I want to serve a .yaml file via a REST endpoint with Spring, I know that it cannot be directly displayed in a browser (just talking about Chrome here), since it doesn't support display of yaml files. I have included what I think is the necessary library for this purpose compile group: 'com.fasterxml.jackson.dataformat', name: 'jackson-dataformat-yaml', version: '2.9.9'
.
If I open the endpoint /v2/api-doc
in the browser, it will prompt me, to download a file named exactly as the endpoint /v2/api-doc
. It contains the correct content.
Question: Is there a way to correctly transfer the .yaml file, so that the user will be prompted to safe myfile.yaml?
@RequestMapping(value = "/v2/api-doc", produces = "application/x-yaml")
public ResponseEntity<String> produceApiDoc() throws IOException {
byte[] fileBytes;
try (InputStream in = getClass().getResourceAsStream("/restAPI/myfile.yaml")) {
fileBytes = IOUtils.toByteArray(in);
}
if (fileBytes != null) {
String data = new String(fileBytes, StandardCharsets.UTF_8);
return new ResponseEntity<>(data, HttpStatus.OK);
} else {
return new ResponseEntity<>(HttpStatus.NOT_FOUND);
}
}
You should set a Content-Disposition
header (and I recommend using ResourceLoader
to load resources in Spring Framework).
Example:
@RestController
public class ApiDocResource {
private final ResourceLoader resourceLoader;
public ApiDocResource(ResourceLoader resourceLoader) {
this.resourceLoader = resourceLoader;
}
@GetMapping(value = "/v2/api-doc", produces = "application/x-yaml")
public ResponseEntity produceApiDoc() throws IOException {
Resource resource = resourceLoader.getResource("classpath:/restAPI/myfile.yaml");
if (resource.exists()) {
return ResponseEntity
.ok()
.contentType(MediaType.parseMediaType("application/x-yaml"))
.header("Content-Disposition", "attachment; filename=myfile.yaml")
.body(new InputStreamResource(resource.getInputStream()));
} else {
return new ResponseEntity<>(HttpStatus.NOT_FOUND);
}
}
}
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