Path problem of file upload and download under Web

Engineering structure

1. Generate a file to the specified folder

 //   This  .setfilename (string.valueof (System.currenttimemillis ())); String Path  = Servletactioncontext.getservletcontext (). Getrealpath ("/template/wordexporttemplate"  //  full path name under Project  String filepath = path + "\ \" + FileName + ". Doc"  //  write to file  Writer out = new  bufferedwriter (new  OutputStreamWriter (new  fileoutputstream (FilePath, true ), "Utf-8"), 10240 ""); 

2. Output the file under the specified file to the client (the output of the file generated above)

 Public throws Exception {        create ();         = Servletactioncontext.getservletcontext (). Getrealpath ("/template/wordexporttemplate");         = path + "\ \" + FileName + ". doc"        ; // the full path name        New File (filepath);         // returns only one input stream        return fileutils.openinputstream (file);    }

3. Save the file uploaded by the client to the specified folder under a specific name:

 PublicString Execute ()throwsexception{ServletContext ServletContext=Servletactioncontext.getservletcontext ();//Filenamefilename is the name of the file when it is uploaded, or you can define a new name with the UUID yourself .String dir = Servletcontext.getrealpath ("/template/excelimporttemplate/" +filenamefilename); System.out.println (dir);//file output stream, written to DIR-specified directory with first nameFileOutputStream OutputStream =NewFileOutputStream (dir);//Open the input stream of the uploaded fileFileInputStream InputStream =NewFileInputStream (fileName); byte[] buffer =New byte[1024]; intLen = 0;//reads a maximum of b.length bytes of data from this input stream into a byte array. The total number of bytes read into the buffer, or 1 if there is no more data because the end of the file has been reached.          while(len = inputstream.read (buffer))!=-1){//writes Len bytes from offset off in the specified byte array to this file output stream. Outputstream.write (buffer, 0, Len);        } inputstream.close ();        Outputstream.close (); returnSUCCESS; }

4. Read the file with the specified name under the specified folder:

ServletContext ServletContext = Servletactioncontext.getservletcontext ();
Filenamefilename is the name of the file when it is uploaded, or you can define a new name with the UUID yourself.
String dir = Servletcontext.getrealpath ("/template/excelimporttemplate/" +filenamefilename);
New File (dir);
FileInputStream Openinputstream = fileutils.openinputstream (file);
Hssfworkbook workbook = new Hssfworkbook (openinputstream);

5. FileInputStream FileOutputStream is a class. Can read and write directly.

Construction method

For example:

 Public throws Exception {        create ();         = Servletactioncontext.getservletcontext (). Getrealpath ("/template/excelexporttemplate");         = path + "\ \" + FileName + ". xls"        ; New File (filepath);         // returns only one input stream        return fileutils.openinputstream (file);    }

String Path = Servletactioncontext.getservletcontext (). Getrealpath ("/template"); File Tempfile = new file (path, templatename + ". xls"); Tempfile.delete (); Tempfile.createnewfile ();
= Fileutils.openoutputstream (tempfile);

1 ServletContext servletcontext = servletactioncontext.getservletcontext (); 2 // Filenamefilename is the name of the file when it is uploaded, or you can define a new name with the UUID yourself . 3  String dir = Servletcontext.getrealpath ("/template/excelimporttemplate/" +filename); 4 FileOutputStream OutputStream = new FileOutputStream (dir);

Path problem of file upload and download under Web