文件数据并从流中提取文件.目标是文件将保存在计算机上的位置的路径.按下表单底部的"上载"按钮会将数据发布到servlet,从而将文件保存在指定的目标位置.

tut-install/examples/web/fileupload/web/index.html中的HTML表单如下:

<!DOCTYPE html>
<html lang="en">
    <head>
        <title>File Upload</title>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
    </head>
    <body>
        <form method="POST" action="upload" enctype="multipart/form-data" >
            File:
            <input type="file" name="file" id="file" /> <br/>
            Destination:
            <input type="text" value="/tmp" name="destination"/>
            </br>
            <input type="submit" value="Upload" name="upload" id="upload" />
        </form>
    </body>
</html>

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当客户端需要将数据作为请求的一部分发送到服务器时,例如上载文件或提交完成的表单时,将使用POST请求方法.相反,GET请求方法仅向服务器发送URL和标头,而POST请求也包括消息正文.这允许将任意类型的任意长度数据发送到服务器.POST请求中的标头字段通常表示邮件正文的Internet媒体类型.

在提交表单时,浏览器将内容组合在一起,将所有部分组合在一起,每个部分代表一个表单的字段.部件以输入元素命名,并使用名为boundary的字符串分隔符彼此分隔.

在选择sample.txt作为将上载到本地文件系统上的tmp目录的文件之后,这就是从fileupload表单提交的数据:

POST /fileupload/upload HTTP/1.1
Host: localhost:8080
Content-Type: multipart/form-data; 
boundary=---------------------------263081694432439
Content-Length: 441
-----------------------------263081694432439
Content-Disposition: form-data; name="file"; filename="sample.txt"
Content-Type: text/plain

Data from sample file
-----------------------------263081694432439
Content-Disposition: form-data; name="destination"

/tmp
-----------------------------263081694432439
Content-Disposition: form-data; name="upload"

Upload
-----------------------------263081694432439--

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servlet FileUploadServlet.java可以在tut-install/examples/web/fileupload/src/java/fileupload /目录中找到.servlet开头如下:

@WebServlet(name = "FileUploadServlet", urlPatterns = {"/upload"})
@MultipartConfig
public class FileUploadServlet extends HttpServlet {

    private final static Logger LOGGER = 
            Logger.getLogger(FileUploadServlet.class.getCanonicalName());

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@WebServlet批注使用urlPatterns属性来定义servlet映射.

@MultipartConfig注释指示servlet期望使用multipart/form-data MIME类型进行请求.

processRequest方法从请求中检索目标和文件部分,然后调用getFileName方法从文件部分检索文件名.然后,该方法创建FileOutputStream并将文件复制到指定的目标.该方法的错误处理部分捕获并处理了无法找到文件的一些最常见原因.processRequest和getFileName方法如下所示:

protected void processRequest(HttpServletRequest request,
        HttpServletResponse response)
        throws ServletException, IOException {
    response.setContentType("text/html;charset=UTF-8");

    // Create path components to save the file
    final String path = request.getParameter("destination");
    final Part filePart = request.getPart("file");
    final String fileName = getFileName(filePart);

    OutputStream out = null;
    InputStream filecontent = null;
    final PrintWriter writer = response.getWriter();

    try {
        out = new FileOutputStream(new File(path + File.separator
                + fileName));
        filecontent = filePart.getInputStream();

        int read = 0;
        final byte[] bytes = new byte[1024];

        while ((read = filecontent.read(bytes)) != -1) {
            out.write(bytes, 0, read);
        }
        writer.println("New file " + fileName + " created at " + path);
        LOGGER.log(Level.INFO, "File{0}being uploaded to {1}", 
                new Object[]{fileName, path});
    } catch (FileNotFoundException fne) {
        writer.println("You either did not specify a file to upload or are "
                + "trying to upload a file to a protected or nonexistent "
                + "location.");
        writer.println("<br/> ERROR: " + fne.getMessage());

        LOGGER.log(Level.SEVERE, "Problems during file upload. Error: {0}", 
                new Object[]{fne.getMessage()});
    } finally {
        if (out != null) {
            out.close();
        }
        if (filecontent != null) {
            filecontent.close();
        }
        if (writer != null) {
            writer.close();
        }
    }
}

private String getFileName(final Part part) {
    final String partHeader = part.getHeader("content-disposition");
    LOGGER.log(Level.INFO, "Part Header = {0}", partHeader);
    for (String content : part.getHeader("content-disposition").split(";")) {
        if (content.trim().startsWith("filename")) {
            return content.substring(
                    content.indexOf('=') + 1).trim().replace("\"", "");
        }
    }
    return null;
}

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