CODEWITHSUNDEEP
spring-boot
OraBo
OraBo

Reputation: 73

Change the name of MultipartFile

I'm trying to change the name of MultipartFile.

I'm using MultipartFile on my controller to call rest service:

@PostMapping("/post")
public ResponseEntity<String> handleFileUpload(@RequestParam("file") MultipartFile file)
{
    ...
}

Have you please any idea about changing the OriginalFilename of the uploaded file ?.

Big thanks.

Upvotes: 4

Views: 13640

Answers (3)

thienchidh
thienchidh

Reputation: 1

if you "forward" your file to another server, you can try the following code:

import jakarta.validation.constraints.NotNull;
import org.springframework.core.io.Resource;
import org.springframework.lang.Nullable;

import java.io.File;
import java.io.IOException;
import java.io.InputStream;
import java.net.URI;
import java.net.URL;
import java.nio.channels.ReadableByteChannel;
import java.nio.charset.Charset;

public class MultipartFileResourceNameDecorator implements Resource {
    private final Resource resource;
    private final String newFileName;


    public MultipartFileResourceNameDecorator(@NotNull Resource resource, @NotNull String newFileName) {
        this.resource = resource;
        this.newFileName = newFileName;
    }

    @Override
    @Nullable
    public String getFilename() {
        return newFileName;
    }

    // delegate another methods
}

Caller:

private String sendByFile(Resource fileResource, String newFileName) {
    var resource = new MultipartFileResourceNameDecorator(fileResource, newFileName);
    log.info("Upload file: {}", resource.getFilename());
    try {
        HttpHeaders headers = new HttpHeaders();
        headers.setContentType(MediaType.MULTIPART_FORM_DATA);
        // Create the request body with the file
        var body = new LinkedMultiValueMap<String, Object>();
        body.add("file", resource);
        // Create the HttpEntity
        var entity = new HttpEntity<>(body, headers);
        // Make the POST request
        var response = restTemplate.postForEntity(urlUpload, entity, String.class);
        
        // Your logic here with response
        
    } catch (Exception e) {
        log.error("Upload fail", e);
        return null;
    }
}

Upvotes: 0

Romil Patel
Romil Patel

Reputation: 13777

You can achieve renaming of a file as below. To verify go the uploadDir and you will have a file with "renameTest".

You can append clientId + uploadTime to a filename to avoid the same filenames in database

@PostMapping(value = "/post")
    public  String renameMultipartFile(@RequestParam("file") MultipartFile file) {
        String uploadDir = "yourPath";
        String filename = "renameTest";
        Path saveTO = Paths.get(uploadDir + filename);
        try {
            try {
                Files.copy(file.getInputStream(), saveTO);
            } catch (Exception e) {
                e.printStackTrace();
                throw new RuntimeException("Error : " + e.getMessage());
            }
            return "File uploaded successfully";
        } catch (Exception e) {
            e.printStackTrace();
            return "Error : " + e.getMessage();
        }
    }

Upvotes: 0

Johna
Johna

Reputation: 1894

You can try the following code.

@PostMapping("/post")
public ResponseEntity<String> handleFileUpload(@RequestParam("file") MultipartFile file)
{
    try {
        String filename = "random_filename.pdf"; // Give a random filename here.
        byte[] bytes = file.getBytes();
        String insPath = <DIRECTORY PATH> + filename // Directory path where you want to save ;
        Files.write(Paths.get(insPath), bytes);

        return ResponseEntity.ok(filename);
    }

    catch (IOException e) {
        // Handle exception here 
    }
}

You have to remember to add a random string to the file name. If you just hard code the file name, every time you upload a file, the previous file will be replaced.

Upvotes: 2

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