首页 \ 问答 \ 使用单个请求发送JSON和图像。(Send JSON and Image with single request. Angular + Spring Boot)

使用单个请求发送JSON和图像。(Send JSON and Image with single request. Angular + Spring Boot)

弹簧控制器

@PostMapping(
    value = "/post",
    produces = MediaType.APPLICATION_JSON_VALUE,
    consumes = {MediaType.APPLICATION_JSON_VALUE, MediaType.MULTIPART_FORM_DATA_VALUE}
)
public ResponseEntity<User> handleFileUpload(@RequestParam("user") User user, @RequestPart("file") MultipartFile file) {
    // do something with User and file
    return ResponseEntity.ok().build();
}

角度服务

@Injectable()
export class UploadFileService {

  constructor(private http: HttpClient) { }
  pushFileToStorage(file: File): Observable<HttpEvent<{}>> {
    let formdata: FormData = new FormData();
    formdata.append('file', file);
    formdata.append('user', JSON.stringify(new User('John', 12)))

    const req = new HttpRequest('POST', '/post', formdata, {
      reportProgress: true,
    });

    return this.http.request(req);
  }
}

当我尝试发送请求时,我得到500 Internal Server Error.

这是一个请求标题

POST /post HTTP/1.1
Host: localhost:4200
Connection: keep-alive
Content-Length: 152881
Accept: application/json, text/plain, */*
Content-Type: multipart/form-data; boundary=----WebKitFormBoundarydaQb5yaWw2xu1V9r
Accept-Encoding: gzip, deflate, br
Accept-Language: en-US,en;q=0.9

请求负载

------WebKitFormBoundarydaQb5yaWw2xu1V9r
Content-Disposition: form-data; name="file"; filename="Screen Shot 2017-10-24 at 8.49.13 PM.png"
Content-Type: image/png


------WebKitFormBoundarydaQb5yaWw2xu1V9r
Content-Disposition: form-data; name="user"

{"name":"John","age":12}
------WebKitFormBoundarydaQb5yaWw2xu1V9r--

注意:如果在Spring rest控制器中,我将参数类型User更改为String,它将起作用。

问题:如何从Angular发送请求,以便在Spring上可以获取User和MultipartFile,而不是String。


Spring Rest Controller

@PostMapping(
    value = "/post",
    produces = MediaType.APPLICATION_JSON_VALUE,
    consumes = {MediaType.APPLICATION_JSON_VALUE, MediaType.MULTIPART_FORM_DATA_VALUE}
)
public ResponseEntity<User> handleFileUpload(@RequestParam("user") User user, @RequestPart("file") MultipartFile file) {
    // do something with User and file
    return ResponseEntity.ok().build();
}

Angular Service

@Injectable()
export class UploadFileService {

  constructor(private http: HttpClient) { }
  pushFileToStorage(file: File): Observable<HttpEvent<{}>> {
    let formdata: FormData = new FormData();
    formdata.append('file', file);
    formdata.append('user', JSON.stringify(new User('John', 12)))

    const req = new HttpRequest('POST', '/post', formdata, {
      reportProgress: true,
    });

    return this.http.request(req);
  }
}

When I try to send the request I get 500 Internal Server Error.

Here's a request header

POST /post HTTP/1.1
Host: localhost:4200
Connection: keep-alive
Content-Length: 152881
Accept: application/json, text/plain, */*
Content-Type: multipart/form-data; boundary=----WebKitFormBoundarydaQb5yaWw2xu1V9r
Accept-Encoding: gzip, deflate, br
Accept-Language: en-US,en;q=0.9

Request payload

------WebKitFormBoundarydaQb5yaWw2xu1V9r
Content-Disposition: form-data; name="file"; filename="Screen Shot 2017-10-24 at 8.49.13 PM.png"
Content-Type: image/png


------WebKitFormBoundarydaQb5yaWw2xu1V9r
Content-Disposition: form-data; name="user"

{"name":"John","age":12}
------WebKitFormBoundarydaQb5yaWw2xu1V9r--

Note: If in Spring rest controller I change parameter type User to String it works.

Question: How to send request from Angular so that on Spring I can get User and MultipartFile, instead of String.


原文:https://stackoverflow.com/questions/47756044
更新时间:2023-06-11 18:06

最满意答案

在SQL Server中

SELECT  PK,
        account, 
        sum_value AS [value]
FROM (
    SELECT  PK,
            account, 
            SUM([value]) as sum_value,
            ROW_NUMBER() OVER (PARTITION BY PK ORDER BY SUM([value]) DESC) as rn
    FROM [table]
    GROUP BY account, PK
) as p
WHERE rn = 1

输出:

PK  account     value
1   40010101    630
2   40010569    300

In SQL Server

SELECT  PK,
        account, 
        sum_value AS [value]
FROM (
    SELECT  PK,
            account, 
            SUM([value]) as sum_value,
            ROW_NUMBER() OVER (PARTITION BY PK ORDER BY SUM([value]) DESC) as rn
    FROM [table]
    GROUP BY account, PK
) as p
WHERE rn = 1

Output:

PK  account     value
1   40010101    630
2   40010569    300

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