tutorial for file upload in spring

View Answers

December 19, 2008 at 12:22 AM

Hi friend,

Code to help in solving the problem :

FileUploadController.java file

package example;

import org.springframework.web.servlet.mvc.SimpleFormController;
import org.springframework.web.servlet.ModelAndView;
import org.springframework.web.servlet.view.RedirectView;

import org.springframework.web.multipart.MultipartHttpServletRequest;
import org.springframework.web.multipart.commons.CommonsMultipartFile;
import org.springframework.web.bind.ServletRequestDataBinder;
import org.springframework.web.multipart.support.ByteArrayMultipartFileEditor;
import org.springframework.util.FileCopyUtils;

import java.io.File;
import java.io.IOException;
import java.util.Map;
import java.util.HashMap;
import java.net.BindException;

import javax.servlet.ServletException;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import org.apache.commons.logging.Log;
import org.apache.commons.logging.LogFactory;

import service.PriceIncrease;

import example.FileUploadBean;

public class FileUploadController extends SimpleFormController
{

protected final Log logger = LogFactory.getLog(getClass());

protected ModelAndView onSubmit(HttpServletRequest request, HttpServletResponse response,
Object command, BindException errors) throws ServletException, IOException , Exception {

// cast the bean
FileUploadBean bean = (FileUploadBean) command;

// let's see if there's content there
byte[] file = bean.getFile();
if (file == null) {
// hmm, that's strange, the user did not upload anything
}

// well, let's do nothing with the bean for now and return
return onSubmit(request, response, command, errors);
}

protected void initBinder(HttpServletRequest request, ServletRequestDataBinder binder)
throws ServletException {
// to actually be able to convert Multipart instance to byte[]
// we have to register a custom editor
binder.registerCustomEditor(byte[].class, new ByteArrayMultipartFileEditor());
// now Spring knows how to handle multipart object and convert them
}

}

package example;

import java.io.Serializable;

/*public class FileUploadBean implements Serializable {

private static final long serialVersionUID = 1L;

private byte[] file;

public FileUploadBean() {
super();
}

public FileUploadBean(byte[] file) {
super();
setFile(file);
}

public void setFile(byte[] file) {
this.file = file;
}

public byte[] getFile() {
return file;
}

}*/

My FileUploadBean.java file

import org.springframework.web.multipart.MultipartFile;

public class FileUploadBean {

private byte[] file;

public void setFile(byte[] file) {
this.file = file;
}

public byte[] getFile() {
return file;
}
}

December 19, 2008 at 12:23 AM

fileuploadform.jsp file

<%@ taglib prefix="form" uri="http://www.springframework.org/tags/form"; %>
<html>
<head>
<title>Upload a file please</title>
</head>
<body>
<h1>Please upload a file</h1>
<form method="post" action="upload.form" enctype="multipart/form-data">
<input type="file" name="file"/>
<input type="submit"/>
</form>


</body>
</html>

the problem is when confire this using in springapp-servelt.xml like below i am getting the requested resource not found 404 error (when i start the tomcat5.50

the bean configuration is :

<bean id="multipartResolver"
class="org.springframework.web.multipart.commons.CommonsMultipartResolver"/>

<bean id="urlMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
<property name="mappings">

<value>
/upload.form=fileUploadController
</value>
</property>
</bean>

<bean id="fileUploadController" class="example.FileUploadController">
<property name="commandClass" value="example.FileUploadBean"/>
<property name="formView" value="fileuploadform"/>
<property name="successView" value="confirmation"/>
</bean>

Thanks

December 19, 2008 at 12:34 AM

thanks for the example dude. But this example uses Simple Form Controller and beans . I want to use a Controller which implements just the Controller interface