我知道这是一个非常普遍的问题,但我在Angular 2中上传文件失败了。 我试过了

1) http://valor-software.com/ng2-file-upload/

2) http://ng2-uploader.com/home

...但失败了。有人在Angular中上传过文件吗?你用了什么方法?怎么做呢?如果提供了任何示例代码或演示链接,将非常感谢。


当前回答

我已经成功地使用了下面的工具。我和primeNg没有利害关系,只是传递我的建议。

http://www.primefaces.org/primeng/#/fileupload

其他回答

在Angular 2+中,让Content-Type为空是非常重要的。如果您将“内容类型”设置为“multipart/form-data”,则上传将无法工作!

upload.component.html

<input type="file" (change)="fileChange($event)" name="file" />

upload.component.ts

export class UploadComponent implements OnInit {
    constructor(public http: Http) {}

    fileChange(event): void {
        const fileList: FileList = event.target.files;
        if (fileList.length > 0) {
            const file = fileList[0];

            const formData = new FormData();
            formData.append('file', file, file.name);

            const headers = new Headers();
            // It is very important to leave the Content-Type empty
            // do not use headers.append('Content-Type', 'multipart/form-data');
            headers.append('Authorization', 'Bearer ' + 'eyJhbGciOiJSUzI1NiIsInR5cCI6IkpXVCJ9....');
            const options = new RequestOptions({headers: headers});

            this.http.post('https://api.mysite.com/uploadfile', formData, options)
                 .map(res => res.json())
                 .catch(error => Observable.throw(error))
                 .subscribe(
                     data => console.log('success'),
                     error => console.log(error)
                 );
        }
    }
}

由于代码示例有点过时,我想我应该分享一个更近期的方法,使用Angular 4.3和新的HttpClient API @angular/common/http

export class FileUpload {

@ViewChild('selectedFile') selectedFileEl;

uploadFile() {
let params = new HttpParams();

let formData = new FormData();
formData.append('upload', this.selectedFileEl.nativeElement.files[0])

const options = {
    headers: new HttpHeaders().set('Authorization', this.loopBackAuth.accessTokenId),
    params: params,
    reportProgress: true,
    withCredentials: true,
}

this.http.post('http://localhost:3000/api/FileUploads/fileupload', formData, options)
.subscribe(
    data => {
        console.log("Subscribe data", data);
    },
    (err: HttpErrorResponse) => {
        console.log(err.message, JSON.parse(err.error).error.message);
    }
)
.add(() => this.uploadBtn.nativeElement.disabled = false);//teardown
}

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

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))

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