Question

当我尝试上传zip文件时，不幸的是它没有在服务器上上传。它给我错误了一些事情！这是我的代码：

const event = this.fileEvent;
console.log('FileEvent --> ', event);
const fileList: FileList = event.target.files;
if (fileList.length > 0) {
   const file: File = fileList[0];
   const formData: FormData = new FormData();
   formData.append('file', file, file.name);
   formData.append('promotion_name' , this.promotionName);
   formData.append('template_id' , this.template_id + '' );
   const headers = new Headers();

   // No need to include Content-Type in Angular 4
   headers.append('Content-Type', 'multipart/form-data');
   headers.append('Accept', 'application/json');
   const options = new RequestOptions({ headers: headers });
      this.http.post(this.apiURL + 'api/v1/advertiser/promotion/createPromotion',

这是我的html代码

<div class="form-group row">
<label class="col-md-3 form-control-label" for="file-input">Select File</label>
    <div class="col-md-9">
        <div class="input-group">
            <input id="file-input" class="form-control" name="marker_image" placeholder="Select File" type="file" (change)="fileChange($event)" accept="application/zip,application/ZIP" />                                         
        </div>
        <div class="col-md-8 file-err-msg">
        <span class="help-block err-color" *ngIf="fileupload">Please upload CSV file</span>
        </div>
    </div>
</div>

这是我的文件选择代码`返回response.Error（c，errors.New（“请提供zip文件”）） }

filenamePrefix := filepath.Join("./assets/promotion/zipfiles/", buildFileName())

// zipFiles name will be stored in filename
filename, err := pickFileName(filenamePrefix)
if err != nil {
    return response.Error(c, errors.New("something went wrong please try agian later"))
}`

Answer 1

您可以使用类似的内容：

<input type="file" formControlName="uploadFile" (change)="uploadFileToServer($event)"/>


import { Component } from '@angular/core';

import { RequestOptions, Headers, Http } from '@angular/http';

@Component({



 selector: 'file-uploader',
  templateUrl: './uploadFile.component.html',
  styleUrls: ['./uploadFile.component.css'],
})
export class FileUploadComponent {

public uploadFileToServer(event) {
  let fileList: FileList = event.target.files;
  if (fileList.length > 0) {
    let file: File = fileList[0];
    let formData: FormData = new FormData();
    formData.append('uploadFile', file, file.name);
    formData.append('fileType', 'zip');
    let headers = new Headers();
    headers.append('Accept', 'application/json');
    let options = new RequestOptions({ headers: headers });
    this.http.post('your url here', formData, options)
      .map(res => res.json())
      .catch(error => Observable.throw(error))
      .subscribe(
      data => console.log('success'),
      error => console.log(error)
      )
  }
} 

}

Angular 2中的zip文件上传问题

1 个答案: