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Thread: File upload (jersey ) Using cURL

  1. #1
    Join Date
    May 2012
    Posts
    3

    Smile File upload (jersey ) Using cURL

    Hello all,

    I have the following program written that uploads a file to a specified location. what I'm trying to do is use cURL instead of an html webpage to upload the file.
    I tried doing the following
    Code:
    curl -F "FileUpload=@Test.txt" http://127.0.0.1/File_Upload/rest/upload
    that returned requested resource not found error, also what can I do to enable multiple uploads at a time ? Use Parts ?

    Any help would be appreciated

    Here is my program along with other info.
    Using Tomcat server 7.026
    Jersey 1.12

    Code:
    package com.file_upload;
    
    import java.io.File;
    import java.io.FileOutputStream;
    import java.io.IOException;
    import java.io.InputStream;
    import java.io.OutputStream;
    import javax.ws.rs.Consumes;
    import javax.ws.rs.POST;
    import javax.ws.rs.Path;
    import javax.ws.rs.core.MediaType;
    import javax.ws.rs.core.Response;
    import com.sun.jersey.core.header.FormDataContentDisposition;
    import com.sun.jersey.multipart.FormDataParam;
    
    @Path("/file")
    public class Upload{
    	@POST
    	@Path("/upload")
    	@Consumes(MediaType.MULTIPART_FORM_DATA)
    	public Response uploadFile(
    			@FormDataParam("file")InputStream uploadedInputStream,
    			@FormDataParam("file") FormDataContentDisposition fileDetail){
    	
    		
    			String uploadedFileLocation = "C:/Program Files (x86)/Apache Software Foundation/Tomcat 7.0/webapps/uploaded" +fileDetail.getFileName();
    			//save the file
    			writeToFile(uploadedInputStream,uploadedFileLocation);
    			
    			String output = "File uploaded to: " + uploadedFileLocation;
    		
    			return Response.status(200).entity(output).build();
    		}
    	//save uploaded file to a new location 
    	private void writeToFile(InputStream uploadedInputStream,
    			String uploadedFileLocation){
    		
    		try{
    			OutputStream out = new FileOutputStream(new File(uploadedFileLocation));
    			int read = 0;
    			byte[]bytes = new byte[1024];
    			
    			out = new FileOutputStream(new File(uploadedFileLocation));
    			while((read = uploadedInputStream.read(bytes)) != -1){
    				out.write(bytes,0,read);
    			}
    			out.flush();
    			out.close();
    		}
    		catch(IOException e){
    			e.printStackTrace();
    		}
    	}
    
    
    
    }
    web.xml
    Code:
    <?xml version="1.0" encoding="UTF-8"?>
    <web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0">
      <display-name>File_Upload</display-name>
      <servlet>
        <servlet-name>Upload</servlet-name>
        <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
        <init-param>
          <param-name>com.sun.jersey.config.property.packages</param-name>
          <param-value>com.file_upload</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
      </servlet>
      <servlet-mapping>
        <servlet-name>Upload</servlet-name>
        <url-pattern>/rest/*</url-pattern>
      </servlet-mapping>
    </web-app>

  2. #2
    Join Date
    Jun 2012
    Posts
    1
    It might be that these two @Path annotations:

    @Path("/file")
    public class Upload{
    @POST
    @Path("/upload")

    put your method at /file/upload, not just upload, so the curl command isn't specifying the resource.

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