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i am new with servlet programming, i need to develop a .WAR API using servlet which will upload only xlsx files to a specific path in server (this API will be using by WebFocus (a programming language for reporting))

the specific path will be in a hidden filed of html <form>

here is my HTML 

<html>  
<body>  
<form action="UploadServlet" method="Post" enctype="multipart/form-data">  
Select File:<input type="file" name="fname"/><br/>

  <input type="hidden" name="path" value="G:/newFolder" />
<input type="submit" value="upload"/>  

</form>  
</body>  
</html>

now what should write inside my UploadServlet.java file in order to upload xlsx files into path defined in path hidden filed

currently in my UploadServlet.java i used MultipartRequest and it uploads the file to destination correctly but i want the distenation to be the value of path hidden field of the <form> 

import java.io.*;  
import javax.servlet.ServletException;  
import javax.servlet.http.*;  
import com.oreilly.servlet.MultipartRequest;  

public class UploadServlet extends HttpServlet {  

public void doPost(HttpServletRequest request, HttpServletResponse response)  
    throws ServletException, IOException {  

response.setContentType("text/html");  
PrintWriter out = response.getWriter();  

MultipartRequest m=new MultipartRequest(request,"g:/newFolder");  
out.print("successfully uploaded");  
}  
}

any help would be much appreciated.

BalusC
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Masoud Mustamandi
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1 Answers1

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Simply replace the path in your java code with the value of the submitted variable:

String uploadPath = request.getParameter("path");
MultipartRequest m = new MultipartRequest(request, uploadPath);

Now your MultipartRequest gets initialized with the value from the hidden field.

Please note that this constructor of MultipartRequest limits the upload size to 1 megabyte.

f1sh
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