我正在尝试使用AngularJS 2 rc5和asp.net mvc上传文件。 我无法找到在angularjs 2和asp.net mvc上传文件的方法。
答案 0 :(得分:8)
最后
解决方案
这项工作对我来说
upload.service.ts文件
import {Injectable}from '@angular/core';
import {Observable} from 'rxjs/Rx';
@Injectable()
export class UploadService {
progress$: any;
progress: any;
progressObserver: any;
constructor() {
this.progress$ = Observable.create(observer => {
this.progressObserver = observer
}).share();
}
makeFileRequest(url: string, params: string[], files: File[]): Observable<any> {
return Observable.create(observer => {
let formData: FormData = new FormData(),
xhr: XMLHttpRequest = new XMLHttpRequest();
for (let i = 0; i < files.length; i++) {
formData.append("uploads[]", files[i], files[i].name);
}
xhr.onreadystatechange = () => {
if (xhr.readyState === 4) {
if (xhr.status === 200) {
observer.next(JSON.parse(xhr.response));
observer.complete();
} else {
observer.error(xhr.response);
}
}
};
xhr.upload.onprogress = (event) => {
this.progress = Math.round(event.loaded / event.total * 100);
this.progressObserver.next(this.progress);
};
xhr.open('POST', url, true);
var serverFileName = xhr.send(formData);
return serverFileName;
});
}
}
和appcomponnet.ts文件
import {Component } from 'angular2/core';
import {UploadService} from './app.service';
@Component({
selector: 'my-app',
template: `
<div>
<input type="file" (change)="onChange($event)"/>
</div>
`,
providers: [ UploadService ]
})
export class AppComponent {
picName: string;
constructor(private service:UploadService) {
this.service.progress$.subscribe(
data => {
console.log('progress = '+data);
});
}
onChange(event) {
console.log('onChange');
var files = event.srcElement.files;
console.log(files);
this.service.makeFileRequest('http://localhost:8182/upload', [], files).subscribe(() => {
console.log('sent');
this.picName = fileName;
});
}
}
和操作方法
public HttpResponseMessage UploadFile()
{
var file = HttpContext.Current.Request.Files[0];
if (file.ContentLength > 0)
{
var fileName = Path.GetFileName(file.FileName);
var path = Path.Combine(Server.MapPath("~/App_Data/uploads"), fileName);
file.SaveAs(path);
var content = JsonConvert.SerializeObject(serverFileName, new JsonSerializerSettings
{
ContractResolver = new CamelCasePropertyNamesContractResolver()
});
var response = Request.CreateResponse(HttpStatusCode.OK);
response.Content = new StringContent(content, Encoding.UTF8, "application/json");
return response;
}