如何解析http标头以获取上传文件并将其保存到磁盘

Question

I am developing a http web server in java using socket which gets post header InputStream and then I processed the header with some String split by the header 'boundary' and '\\r\\n' and got all Headers, Cookies in HashMap(s) and got the contents of the file in a String and saved that String to a file on the server. It works fine when I upload text file or java source file to the server but in case of doc, pdf and image it shows corrupted file and corrupted image.

    PrintWriter out;
        try {
            out = new PrintWriter(new OutputStreamWriter(
                    new FileOutputStream(UploadPath + "\\" + FileName)));
            out.print(FileData);
            out.close();
        } catch (Exception e) {

        }

Above code will save contents of 'FileData' at 'UploadPath' with 'FileName'.

In case of jpg or doc file String FileData is having binary contents of the uploaded file which saved by the above code and also I checked both files for their size in bytes and both were having equal size in byte and I also matched contents of the actual file and content FileData String by debugging the application.

I also checked actual uploaded image file and the FileData String and both matches byte by byte but the image uploaded is totally corrupted.

After searching on internet for this complete day I am not able to find the solution for this. Please help.

I do not want to use apache commons which was suggested on most of the pages.

If you want to see more codes then I will post them.

Answer 1

As you are dealing with binary data, you should use byte and OutputStream instead of String and Writer : If you put some bytes in a string, they are decoded

So if you have found the boundaries of the binary data in your request (represented by a byte array), copy the content byte-wise directly to an output stream.

This only works, if your request is already completely in memory. Regarding file upload, this is not always possible, because you can run out of memory, if you have large files.

So the best way to implement a file upload is to read only the next byte from the stream: This is the difference between splitting and parsing . Actually you need a real parser for multipart form data. Now things get complex, and this is the reason why everybody uses commons-fileupload: It's not that easy to detect the boundaries, if your "look ahead" is just some bytes.

I had to implement a clean-room implementation for legal reasons. If that is not your situation, look in the the source of commons-fileupload. And have a look at the RFC

Answer 2

Since you use Java 7, this is quite easy: use Files.copy() .

Also, DO NOT store file contents as String s, those will only ever be valid for text files. Use classical InputStream / OutputStream s to read/write.

Answer 3

You could read it using an array of bytes like the following

InputStream is = ...
ByteArrayOutputStream buffer = new ByteArrayOutputStream();

int nRead;
byte[] data = new byte[16384];

while ((nRead = is.read(data, 0, data.length)) != -1) {
  buffer.write(data, 0, nRead);
}

buffer.flush();

return buffer.toByteArray();

Answer 4

I solved my problem like this,

    while (inputRequest.available()>0) {
            try {
                int t = inputRequest.read();
                ch = (char) t;
                //here i checked each byte data
            } catch (IOException e) {
            }
    }

Problem was that the input stream was having http header fields along with the file content located anywhere in the stream, so I firstly stored the bytes in a temp String until i get '\\r' and '\\n' in the stream. In this way I got the boundary for multipart/form-data HTTP header and then I compared the temp String until I found the boundary and other known header contents and then I sent the input-stream to file output-stream. But in some cases header may contain other contents after file content so and definitely it will have a ending boundary so I was continuously keeping track of each byte that I have read and then I sent each byte individually to the file output-stream. Here is the sample http header-

   Host: localhost
   User-Agent: Mozilla/5.0 (Windows NT 6.2; WOW64; rv:21.0) Gecko/20100101 Firefox/21.0
   Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8
   Accept-Language: en-US,en;q=0.5
   Accept-Encoding: gzip, deflate
   DNT: 1
   Referer: http://localhost/index.html
   Connection: keep-alive
   Content-Type: multipart/form-data; boundary=---------------------------274761981030199
   Content-Length: 1405

   -----------------------------274761981030199
   Content-Disposition: form-data; name="name1"

   pppppp
   -----------------------------274761981030199
   Content-Disposition: form-data; name="name2"

   rrrrrrrrr
   -----------------------------274761981030199
   Content-Disposition: form-data; name="name3"

   eeeeeeee
   -----------------------------274761981030199
   Content-Disposition: form-data; name="name4"

   2
   -----------------------------274761981030199
   Content-Disposition: form-data; name="name5"; filename="CgiPost.java"
   Content-Type: text/x-java-source

   import java.io.*;

   // This appears in Core Web Programming from
   // Prentice Hall Publishers, and may be freely used
   // or adapted. 1997 Marty Hall, hall@apl.jhu.edu.


   public class CgiPost extends CgiGet 
   {

   public static void main(String[] args) 
   {

   try 
   {

   DataInputStream in
    = new DataInputStream(System.in);

   String[] data = { in.readLine() };

   CgiPost app = new CgiPost("CgiPost", data, "POST");

   app.printFile();
       } catch(IOException ioe) {
         System.out.println
           ("IOException reading POST data: " + ioe);

   }
     }

     public CgiPost(String name, String[] args,
     String type) {
       super(name, args, type);
     }
   }

   -----------------------------274761981030199
   Content-Disposition: form-data; name="name6"

   pppppppppp
   -----------------------------274761981030199--

NOTE: In some cases there are chances that your application code reaches to inputRequest.available() but the browser haven't sent the request yet, in this case inputRequest.available() will always return 0 and your while loop will exit immediately. To avoid this first read one byte using inputRequest.read() and then execute code because you can guess the first byte from others in case of http header.

If you are using some count int then use long instead of int, because stream stops in some cases where int variable reaches its limit.

Try to transfer the int value returned from int t = inputRequest.read() to fileoutputstream.write(t).

inputRequest.available() keeps decreasing as you are reading byte form inputstream, it returns number of bytes available in the stream.

In this way you can upload files of large size without any corruption in it.

Leave your comment if anyone needs more details about this.