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()
The HTTP 403 Forbidden response status code indicates that the server understands the request but refuses to authorize it.
The simple answer is; “You need to be given the correct access”. Without being given the correct access you'd technically be hacking the server, as it is specifically set up to restrict said access.
This is covered in the CSRF - Multipart (File Upload) section of the Spring Security reference. You have two options:
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.
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>
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