Spring MVC save uploaded MultipartFile to specific folder

Question

I want to save uploaded images to a specific folder in a Spring 3 MVC application deployed on Tomcat

My problem is that I cannot save the uploaded images files to the host where the appliciation is running.

Here is what I tried:

private void saveFile(MultipartFile multipartFile, int id) throws Exception {
    String destination = "/images/" + id + "/"  + multipartFile.getOriginalFilename();
    File file = new File(destination);
    multipartFile.transferTo(file);
}

Result: FileNotFoundException - Yes sure, I do want create this file!?!

I tried it using the context.getRealPath or getResources("destination"), but without any success.

How can I create a new file in a specific folder of my app with the content of my multipart file?

What errors did you get when you tried context.getRealPath or getResources("destination") see http://stackoverflow.com/questions/2970643/create-a-file-and-store-in-java-web-application-folder — Rob Kielty, Jun 01 '12 at 10:41
Ah okay thanks, getRealPath didn't work, because it returns null if the app got deployed into a war file and was not extracted! Its working outside of the war container! But are there any options to get the path to a file inside of a war? — Alexander, Jun 01 '12 at 11:37
I wouldn't say so. It doesn't really make sense to add files into a deployed war. Is that what you want to do? — Rob Kielty, Jun 01 '12 at 11:56
No not really, but I read that some webservers do not extract the war file so that I thought to make the above example possible, there must be a way to write in that war directory. Therfore getRealPath is not the proper method cause it'll return null. Fortunately it is working on my tomcat, but just to briden my horizont I would like to know if there is another solution — Alexander, Jun 02 '12 at 09:02
FileNotFoundException means you don't have folder /images/{id}/, you should create folder first. — teik, Apr 20 '21 at 06:52

score 43 · Answer 1 · edited Oct 10 '14 at 12:19

43

This code will surely help you.

String filePath = request.getServletContext().getRealPath("/"); 
multipartFile.transferTo(new File(filePath));

edited Oct 10 '14 at 12:19

vault

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answered Dec 12 '13 at 13:19

Ravi Maroju

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but without using the http servlet request how can we get the path..? – Jijesh Kumar Apr 23 '15 at 13:20
There are many way: You can use request or session to get the real path. Thanks Ravi. M – Phuong Oct 05 '15 at 04:24
@JijeshKumar easiest way to send a parameter of type HttpServletRequest in the controller – Long tran Aug 26 '23 at 04:21

Yuliia Ashomok · Answer 2 · 2015-12-08T18:00:24.313

Let's create the uploads directory in webapp and save files in webapp/uploads:

@RestController
public class GreetingController {

    private final static Logger log = LoggerFactory.getLogger(GreetingController.class);

    @Autowired
    private HttpServletRequest request;


    @RequestMapping(value = "/uploadfile", method = RequestMethod.POST)
        public
        @ResponseBody
        ResponseEntity handleFileUpload(@RequestParam("file") MultipartFile file) {
            if (!file.isEmpty()) {
                try {
                    String uploadsDir = "/uploads/";
                    String realPathtoUploads =  request.getServletContext().getRealPath(uploadsDir);
                    if(! new File(realPathtoUploads).exists())
                    {
                        new File(realPathtoUploads).mkdir();
                    }

                    log.info("realPathtoUploads = {}", realPathtoUploads);


                    String orgName = file.getOriginalFilename();
                    String filePath = realPathtoUploads + orgName;
                    File dest = new File(filePath);
                    file.transferTo(dest);

the code
String realPathtoUploads = request.getServletContext().getRealPath(uploadsDir);

returns me next path if I run the app from Idea IDE
C:\Users\Iuliia\IdeaProjects\ENumbersBackend\src\main\webapp\uploads\

and next path if I make .war and run it under Tomcat:
D:\Programs_Files\apache-tomcat-8.0.27\webapps\enumbservice-0.2.0\uploads\

My project structure:

The last line of the code ' file.transferTo(dest);' gives me IO exception and the error message is "The system cannot find the path specified". Any idea why this is happening? — MR AND, Apr 17 '17 at 20:20
@AND, are you sure the path is exist in your file system? You need to create directories firstly (manually or programmatically). — Yuliia Ashomok, Apr 18 '17 at 07:22
This was very helpful. While its effectively the same answer as Ravi's, this is a more thorough answer. This answer is still valid in Spring Boot 1.5.10 — Stephan, Mar 26 '18 at 20:39

Diego Victor de Jesus · Answer 3 · 2021-06-16T13:48:05.293

13

You can get the inputStream from multipartfile and copy it to any directory you want.

public String write(MultipartFile file, String fileType) throws IOException {
    String date = LocalDateTime.now().format(DateTimeFormatter.ofPattern("yyMMddHHmmss-"));
    String fileName = date + file.getOriginalFilename();
    
    // folderPath here is /sismed/temp/exames
    String folderPath = SismedConstants.TEMP_DIR + fileType;
    String filePath = folderPath + File.separator + fileName;
    
    // Copies Spring's multipartfile inputStream to /sismed/temp/exames (absolute path)
    Files.copy(file.getInputStream(), Paths.get(filePath), StandardCopyOption.REPLACE_EXISTING);
    return filePath;
}

This works for both Linux and Windows.

edited Jun 16 '21 at 13:48

answered Feb 05 '19 at 06:37

Diego Victor de Jesus

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good example. just replaced "/" with File.separator – Simone Celia Aug 26 '20 at 15:31
But not working in Mac – Hafiz Hamza Apr 04 '22 at 07:14

score 1 · Answer 4 · answered Jan 26 '16 at 23:20

I saw a spring 3 example using xml configuration (note this does not wok for spring 4.2.*): http://www.devmanuals.com/tutorials/java/spring/spring3/mvc/spring3-mvc-upload-file.html `

<bean id="multipartResolver"
class="org.springframework.web.multipart.commons.CommonsMultipartResolver">
<property name="maxUploadSize" value="100000" />
<property name="uploadTempDir" ref="uploadDirResource" />
</bean>

<bean id="uploadDirResource" class="org.springframework.core.io.FileSystemResource">
<constructor-arg>
<value>C:/test111</value>
</constructor-arg>
</bean>

score 0 · Answer 5 · answered Dec 13 '16 at 05:20

0

String ApplicationPath = 
        ContextLoader.getCurrentWebApplicationContext().getServletContext().getRealPath("");

This is how to get the real path of App in Spring (without using response, session ...)

answered Dec 13 '16 at 05:20

2

you shouldn't use `getRealPath()` to save files.. see [this answer](https://stackoverflow.com/a/12160863/3891638) – cozyconemotel May 29 '17 at 03:28

score 0 · Answer 6 · edited Jun 07 '18 at 16:14

0

The following worked for me on ubuntu:

String filePath = request.getServletContext().getRealPath("/");
File f1 = new File(filePath+"/"+multipartFile.getOriginalFilename());
multipartFile.transferTo(f1);

edited Jun 07 '18 at 16:14

Alan Birtles

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answered Jun 07 '18 at 14:19

Vijay

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Spring MVC save uploaded MultipartFile to specific folder

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