Java Spring end URL with a file extension










0














I have a service that returns a string. String is in the form of XML. The URL is /monitoring. But the client now wants, that the URL ends with /status.xml. How do I do that? I tried to change the value in the GetMapping, but then it stopped working. Is there anything special in URL ending with status.xml and how can I achieve this?



@GetMapping(value = "/monitoring")
public ResponseEntity getStatuses() throws TransformerException, UnsupportedEncodingException, JAXBException
return indexService.getStatusEntity();


public ResponseEntity getStatusEntity() throws JAXBException, UnsupportedEncodingException, TransformerException
byte xml = getStatus();
String xmlStr = new String(xml, "UTF-8");
xmlStr = xmlStr.replaceAll("<", "<");
xmlStr = xmlStr.replaceAll(">", ">");
Source src = new StreamSource(new ByteArrayInputStream(xmlStr.getBytes("UTF-8")));

TransformerFactory transformerFactory = TransformerFactory.newInstance();
Transformer transformer = transformerFactory.newTransformer();
ByteArrayOutputStream bos=new ByteArrayOutputStream();
StreamResult result = new StreamResult(bos);
transformer.transform(src, result);

return setDocumentHeaderAndReturnOK(bos.toByteArray());


private byte getStatus() throws JAXBException
String tomcat = ServerInfo.getServerInfo();
String pieces = tomcat.split("/");
App status = new App(applicationInfoService.getVersion().getBody(), "name", "OK", "",
new ServerPlatform(pieces[1], pieces[0]), new RunTimeEnvironment(System.getProperty("java.version"), "JAVA"),
checkExternalDependencies());

QName main = new QName("app");
JAXBElement<App> root = new JAXBElement<>(main, App.class, status);

ByteArrayOutputStream out = new ByteArrayOutputStream();
JAXBContext jaxbContext = JAXBContext.newInstance(App.class);
Marshaller jaxbMarshaller = jaxbContext.createMarshaller();

jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);

jaxbMarshaller.marshal(root, out);
jaxbMarshaller.marshal(root, System.out);

return out.toByteArray();


@Transactional
public ResponseEntity setDocumentHeaderAndReturnOK(byte contents) {

HttpHeaders headers = new HttpHeaders();

String fileType = "xml";

String filename = "status.xml";
headers.setContentType(MediaType.parseMediaType("application/" + fileType));

headers.add("Content-Disposition", "inline; filename=" + filename);
headers.setCacheControl("must-revalidate, post-check=0, pre-check=0");

return new ResponseEntity(contents, headers, HttpStatus.OK);


EDIT:



I tried generating the classes with jaxb and the use those to make the XML. But that does not change the url. I have the XML content generated correctly, i just need a way to get the GetMapping from /monitoring to /status.xml. What does it even mean to have a url in the form of status.xml ?










share|improve this question























  • Hi @Marko, would you add your code?
    – Reg
    Nov 13 '18 at 7:51










  • @Reg i added the code.
    – Marko Taht
    Nov 13 '18 at 8:02










  • Do you have any resource mappers?
    – Reg
    Nov 13 '18 at 8:34










  • For swagger. Cant seem to find anything else.
    – Marko Taht
    Nov 13 '18 at 10:29










  • @Reg what exactly do you mean under resource mappers?
    – Marko Taht
    Nov 27 '18 at 14:29















0














I have a service that returns a string. String is in the form of XML. The URL is /monitoring. But the client now wants, that the URL ends with /status.xml. How do I do that? I tried to change the value in the GetMapping, but then it stopped working. Is there anything special in URL ending with status.xml and how can I achieve this?



@GetMapping(value = "/monitoring")
public ResponseEntity getStatuses() throws TransformerException, UnsupportedEncodingException, JAXBException
return indexService.getStatusEntity();


public ResponseEntity getStatusEntity() throws JAXBException, UnsupportedEncodingException, TransformerException
byte xml = getStatus();
String xmlStr = new String(xml, "UTF-8");
xmlStr = xmlStr.replaceAll("<", "<");
xmlStr = xmlStr.replaceAll(">", ">");
Source src = new StreamSource(new ByteArrayInputStream(xmlStr.getBytes("UTF-8")));

TransformerFactory transformerFactory = TransformerFactory.newInstance();
Transformer transformer = transformerFactory.newTransformer();
ByteArrayOutputStream bos=new ByteArrayOutputStream();
StreamResult result = new StreamResult(bos);
transformer.transform(src, result);

return setDocumentHeaderAndReturnOK(bos.toByteArray());


private byte getStatus() throws JAXBException
String tomcat = ServerInfo.getServerInfo();
String pieces = tomcat.split("/");
App status = new App(applicationInfoService.getVersion().getBody(), "name", "OK", "",
new ServerPlatform(pieces[1], pieces[0]), new RunTimeEnvironment(System.getProperty("java.version"), "JAVA"),
checkExternalDependencies());

QName main = new QName("app");
JAXBElement<App> root = new JAXBElement<>(main, App.class, status);

ByteArrayOutputStream out = new ByteArrayOutputStream();
JAXBContext jaxbContext = JAXBContext.newInstance(App.class);
Marshaller jaxbMarshaller = jaxbContext.createMarshaller();

jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);

jaxbMarshaller.marshal(root, out);
jaxbMarshaller.marshal(root, System.out);

return out.toByteArray();


@Transactional
public ResponseEntity setDocumentHeaderAndReturnOK(byte contents) {

HttpHeaders headers = new HttpHeaders();

String fileType = "xml";

String filename = "status.xml";
headers.setContentType(MediaType.parseMediaType("application/" + fileType));

headers.add("Content-Disposition", "inline; filename=" + filename);
headers.setCacheControl("must-revalidate, post-check=0, pre-check=0");

return new ResponseEntity(contents, headers, HttpStatus.OK);


EDIT:



I tried generating the classes with jaxb and the use those to make the XML. But that does not change the url. I have the XML content generated correctly, i just need a way to get the GetMapping from /monitoring to /status.xml. What does it even mean to have a url in the form of status.xml ?










share|improve this question























  • Hi @Marko, would you add your code?
    – Reg
    Nov 13 '18 at 7:51










  • @Reg i added the code.
    – Marko Taht
    Nov 13 '18 at 8:02










  • Do you have any resource mappers?
    – Reg
    Nov 13 '18 at 8:34










  • For swagger. Cant seem to find anything else.
    – Marko Taht
    Nov 13 '18 at 10:29










  • @Reg what exactly do you mean under resource mappers?
    – Marko Taht
    Nov 27 '18 at 14:29













0












0








0







I have a service that returns a string. String is in the form of XML. The URL is /monitoring. But the client now wants, that the URL ends with /status.xml. How do I do that? I tried to change the value in the GetMapping, but then it stopped working. Is there anything special in URL ending with status.xml and how can I achieve this?



@GetMapping(value = "/monitoring")
public ResponseEntity getStatuses() throws TransformerException, UnsupportedEncodingException, JAXBException
return indexService.getStatusEntity();


public ResponseEntity getStatusEntity() throws JAXBException, UnsupportedEncodingException, TransformerException
byte xml = getStatus();
String xmlStr = new String(xml, "UTF-8");
xmlStr = xmlStr.replaceAll("<", "<");
xmlStr = xmlStr.replaceAll(">", ">");
Source src = new StreamSource(new ByteArrayInputStream(xmlStr.getBytes("UTF-8")));

TransformerFactory transformerFactory = TransformerFactory.newInstance();
Transformer transformer = transformerFactory.newTransformer();
ByteArrayOutputStream bos=new ByteArrayOutputStream();
StreamResult result = new StreamResult(bos);
transformer.transform(src, result);

return setDocumentHeaderAndReturnOK(bos.toByteArray());


private byte getStatus() throws JAXBException
String tomcat = ServerInfo.getServerInfo();
String pieces = tomcat.split("/");
App status = new App(applicationInfoService.getVersion().getBody(), "name", "OK", "",
new ServerPlatform(pieces[1], pieces[0]), new RunTimeEnvironment(System.getProperty("java.version"), "JAVA"),
checkExternalDependencies());

QName main = new QName("app");
JAXBElement<App> root = new JAXBElement<>(main, App.class, status);

ByteArrayOutputStream out = new ByteArrayOutputStream();
JAXBContext jaxbContext = JAXBContext.newInstance(App.class);
Marshaller jaxbMarshaller = jaxbContext.createMarshaller();

jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);

jaxbMarshaller.marshal(root, out);
jaxbMarshaller.marshal(root, System.out);

return out.toByteArray();


@Transactional
public ResponseEntity setDocumentHeaderAndReturnOK(byte contents) {

HttpHeaders headers = new HttpHeaders();

String fileType = "xml";

String filename = "status.xml";
headers.setContentType(MediaType.parseMediaType("application/" + fileType));

headers.add("Content-Disposition", "inline; filename=" + filename);
headers.setCacheControl("must-revalidate, post-check=0, pre-check=0");

return new ResponseEntity(contents, headers, HttpStatus.OK);


EDIT:



I tried generating the classes with jaxb and the use those to make the XML. But that does not change the url. I have the XML content generated correctly, i just need a way to get the GetMapping from /monitoring to /status.xml. What does it even mean to have a url in the form of status.xml ?










share|improve this question















I have a service that returns a string. String is in the form of XML. The URL is /monitoring. But the client now wants, that the URL ends with /status.xml. How do I do that? I tried to change the value in the GetMapping, but then it stopped working. Is there anything special in URL ending with status.xml and how can I achieve this?



@GetMapping(value = "/monitoring")
public ResponseEntity getStatuses() throws TransformerException, UnsupportedEncodingException, JAXBException
return indexService.getStatusEntity();


public ResponseEntity getStatusEntity() throws JAXBException, UnsupportedEncodingException, TransformerException
byte xml = getStatus();
String xmlStr = new String(xml, "UTF-8");
xmlStr = xmlStr.replaceAll("<", "<");
xmlStr = xmlStr.replaceAll(">", ">");
Source src = new StreamSource(new ByteArrayInputStream(xmlStr.getBytes("UTF-8")));

TransformerFactory transformerFactory = TransformerFactory.newInstance();
Transformer transformer = transformerFactory.newTransformer();
ByteArrayOutputStream bos=new ByteArrayOutputStream();
StreamResult result = new StreamResult(bos);
transformer.transform(src, result);

return setDocumentHeaderAndReturnOK(bos.toByteArray());


private byte getStatus() throws JAXBException
String tomcat = ServerInfo.getServerInfo();
String pieces = tomcat.split("/");
App status = new App(applicationInfoService.getVersion().getBody(), "name", "OK", "",
new ServerPlatform(pieces[1], pieces[0]), new RunTimeEnvironment(System.getProperty("java.version"), "JAVA"),
checkExternalDependencies());

QName main = new QName("app");
JAXBElement<App> root = new JAXBElement<>(main, App.class, status);

ByteArrayOutputStream out = new ByteArrayOutputStream();
JAXBContext jaxbContext = JAXBContext.newInstance(App.class);
Marshaller jaxbMarshaller = jaxbContext.createMarshaller();

jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);

jaxbMarshaller.marshal(root, out);
jaxbMarshaller.marshal(root, System.out);

return out.toByteArray();


@Transactional
public ResponseEntity setDocumentHeaderAndReturnOK(byte contents) {

HttpHeaders headers = new HttpHeaders();

String fileType = "xml";

String filename = "status.xml";
headers.setContentType(MediaType.parseMediaType("application/" + fileType));

headers.add("Content-Disposition", "inline; filename=" + filename);
headers.setCacheControl("must-revalidate, post-check=0, pre-check=0");

return new ResponseEntity(contents, headers, HttpStatus.OK);


EDIT:



I tried generating the classes with jaxb and the use those to make the XML. But that does not change the url. I have the XML content generated correctly, i just need a way to get the GetMapping from /monitoring to /status.xml. What does it even mean to have a url in the form of status.xml ?







java spring






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 28 '18 at 8:48







Marko Taht

















asked Nov 13 '18 at 7:15









Marko TahtMarko Taht

3541419




3541419











  • Hi @Marko, would you add your code?
    – Reg
    Nov 13 '18 at 7:51










  • @Reg i added the code.
    – Marko Taht
    Nov 13 '18 at 8:02










  • Do you have any resource mappers?
    – Reg
    Nov 13 '18 at 8:34










  • For swagger. Cant seem to find anything else.
    – Marko Taht
    Nov 13 '18 at 10:29










  • @Reg what exactly do you mean under resource mappers?
    – Marko Taht
    Nov 27 '18 at 14:29
















  • Hi @Marko, would you add your code?
    – Reg
    Nov 13 '18 at 7:51










  • @Reg i added the code.
    – Marko Taht
    Nov 13 '18 at 8:02










  • Do you have any resource mappers?
    – Reg
    Nov 13 '18 at 8:34










  • For swagger. Cant seem to find anything else.
    – Marko Taht
    Nov 13 '18 at 10:29










  • @Reg what exactly do you mean under resource mappers?
    – Marko Taht
    Nov 27 '18 at 14:29















Hi @Marko, would you add your code?
– Reg
Nov 13 '18 at 7:51




Hi @Marko, would you add your code?
– Reg
Nov 13 '18 at 7:51












@Reg i added the code.
– Marko Taht
Nov 13 '18 at 8:02




@Reg i added the code.
– Marko Taht
Nov 13 '18 at 8:02












Do you have any resource mappers?
– Reg
Nov 13 '18 at 8:34




Do you have any resource mappers?
– Reg
Nov 13 '18 at 8:34












For swagger. Cant seem to find anything else.
– Marko Taht
Nov 13 '18 at 10:29




For swagger. Cant seem to find anything else.
– Marko Taht
Nov 13 '18 at 10:29












@Reg what exactly do you mean under resource mappers?
– Marko Taht
Nov 27 '18 at 14:29




@Reg what exactly do you mean under resource mappers?
– Marko Taht
Nov 27 '18 at 14:29












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