英文:
Upload a file using Java 11 HttpClient to SpringBoot Server
问题
我的客户端上传方法:
public static void addPhoto(File photo) throws ParseException, IOException, InterruptedException {
    HttpClient client = HttpClient.newBuilder().build();
    HttpRequest request = HttpRequest.newBuilder()
            .header("Content-Type","image/jpg")
            .uri(URI.create(baseUrl + "data/addPhoto?date=4&category=temp&jwt="+jwt))
            .PUT(HttpRequest.BodyPublishers.ofFile(photo.toPath()))
            .build();
    client.send(request, HttpResponse.BodyHandlers.ofString());
}
我的Spring Boot接收文件的方法:
@PutMapping(path = "/addPhoto")
public @ResponseBody
boolean addPhoto(@RequestParam(name = "jwt") String jwt,
                 @RequestParam("file") MultipartFile file,
                 @RequestParam(name = "date") long date,
                 @RequestParam(name = "category") String category) {
    return crudService.addPhoto(jwt, date, file, category);
}
当前的错误:
2020-09-17 16:29:02.313 ERROR 8636 --- [nio-5000-exec-9] o.a.c.c.C.[.[.[/].[dispatcherServlet]    : Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exception [Request processing failed; nested exception is org.springframework.web.multipart.MultipartException: Current request is not a multipart request] with root cause
我应该添加什么样的标头以确保我的Spring Boot服务器可以无错误地接收文件?
英文:
My client upload method:
public static void addPhoto(File photo) throws ParseException, IOException, InterruptedException {
    HttpClient client = HttpClient.newBuilder().build();
    HttpRequest request = HttpRequest.newBuilder()
            .header("Content-Type","image/jpg")
            .uri(URI.create(baseUrl + "data/addPhoto?date=4&category=temp&jwt="+jwt))
            .PUT(HttpRequest.BodyPublishers.ofFile(photo.toPath()))
            .build();
    client.send(request, HttpResponse.BodyHandlers.ofString());
}
My Spring Boot method that receives the file:
@PutMapping(path = "/addPhoto")
public @ResponseBody
boolean addPhoto(@RequestParam(name = "jwt") String jwt,
                 @RequestParam("file") MultipartFile file,
                 @RequestParam(name = "date") long date,
                 @RequestParam(name = "category") String category) {
    return crudService.addPhoto(jwt, date, file, category);
}
The current error:
2020-09-17 16:29:02.313 ERROR 8636 --- [nio-5000-exec-9] o.a.c.c.C.[.[.[/].[dispatcherServlet]    : Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exception [Request processing failed; nested exception is org.springframework.web.multipart.MultipartException: Current request is not a multipart request] with root cause
What kind of headers can I add to ensure my Spring Boot server receives the file without errors?
答案1
得分: 3
MultipartException: Current request is not a multipart request
这是在告诉你出了什么问题。
在你的代码中:.PUT(HttpRequest.BodyPublishers.ofFile(photo.toPath())),你正在进行一个带有文件字节数组的PUT请求,将其放在了请求体中。但是在你的服务器端,你却期望它是一个MultipartFile(多部分文件)。MultipartFile是一个在POST请求体中附带额外数据的上传文件表示。https://en.wikipedia.org/wiki/MIME#Multipart_messages
你可以简单地按照以下方式上传你的文件:
参考:https://ganeshtiwaridotcomdotnp.blogspot.com/2020/09/java-httpclient-tutorial-with-file.html
在请求中发送文件名:
.uri(URI.create("http://localhost:8085/addPhoto?fileName=" + photo.getName()))
在RequestBody中接收字节数组,并在RequestParam中接收文件名:
@PostMapping(path = "/addPhoto")
public void addPhoto(@RequestBody byte[] barr,
                     @RequestParam(name = "fileName") String fileName) throws Exception {
    System.out.println(" received file " + fileName + " length " + barr.length);
    try (OutputStream os = new FileOutputStream(new File("UPL" + fileName))) {
        os.write(barr);
    }
}
如果你必须使用MultipartFile,那么你可以类似于以下方式处理:
- https://golb.hplar.ch/2019/01/java-11-http-client.html#upload-with-multipart
 - 或者 https://github.com/mizosoft/methanol#multipart-bodies
 
英文:
MultipartException: Current request is not a multipart request
This is telling you what's wrong.
In your code: .PUT(HttpRequest.BodyPublishers.ofFile(photo.toPath())), you are doing a PUT request with file's byte array in BODY. But in your server, you are expecting it as MultipartFile. MultipartFile is a representation of uploaded file with additional data in the POST request body. https://en.wikipedia.org/wiki/MIME#Multipart_messages
You can simply do the following to upload your file:
Ref: https://ganeshtiwaridotcomdotnp.blogspot.com/2020/09/java-httpclient-tutorial-with-file.html
Send filename in request:
.uri(URI.create("http://localhost:8085/addPhoto?fileName=" + photo.getName()))
Receive byte array in RequestBody and fileName in RequestParam
@PostMapping(path = "/addPhoto")
public void addPhoto(@RequestBody byte[] barr,
                     @RequestParam(name = "fileName") String fileName) throws Exception {
    System.out.println(" received file " + fileName + " length " + barr.length);
    try (OutputStream os = new FileOutputStream(new File("UPL" + fileName))) {
        os.write(barr);
    }
}
If you must use MultipartFile then you can do sth similar to:
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