如何提供昂首阔步的注释MultipartFormDataInput在RestEasy与QUUKUS

Question

在使用RestEasy框架处理Quarkus时，我可以使用MultipartFormDataInput上传文件。此功能按预期工作，但我无法为swagger UI提供适当的OpenAPI注释。我尝试了多种选择和组合，但没有结果。请帮帮我。我在下面提供示例html" target="_blank">代码。

@Operation(summary = "Upload a single file", description = "Upload a single file")
    @APIResponses({
            @APIResponse(responseCode = "200", description = "Upload file successfully"),
            @APIResponse(name = "500", responseCode = "500", description = "Internal service error") })
    @RequestBody(content = @Content(
            mediaType = MediaType.MULTIPART_FORM_DATA,
            schema = @Schema(type = SchemaType.STRING, format = "binary"),
            encoding = @Encoding(name = "attachment", contentType = "application/octet-stream")))
    @POST
    @Path("/singleFile")
    @Consumes(MediaType.MULTIPART_FORM_DATA)
    @Produces(MediaType.TEXT_PLAIN)
    public Response handleFileUpload(@MultipartForm MultipartFormDataInput input) {
        String fileName = null;

        Map<String, List<InputPart>> uploadForm = input.getFormDataMap();
        // Get file data to save
        List<InputPart> inputParts = uploadForm.get("attachment");
        for (InputPart inputPart : inputParts) {
            try {
                MultivaluedMap<String, String> header = inputPart.getHeaders();
                fileName = getFileName(header);
                InputStream inputStream = inputPart.getBody(InputStream.class, null);
                byte[] bytes = IOUtils.toByteArray(inputStream);
                File customDir = new File(UPLOAD_DIR);
                if (!customDir.exists()) {
                    customDir.mkdir();
                }
                fileName = customDir.getCanonicalPath() + File.separator + fileName;
                Files.write(Paths.get(fileName), bytes, StandardOpenOption.CREATE);
                return Response.status(200).entity("Uploaded file name : " + fileName).build();
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
        return Response.status(200).entity("Uploaded file name : " + fileName).build();
    }

我还提到了以下链接。

https://community.smartbear.com/t5/Swagger-Open-Source-Tools/How-to-swagger-annotate-multipart-form-data-with-resteasy/td-p/178776

https://github.com/swagger-api/swagger-core/issues/3050

如果我使用@Schema（type=SchemaType.STRING，format=“binary”）和@PartType（MediaType.APPLICATION\u OCTET\u STREAM）注释创建一个名为MultipartBody的单独类，我就能够生成swagger UI。但我的要求是只使用MultipartFormDataInput。

Answer 1

您几乎做到了：）只需在RequestBody/Schema中使用一个专用类，并告诉OpenAPI忽略方法的参数。

@POST
@Consumes(MediaType.MULTIPART_FORM_DATA)
@RequestBody(content = @Content(mediaType = MediaType.MULTIPART_FORM_DATA,
    schema = @Schema(implementation = MultipartBody.class))
)
@Operation(operationId = "uploadFile")
public Response uploadFile(@Parameter(hidden = true) MultipartFormDataInput input) {
//... your logic here
}

需要注意的两件事：@Parameter（隐藏=true）告诉Smallrye OpenAPI在生成模式模型时不要考虑您的MultipartFormDataInput。然后你需要告诉显式描述模式，使用@recestBody，其中MultipartBody是一个类，它描述了你所有的输入参数（你可以在那里添加更多的参数，如果你想例如传递其他道具和文件负载）

public class MultipartBody {

    @FormParam("file")
    @Schema(type = SchemaType.STRING, format = "binary", description = "file data")
    public String file;


    @FormParam("fileName")
    @PartType(MediaType.TEXT_PLAIN)
    public String fileName;
}

只要确保MultipartBody中的@FormParam注释字段与您期望在MultipartFormDataInput中找到的，，部分相匹配-例如，在您的情况下，文件属性应该具有@FormParam（附件）