Uploading File Returns 403 Error - Spring MVC

Question

In my Spring MVC project I am trying to upload a file via a simple form.

HTML Form:

<form method="POST" enctype="multipart/form-data" action="/upload">
    <label>Select File</label> 
    <input type="file" name="file"/>
</form>

My Controller:

@Controller
public class FileController {
    @RequestMapping(value="/upload", method=RequestMethod.POST)
        public @ResponseBody String handleFileUpload(
                @RequestParam("name") String name,
                @RequestParam("file") MultipartFile file){
            if (!file.isEmpty()) {
                try {
                    //do stuff
                } catch (Exception e) {
                    return "You failed to upload " + name + " => " + e.getMessage();
                }
            } else {
                return "You failed to upload " + name + " because the file was empty.";
            }
        }
}

Security Config:

@Configuration
@EnableWebMvcSecurity
public class WebSecurityConfig extends WebSecurityConfigurerAdapter {

    @Override
    protected void configure(HttpSecurity http) throws Exception {
        http.authorizeRequests()
            .antMatchers("/upload").permitAll()
            .and()
               .exceptionHandling().accessDeniedPage("/403")
    }
}

However I get a 403: Forbidden error and am redirected to my 403.html view every time

So far I've tried specifying the MultipartFilter before the Spring Security filter is initialized in a separate class, but no luck

public class SecurityApplicationInitializer extends AbstractSecurityWebApplicationInitializer {

    @Override
    protected void beforeSpringSecurityFilterChain(ServletContext servletContext) {
        insertFilters(servletContext, new MultipartFilter());
    }
}

Any ideas?

UPDATE: Including my WebAppInitializer

@Configuration
@Import({ WebSecurityConfig.class })
public class WebAppInitializer implements WebApplicationInitializer {

    @Override
    public void onStartup(ServletContext servletContext) throws ServletException {
        System.out.println(":::Starting My App:::");
        AnnotationConfigWebApplicationContext context = new AnnotationConfigWebApplicationContext();
        context.register(WebMVCConfig.class);
        context.setServletContext(servletContext);
        context.setConfigLocation("com.myApp.configuration");
    }

}

I have a list of servlet request attributes that returns the following with the 403 error:

javax.servlet.forward.request_uri
javax.servlet.forward.context_path
javax.servlet.forward.servlet_path
__spring_security_scpf_applied
org.springframework.web.servlet.DispatcherServlet.THEME_SOURCE
SPRING_SECURITY_403_EXCEPTION
org.springframework.web.servlet.DispatcherServlet.THEME_RESOLVER
springMacroRequestContext
themes
thymeleafEvaluationContext
org.springframework.security.web.FilterChainProxy.APPLIED
_csrf
org.springframework.security.web.context.request.async.WebAsyncManagerIntegrationFilter.FILTERED
org.springframework.security.web.csrf.CsrfFilter@539743f9.FILTERED
beans
springRequestContext
org.springframework.web.servlet.HandlerMapping.introspectTypeLevelMapping
org.springframework.web.servlet.DispatcherServlet.FLASH_MAP_MANAGER
org.springframework.web.servlet.DispatcherServlet.CONTEXT
org.springframework.core.convert.ConversionService
execInfo
org.springframework.web.servlet.HandlerMapping.pathWithinHandlerMapping
org.springframework.web.context.request.async.WebAsyncManager.WEB_ASYNC_MANAGER
org.springframework.web.servlet.resource.ResourceUrlProvider
org.springframework.web.servlet.DispatcherServlet.OUTPUT_FLASH_MAP
org.springframework.web.servlet.HandlerMapping.bestMatchingPattern
org.springframework.security.web.csrf.CsrfToken
org.springframework.web.servlet.DispatcherServlet.LOCALE_RESOLVER

Update #2: This is surely a CSRF issue; when I include the following in my WebSecurityConfig I get no 403

.csrf().disable()

Answer 1

For me even with the csrf().disable() its didn't work. Once i disable it i got 200 but the file wasn't uploaded and i didn't see any error. Once i set the debug flag logging.level.org.springframework.web: DEBUG i saw the root cause:

[org.springframework.web.multipart.MultipartException: Failed to parse multipart servlet request; nested exception is java.io.IOException: The temporary upload location [/target/tomcat/work/Tomcat/localhost/ROOT] is not valid]","exception":""}

I tried setting the location on application.yaml: spring.servlet.http.multipart.location: "/tmp" but it didn't change the location so what i did is adding the below code and it did the trick:

@Bean

MultipartConfigElement multipartConfigElement() {
    MultipartConfigFactory factory = new MultipartConfigFactory();
    String location = "/tmp";
    LOGGER.debug("Multipart location file:" + location);
    File tmpFile = new File(location);
    if (!tmpFile.exists()) {
        tmpFile.mkdirs();
    }
    factory.setLocation(location);
    return factory.createMultipartConfig();
}

Answer 2

The solution work for me is to disable csrf() in my WebSecurityConfig like :

 @EnableWebSecurity
 @Configuration
public class WebSecurityConfig extends WebSecurityConfigurerAdapter {

    @Override
    protected void configure(HttpSecurity http) throws Exception {
        http.cors().and().csrf().disable();
    }
}

And now i can simply upload multipart file :)

Answer 3

The fast solution for me was the following

<%@ taglib uri="http://java.sun.com/jsp/jstl/core" prefix="c" %>
<%@ page session="false" %>
<html>
<head>
<title>Upload File Request Page</title>
</head>
<body>
    <form method="POST" action="file/uploadFile?${_csrf.parameterName}=${_csrf.token}" enctype="multipart/form-data">
        File to upload: <input type="file" name="file"><br /> 
        Name: <input type="text" name="name"><br /> <br /> 
        <input type="submit" value="Upload"> Press here to upload the file!
    </form> 
</body>
</html>

The controller code is the following:

package com.student.controller;

import java.io.BufferedOutputStream;
import java.io.File;
import java.io.FileOutputStream;
import java.security.Principal;
import javax.servlet.http.HttpServletRequest;

import org.springframework.stereotype.Controller;
import org.springframework.ui.ModelMap;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.bind.annotation.RequestParam;
import org.springframework.web.bind.annotation.ResponseBody;
import org.springframework.web.multipart.MultipartFile;

@Controller
@RequestMapping("/file")
public class FileUploadController {

    @RequestMapping(value = "", method = RequestMethod.GET)
    public String index(ModelMap modelMap,Principal principal,HttpServletRequest request) {
        return "uploadfile";
    }

    @RequestMapping(value = "/uploadFile", method = RequestMethod.POST)
    public @ResponseBody String uploadFileHandler(@RequestParam("name") String name, @RequestParam("file") MultipartFile file) {

        if (!file.isEmpty()) {
            try {
                byte[] bytes = file.getBytes();

                // Creating the directory to store file
                String rootPath = System.getProperty("catalina.home");
                File dir = new File(rootPath + File.separator + "tmpFiles");
                if (!dir.exists())
                    dir.mkdirs();

                // Create the file on server
                File serverFile = new File(dir.getAbsolutePath()
                        + File.separator + name);
                BufferedOutputStream stream = new BufferedOutputStream(
                        new FileOutputStream(serverFile));
                stream.write(bytes);
                stream.close();



                return "You successfully uploaded file=" + rootPath+name;
            } catch (Exception e) {
                return "You failed to upload " + name + " => " + e.getMessage();
            }
        } else {
            return "You failed to upload " + name
                    + " because the file was empty.";
        }
    }

}

I added the following code in spring dispatcher file

<!-- upload files -->
    <bean id="multipartResolver" class="org.springframework.web.multipart.commons.CommonsMultipartResolver">

         <!-- setting maximum upload size -->
        <property name="maxUploadSize" value="100000" />

    </bean>

Answer 4

This is covered in the CSRF - Multipart (File Upload) section of the Spring Security reference. You have two options:

• Placing MultipartFilter before Spring Security
• Include CSRF token in action

Placing MultipartFilter before Spring Security

The first option is to ensure that the MultipartFilter is specified before the Spring Security filter. Specifying the MultipartFilter before the Spring Security filter means that there is no authorization for invoking the MultipartFilter which means anyone can place temporary files on your server. However, only authorized users will be able to submit a File that is processed by your application. In general, this is the recommended approach because the temporary file upload should have a negligble impact on most servers.

To ensure MultipartFilter is specified before the Spring Security filter with java configuration, users can override beforeSpringSecurityFilterChain as shown below:

public class SecurityApplicationInitializer extends AbstractSecurityWebApplicationInitializer {

    @Override
    protected void beforeSpringSecurityFilterChain(ServletContext servletContext) {
        insertFilters(servletContext, new MultipartFilter());
    }
}

To ensure MultipartFilter is specified before the Spring Security filter with XML configuration, users can ensure the element of the MultipartFilter is placed before the springSecurityFilterChain within the web.xml as shown below:

<filter>
    <filter-name>MultipartFilter</filter-name>
    <filter-class>org.springframework.web.multipart.support.MultipartFilter</filter-class>
</filter>
<filter>
    <filter-name>springSecurityFilterChain</filter-name>
    <filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
</filter>
<filter-mapping>
    <filter-name>MultipartFilter</filter-name>
    <url-pattern>/*</url-pattern>
</filter-mapping>
<filter-mapping>
    <filter-name>springSecurityFilterChain</filter-name>
    <url-pattern>/*</url-pattern>
</filter-mapping>

Include CSRF token in action

If allowing unauthorized users to upload temporariy files is not acceptable, an alternative is to place the MultipartFilter after the Spring Security filter and include the CSRF as a query parameter in the action attribute of the form. An example with a jsp is shown below

<form action="./upload?${_csrf.parameterName}=${_csrf.token}" 
      method="post" 
      enctype="multipart/form-data">

The disadvantage to this approach is that query parameters can be leaked. More genearlly, it is considered best practice to place sensitive data within the body or headers to ensure it is not leaked. Additional information can be found in RFC 2616 Section 15.1.3 Encoding Sensitive Information in URI’s.