Spring Web反应客户端

Question

我试图使用Spring反应式WebClient将文件上传到Spring控制器。控制器非常简单，看起来像这样：

@PostMapping(value = "/upload", consumes = MediaType.MULTIPART_FORM_DATA_VALUE)
public ResponseEntity<String> uploadFile(
        @RequestParam("multipartFile") MultipartFile multipartFile,
        @RequestParam Map<String, Object> entityRequest
        ) {
    entityRequest.entrySet().forEach(System.out::println);
    System.out.println(multipartFile);
    return ResponseEntity.ok("OK");
}

当我使用这个控制器与cURL一切正常

curl -X POST http://localhost:8080/upload -H 'content-type: multipart/form-data;' -F fileName=test.txt -F randomKey=randomValue -F multipartFile=@document.pdf

multipartFile转到正确的参数，其他参数进入Map。

当我尝试从WebClient做同样的事情时，我被卡住了。我的代码如下所示：

    WebClient client = WebClient.builder().baseUrl("http://localhost:8080").build();

    MultiValueMap<String, Object> map = new LinkedMultiValueMap<>();
    map.set("multipartFile", new ByteArrayResource(Files.readAllBytes(Paths.get("/path/to/my/document.pdf"))));
    map.set("fileName", "test.txt");
    map.set("randomKey", "randomValue");
    String result = client.post()
            .uri("/upload")
            .contentType(MediaType.MULTIPART_FORM_DATA)
            .syncBody(map)
            .exchange()
            .flatMap(response -> response.bodyToMono(String.class))
            .flux()
            .blockFirst();
    System.out.println("RESULT: " + result);

这会导致400错误

{
  "timestamp":1510228507230,
  "status":400,
  "error":"Bad Request",
  "message":"Required request part 'multipartFile' is not present",
  "path":"/upload"
}

有人知道如何解决这个问题吗？

Answer 1

提供内容处置的更简单方法

MultipartBodyBuilder builder = new MultipartBodyBuilder();
String header = String.format("form-data; name=%s; filename=%s", "paramName", "fileName.pdf");
builder.part("paramName", new ByteArrayResource(<file in byte array>)).header("Content-Disposition", header);

// in the request use
webClient.post().body(BodyInserters.fromMultipartData(builder.build()))

Answer 2

您需要在文件部分包含一个文件名以成功上传，并结合< code>asyncPart()以避免缓冲所有文件内容，然后您可以编写如下代码:

WebClient client = WebClient.builder().baseUrl("http://localhost:8080").build();

Mono<String> result = client.post()
        .uri("/upload")
        .contentType(MediaType.MULTIPART_FORM_DATA)
        .body((outputMessage, context) ->
                Mono.defer(() -> {
                    MultipartBodyBuilder bodyBuilder = new MultipartBodyBuilder();

                    Flux<DataBuffer> data = DataBufferUtils.read(
                            Paths.get("/tmp/file.csv"), outputMessage.bufferFactory(), 4096);

                    bodyBuilder.asyncPart("file", data, DataBuffer.class)
                            .filename("filename.csv");

                    return BodyInserters.fromMultipartData(bodyBuilder.build())
                            .insert(outputMessage, context);
                }))
        .exchange()
        .flatMap(response -> response.bodyToMono(String.class));

System.out.println("RESULT: " + result.block());

Answer 3

所以我自己找到了解决办法。事实证明，Spring确实需要Content Disposition头来包含一个文件名，以便将上传序列化为控制器中的多部分文件。

为此，我必须创建一个支持设置文件名的ByteArrayResource子类

public class MultiPartResource extends ByteArrayResource {

  private String filename;

  public MultiPartResource(byte[] byteArray) {
    super(byteArray);
  }

  public MultiPartResource(byte[] byteArray, String filename) {
    super(byteArray);
    this.filename = filename;
  }

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

  public void setFilename(String filename) {
    this.filename = filename;
  }
}

然后可以在客户机中使用这些代码

WebClient client = WebClient.builder().baseUrl("http://localhost:8080").build();

MultiValueMap<String, Object> map = new LinkedMultiValueMap<>();

map.set("fileName", "test.txt");
map.set("randomKey", "randomValue");
ByteArrayResource resource = new MultiPartResource(Files.readAllBytes(Paths.get("/path/to/my/document.pdf")), "document.pdf");

String result = client.post()
        .uri("/upload")
        .contentType(MediaType.MULTIPART_FORM_DATA)
        .body(BodyInserters.fromMultipartData(map))
        .exchange()
        .flatMap(response -> response.bodyToMono(String.class))
        .flux()
        .blockFirst();
System.out.println("RESULT: " + result);