感谢@Eswar。这段代码非常适合我。我想在解决方案中添加一些东西:

我得到错误:java.io.IOException: RESTEASY007550:无法获得多部分的边界

为了解决这个错误，你应该删除“Content-Type”“multipart/form-data”。它解决了我的问题。

感谢@Eswar。这段代码非常适合我。我想在解决方案中添加一些东西:

我得到错误:java.io.IOException: RESTEASY007550:无法获得多部分的边界

为了解决这个错误，你应该删除“Content-Type”“multipart/form-data”。它解决了我的问题。

上传带有表单字段的图像

SaveFileWithData(article: ArticleModel,picture:File): Observable<ArticleModel> 
{

    let headers = new Headers();
    // headers.append('Content-Type', 'multipart/form-data');
    // headers.append('Accept', 'application/json');

let requestoptions = new RequestOptions({
  method: RequestMethod.Post,
  headers:headers
    });



let formData: FormData = new FormData();
if (picture != null || picture != undefined) {
  formData.append('files', picture, picture.name);
}
 formData.append("article",JSON.stringify(article));

return this.http.post("url",formData,requestoptions)
  .map((response: Response) => response.json() as ArticleModel);
} 

在我的情况下，我需要。net Web Api在c#

// POST: api/Articles
[ResponseType(typeof(Article))]
public async Task<IHttpActionResult> PostArticle()
{
    Article article = null;
    try
    {

        HttpPostedFile postedFile = null;
        var httpRequest = HttpContext.Current.Request;

        if (httpRequest.Files.Count == 1)
        {
            postedFile = httpRequest.Files[0];
            var filePath = HttpContext.Current.Server.MapPath("~/" + postedFile.FileName);
            postedFile.SaveAs(filePath);
        }
        var json = httpRequest.Form["article"];
         article = JsonConvert.DeserializeObject <Article>(json);

        if (!ModelState.IsValid)
        {
            return BadRequest(ModelState);
        }

        article.CreatedDate = DateTime.Now;
        article.CreatedBy = "Abbas";

        db.articles.Add(article);
        await db.SaveChangesAsync();
    }
    catch (Exception ex)
    {
        int a = 0;
    }
    return CreatedAtRoute("DefaultApi", new { id = article.Id }, article);
}

这是一个有用的教程，如何使用ng2-file-upload和不使用ng2-file-upload上传文件。

对我来说很有帮助。

目前，教程包含几个错误:

1-客户端应具有与服务器相同的上传url 在app.component.ts中更改行

const URL = 'http://localhost:8000/api/upload';

to

const URL = 'http://localhost:3000';

2-服务器发送响应为'text/html'，所以在app.component.ts更改

.post(URL, formData).map((res:Response) => res.json()).subscribe(
  //map the success function and alert the response
  (success) => {
    alert(success._body);
  },
  (error) => alert(error))

来

.post(URL, formData)  
.subscribe((success) => alert('success'), (error) => alert(error));

下面的代码以最简单的形式在Angular 6/7中工作

this.http.post("http://destinationurl.com/endpoint", fileFormData)
  .subscribe(response => {
    //handle response
  }, err => {
    //handle error
  });

下面是完整的实现

这个简单的解决方案对我很有效:file-upload.component.html

<div>
  <input type="file" #fileInput placeholder="Upload file..." />
  <button type="button" (click)="upload()">Upload</button>
</div>

然后在组件中直接使用XMLHttpRequest进行上传。

import { Component, OnInit, ViewChild } from '@angular/core';

@Component({
  selector: 'app-file-upload',
  templateUrl: './file-upload.component.html',
  styleUrls: ['./file-upload.component.css']
})
export class FileUploadComponent implements OnInit {

  @ViewChild('fileInput') fileInput;

  constructor() { }

  ngOnInit() {
  }

  private upload() {
    const fileBrowser = this.fileInput.nativeElement;
    if (fileBrowser.files && fileBrowser.files[0]) {
      const formData = new FormData();
      formData.append('files', fileBrowser.files[0]);
      const xhr = new XMLHttpRequest();
      xhr.open('POST', '/api/Data/UploadFiles', true);
      xhr.onload = function () {
        if (this['status'] === 200) {
            const responseText = this['responseText'];
            const files = JSON.parse(responseText);
            //todo: emit event
        } else {
          //todo: error handling
        }
      };
      xhr.send(formData);
    }
  }

}

如果使用dotnet core，参数名必须与from字段名匹配。本例中的文件:

[HttpPost("[action]")]
public async Task<IList<FileDto>> UploadFiles(List<IFormFile> files)
{
  return await _binaryService.UploadFilesAsync(files);
}

这个答案抄袭了http://blog.teamtreehouse.com/uploading-files-ajax

编辑: 上传后，您必须清除文件上传，以便用户可以选择一个新文件。与其使用XMLHttpRequest，不如使用fetch:

private addFileInput() {
    const fileInputParentNative = this.fileInputParent.nativeElement;
    const oldFileInput = fileInputParentNative.querySelector('input');
    const newFileInput = document.createElement('input');
    newFileInput.type = 'file';
    newFileInput.multiple = true;
    newFileInput.name = 'fileInput';
    const uploadfiles = this.uploadFiles.bind(this);
    newFileInput.onchange = uploadfiles;
    oldFileInput.parentNode.replaceChild(newFileInput, oldFileInput);
  }

  private uploadFiles() {
    this.onUploadStarted.emit();
    const fileInputParentNative = this.fileInputParent.nativeElement;
    const fileInput = fileInputParentNative.querySelector('input');
    if (fileInput.files && fileInput.files.length > 0) {
      const formData = new FormData();
      for (let i = 0; i < fileInput.files.length; i++) {
        formData.append('files', fileInput.files[i]);
      }

      const onUploaded = this.onUploaded;
      const onError = this.onError;
      const addFileInput = this.addFileInput.bind(this);
      fetch('/api/Data/UploadFiles', {
        credentials: 'include',
        method: 'POST',
        body: formData,
      }).then((response: any) => {
        if (response.status !== 200) {
          const error = `An error occured. Status: ${response.status}`;
          throw new Error(error);
        }
        return response.json();
      }).then(files => {
        onUploaded.emit(files);
        addFileInput();
      }).catch((error) => {
        onError.emit(error);
      });
    }

https://github.com/yonexbat/cran/blob/master/cranangularclient/src/app/file-upload/file-upload.component.ts