我正在制作接收文件下載請求的 Spring Boot Web 服務。此服務的域是a.com并且此服務從b.com域中帶來請求的文件。此服務需要從外部域請求文件。我想知道在這種情況下如何向客戶端返回文件下載請求的響應。下面的代碼是我制作的。@RequestMapping(path = "/downloadFile", method = RequestMethod.GET)public ResponseEntity<Resource> download(String param) throws IOException { String testFileName = "https://www.google.co.jp/images/branding/googlelogo/2x/googlelogo_color_272x92dp.png" URL url = new URL(testFileName); File file = new File(url.getFile()); HttpHeaders headers = new HttpHeaders(); headers.add("Cache-Control", "no-cache, no-store, must-revalidate"); headers.add("Pragma", "no-cache"); headers.add("Expires", "0"); Path path = Paths.get(file.getAbsolutePath()); ByteArrayResource resource = new ByteArrayResource(Files.readAllBytes(path)); return ResponseEntity.ok() .headers(headers) .contentLength(file.length()) .contentType(MediaType.parseMediaType("application/octet-stream")) .body(resource);}但由于文件路徑，它不起作用。它的路徑是這樣顯示的。/Users/user/devel/acomapplication/https:/www.google.co.jp/images/branding/googlelogo/2x/googlelogo_color_272x92dp.png
2 回答
慕萊塢森
TA貢獻1810條經驗 獲得超4個贊
您需要使用 HTTP 客戶端下載您要代理的文件。下面是一些如何做到這一點的示例。此示例使用 Apache HTTP 客戶端 4.5。
@RequestMapping(path = "/downloadFile", method = RequestMethod.GET)
public ResponseEntity download(String param) throws IOException {
String testFileName = "https://www.google.co.jp/images/branding/googlelogo/2x/googlelogo_color_272x92dp.png"
HttpClient client = HttpClientBuilder.create().build();
HttpGet request = new HttpGet(testFileName);
request.setHeader("Cache-Control", "no-cache, no-store, must-revalidate");
request.setHeader("Pragma", "no-cache");
request.setHeader("Expires", "0");
HttpResponse response = client.execute(request);
if (response.getStatusLine().getStatusCode()==200) {
ByteArrayResource resource = new ByteArrayResource(Files.readAllBytes(path));
return ResponseEntity.ok()
.headers(headers)
.contentLength(file.length())
.contentType(MediaType.parseMediaType("application/octet-stream"))
.body(response.getEntity().getContent());
} else {
return ResponseEntity.notFound();
}
}
回首憶惘然
TA貢獻1847條經驗 獲得超11個贊
Path path = Path.resolve(fileName).normalize(); Resource resource = new UrlResource(path.toUri());
嘗試使用這個。
添加回答
舉報
0/150
提交
取消