我想从角度4上传.zip文件到服务器(弹簧控制器)。 请建议怎么做?
提前致谢
答案 0 :(得分:2)
1)看看这里: https://angular.io/guide/http#making-a-post-request
因此,您可以构建一个服务,当您在表单中按提交时会触发该服务,该表单会附加文件,无论是zip,img还是其他任何内容都可以用于POST请求。
2)在你的模板中,你可以使用类似的东西:
<form>
<input type="file" accept=".zip,application/octet-stream,application/zip,application/x-zip,application/x-zip-compressed">
<input type="submit">
</form>
3)看看这里强制文件扩展: https://www.hongkiat.com/blog/css3-attribute-selector/
答案 1 :(得分:2)
经过一番学习,我找到了如何将文件(.zip / .txt /任何其他文件格式)从角度(4/5)上传到弹簧/静止控制器的答案。在下面写下我对以下同样事情的学习。:)
前端编码::
<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('domain/urservice', formData, options)
.map(res => res.json())
.catch(error => Observable.throw(error))
.subscribe(
data => console.log('success'),
error => console.log(error)
)
}
}
}
(注意 - 此服务器通信调用应存在于某些服务中,而不是在组件中,但为了简单起见,我将其写入组件中)
服务器端编码::
@RequestMapping(value = "/urservice", method = RequestMethod.POST)
public void uploadFile(MultipartHttpServletRequest request) throws IOException {
Iterator<String> itr = request.getFileNames();
// directory to save file
String tempDir = System.getProperty("jboss.server.temp.dir");
MultipartFile file = request.getFile(itr.next());
String fileType = request.getParameter("fileType");
String fileName = file.getOriginalFilename();
File dir = new File(tempDir);
File fileToImport = null;
if (dir.isDirectory()) {
try {
fileToImport = new File(dir + File.separator + fileName);
BufferedOutputStream stream = new BufferedOutputStream(new FileOutputStream(fileToImport));
stream.write(file.getBytes());
stream.close();
} catch (Exception e) {
logger.error("Got error in uploading file.");
}
}