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What's the best way to encode and decode parameter in springboot?

I use #RequestParam to get the parameter value,but I find the if I pass the value like 'name=abc&def&id=123',I will get the name value 'abc' instead of 'abc&def'. I find the encode and decode the parameter value can solve my problem.But I have to write the encode and decode mehtod in every controller method,Do spring have the global mehtod that decode every #RequestParam value?When using #RequestParam, is it necessary to encode and decode every value?
Here is my code:
#PostMapping("/getStudent")
public Student getStudent(
#RequestParam String name,
#RequestParam String id) {
name= URLDecoder.decode(name, "UTF-8");
//searchStudent
return Student;
}
#PostMapping("/getTeacher")
public teacher getTeacher(
#RequestParam String name,
#RequestParam String teacherNo) {
name= URLDecoder.decode(name, "UTF-8");
//searchTeacher
return teacher;
}
Somebody say the the Spring will have already done this,but I have try,the result is not right.Only use curl cmd is ok,but java code is not ok.
#PostMapping(value = "/example")
public String handleUrlDecode1(#RequestParam String param) {
//print ello%26test
System.out.println("/example?param received: " + param);
return "success";
}
#GetMapping(value = "/request")
public String request() {
String url = "http://127.0.0.1:8080/example?param=ello%26test";
System.out.println(url);
RestTemplate restTemplate = new RestTemplate();
return restTemplate.postForObject(url, null, String.class);
}

You must create an HTTP entity and send the headers and parameter in body.
#GetMapping(value = "/request")
public String request() {
String url = "http://127.0.0.1:8080/example";
System.out.println(url);
RestTemplate restTemplate = new RestTemplate();
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_FORM_URLENCODED);
MultiValueMap<String, String> map= new LinkedMultiValueMap<String, String>();
map.add("param","ello&test");
map.add("id","ab&c=def");
HttpEntity<MultiValueMap<String, String>> request = new HttpEntity<MultiValueMap<String, String>>(map, headers);
return restTemplate.postForObject(url, request, String.class);
}

As you can read here, the escape character for & is %26.
So you should use the following
name=abc%26def&id=123
If you don't use an escape character according to URL standards, Spring will try to use what follows & and try to match it as a new query parameter.

No need to manually use URLDecoder, SpringBoot controllers will handle it for you.
#RestController
public class UrlDecodeController {
#GetMapping(value = "/example")
public String handleUrlDecode(#RequestParam String param) {
System.out.println("/example?param received: " + param);
return "success";
}
#PostMapping(value = "/example2")
public String handleUrlDecodeInPostRequest(#RequestParam String param1, ExamplePayload payload) {
System.out.println("/example2?param1 received: " + param1);
System.out.println("request body - value1: " + payload.getValue1());
return "success";
}
#GetMapping(value = "/request")
public String request() {
String url = "http://localhost:8080/example2?param1=test1&test2";
System.out.println(url);
RestTemplate restTemplate = new RestTemplate();
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_FORM_URLENCODED);
MultiValueMap<String, String> map = new LinkedMultiValueMap<String, String>();
map.add("value1","test1&test2");
HttpEntity<MultiValueMap<String, String>> request = new HttpEntity<MultiValueMap<String, String>>(map, headers);
return restTemplate.postForObject(url, request, String.class);
}
class ExamplePayload{
private String value1;
private String value2;
//getters and setters
public ExamplePayload() {
}
}
}
Call with GET /example?param=hello%26test and the System.out.println outputs:
/example?param received: hello&test
Call the POST using curl as an example:
curl -X POST "http://localhost:8080/example2?param1=test1%26test2" -d "value1=test3%26test4"
Prints:
/example2?param1 received: test1&test2
request body - value1: test3&test4
Added GET /request to show using RestTemplate with the application/x-www-form-urlencoded Content-Type. Note that RestTemplate will automatically url encode any values passed as request parameters or in the request body. If you pass a String value of "%26" it will pass it as is, this is what you are seeing in your example. If you pass "&" it will url encode it to "%26" for you, and the Controller decodes it automatically on the other side.

Spring boot - restTemplate.postForObject - params are null

I have 2 spring boot apps running one as a frontend and another as a backend service. From the frontend i make an api call to the backend service and both the parameters that i send show up as null. I think the problem is in the rest template.
UPDATE
So i have noticed if i omit the content value then it works. Since content is the content of a file that is larger than 1mb I added the following to application.yml:
spring.servlet.multipart.max-file-size: 10MB
spring.servlet.multipart.max-request-size: 10MB
Here is my code which I updated from one posted in this issue:
How to POST form data with Spring RestTemplate?
But i still don't get the value in the backend controller instead both values are null.
public void upload(byte[] content, String name) {
String encodedString = Base64.getEncoder().encodeToString(content);
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_FORM_URLENCODED);
MultiValueMap<String, String> map = new LinkedMultiValueMap<String, String>();
map.add("fileName", name);
map.add("content", encodedString);
HttpEntity<MultiValueMap<String, String>> request = new HttpEntity<MultiValueMap<String, String>>(map, headers);
ResponseEntity<String> response = restTemplate.postForEntity(backendUrl + "/upload", request, String.class);
log.debug("Response from upload: " + response);
}
And here is the controller in the backend. Both fileName and content are null:
#CrossOrigin
#SneakyThrows
#ResponseBody
#PostMapping(value = "/upload")
public ResponseEntity<String> upload(#ModelAttribute FormModel form) {
byte[] decodedBytes = Base64.getDecoder().decode(form.getContent());
uploadService.upload(decodedBytes, form.getFileName());
HttpHeaders responseHeaders = new HttpHeaders();
responseHeaders.set("Content-Type", "application/json");
return ResponseEntity.ok().headers(responseHeaders).body("Uploaded");
}
Can anyone please see what is wrong with this code?
Thanks in advance.

I guess the problem is that you are trying to use restTemplate.postForObject but with #RequestParam and not a #RequestBody.
In #RequestParam you are expecting the data to be received in the query params /upload?fileName=&content=. But you are actually sending it in the body with the restTemplate.postForObject(backendService+ "/upload", map, String.class);.
So my suggestion is to change
public ResponseEntity<String> upload(#RequestParam(value = "fileName") String fileName, #RequestParam(value = "content") String content)
to
public ResponseEntity<String> upload(#RequestBody Map<String, String> body)
and then get fileName and fileContent from the body.

Ok i could fix it by sending and receiving bytes instead of bytes encoded as string.
So in the resttemplate:
public void upload(byte[] bytes, String name) {
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.MULTIPART_FORM_DATA);
MultiValueMap<String, Object> map = new LinkedMultiValueMap<String, Object>();
map.add("fileName", name);
map.add("bytes", bytes);
HttpEntity<MultiValueMap<String, Object>> request = new HttpEntity<MultiValueMap<String, Object>>(map, headers);
log.debug("map values: " + map.toString());
ResponseEntity<String> response = restTemplate.postForEntity(backendUrl + "/upload", request, String.class);
log.debug("Response from upload: " + response);
}
And in the controller:
public ResponseEntity<String> upload(#ModelAttribute FormModel form) {
byte[] bytes = form.getBytes();
uploadService.upload(bytes, form.getFileName());
HttpHeaders responseHeaders = new HttpHeaders();
responseHeaders.set("Content-Type", "application/json");
return ResponseEntity.ok().headers(responseHeaders).body("Uploaded");
}
Still it would be good to know why the previous version didn't work.

Spring Web: Download a File from a service via a Spring Service

I want to be able to download a file from a legacy service through a middle-layer Spring Web service. The problem currently is that I am returning the contents of the file and not the file itself.
I've used FileSystemResource before, but I do not want to do this, since I want Spring to only redirect and not create any files on the server itself.
Here is the method:
#Override
public byte[] downloadReport(String type, String code) throws Exception {
final String usernamePassword = jasperReportsServerUsername + ":" + jasperReportsServerPassword;
final String credentialsEncrypted = Base64.getEncoder().encodeToString((usernamePassword).getBytes("UTF-8"));
final HttpHeaders httpHeaders = new HttpHeaders();
httpHeaders.add("Accept", MediaType.APPLICATION_JSON_VALUE);
httpHeaders.add("Authorization", "Basic " + credentialsEncrypted);
httpHeaders.setAccept(Arrays.asList(MediaType.APPLICATION_OCTET_STREAM));
final HttpEntity httpEntity = new HttpEntity(httpHeaders);
final String fullUrl = downloadUrl + type + "?code=" + code;
return restTemplate.exchange(fullUrl, HttpMethod.GET, httpEntity, byte[].class, "1").getBody();
}

Turns out I was missing this annotation parameter in my *Controller class:
produces = MediaType.APPLICATION_OCTET_STREAM_VALUE
The whole method of the controller should look like this:
#RequestMapping(value = "/download/{type}/{id}", method = RequestMethod.GET, produces = MediaType.APPLICATION_OCTET_STREAM_VALUE)
public ResponseEntity<?> downloadReport(#PathVariable String type, #PathVariable String id) throws Exception {
return new ResponseEntity<>(reportService.downloadReport(type, id), HttpStatus.OK);
}

How to send JSON as a Input parameter from one Microservice to another using RestTemplate in Spring Boot

I want to send JSON as an input from Microservice M1 to a Microservice M2.
M1 and M2 both are on different machines.
I am new to Spring Boot,
I found some code but I am unable to get it.
Please help.

make a class on both microservices or make a jar of that class and add to both microservices so that they both can access the same data.
Lets say the class is
class TestData{
private String name;
private String id;
// getters and setters
}
Now you can send data from M1 to M2 as following
RestTemplate restTemplate = new RestTemplate();
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_JSON);
TestData data = new TestData();
HttpEntity<?> entity = new HttpEntity<Object>(data,headers);
ResponseEntity<Object> responseEntity = restTemplate.exchange("url", HttpMethod.POST, entity, Object.class);
In Microservice M2 you can write a controller to get the data and process it as follows
#RequestMapping(value="/url",method=RequestMethod.POST)
public Object do(#RequestBody TestData data){
// do something
return //something
}

Let's Say Your Have MicroService1 which needs to send JSONObject => JsonObject to another MicroService2 which is on different Machine but on same network .
Sender Side:
RestTemplate restTemplate = new RestTemplate();
String jsonString = restTemplate.postForObject("http://10.177.7.128:8080/user/insertJsonObject",jsonObject,String.class);
Syntax for restTemplate.postForObject is:
ResponseType var1 = restTemplate.postForObject("network ip Address:portnumber/path",JSONObject,ResponseType)
To Know the URI go to System Preferences > Network
To Receive the object at the receiver Side
#RequestMapping(value="/user/insertJsonObject", method=RequestMethod.POST)
public String updateProductSold(#RequestBody JSONObject jsonObject) {
...Body
...
...
return responseStatus;

Here is the sample code
public class Test {
public static void main(String[] args) {
String jsonString = "{\"id\" : \"123\",\"name\" : \"Tom\",\"class\" : {\"subject\" : \"Math\",\"teacher\" : \"Jack\"}}";
RestTemplate restTemplate = new RestTemplate();
String url = "http://192.1168.1.190:8080/test" // url for second service
System.out.println(responserEntityValue(jsonString,restTemplate,url,HttpMethod.POST,String.class));
}
public ResponseEntity<String> responserEntityValue(final String body, final RestTemplate restTemplate,
final String uRL, final HttpMethod requestMethod, final Class<String> stringClass) {
HttpHeaders headers = new HttpHeaders();
// Set all headers
headers.add(DatabaseConstants.CONTENT_TYPE, "application/json");
HttpEntity<String> request = new HttpEntity<>(body, headers);
return restTemplate.exchange(uRL, requestMethod, request, stringClass);
}

How to pass raw XML into RESTTemplate POST or PUT method without converting it?

I am trying to update or create xml file if not present. Then I use code below to send the file using PUT method of a service.
public void importClusterProperties(RestManPropertyHolder propertyHolder,File file,String id) throws RestManServiceException {
testRestTemplate = new TestRestTemplate(propertyHolder.getSbusUserName(), propertyHolder.getSbusUserPassword());
String sbusUrl = utils.prepareGatewayURI(propertyHolder);
try {
HttpHeaders requestHeaders = new HttpHeaders();
List <MediaType> mediaTypeList = new ArrayList<MediaType>();
mediaTypeList.add(MediaType.APPLICATION_ATOM_XML);
requestHeaders.setAccept(mediaTypeList);
requestHeaders.setContentType(MediaType.APPLICATION_ATOM_XML);
HttpEntity<String> requestEntity = new HttpEntity<String>(requestHeaders);
// Create the HTTP PUT request,
ResponseEntity<String> response = testRestTemplate.exchange(sbusUrl + "/clusterproperty?",HttpMethod.PUT, requestEntity,String.class);
if (null != response) {
System.out.println("RESPONSE::" + response.toString());
}
} catch (RestClientException rce) {
System.out.println("REST EXCEPTION:::" + rce.getMessage());
}
}
How to pass raw xml file into RestTemplate without converting it first into a java object?
enter image description here

Convert file to byte array and send it using ByteArrayHttpMessageConverter.
RestTemplate restTemplate = new RestTemplate();
// Add ByteArrayHttpMessageConverter if not present by default.
restTemplate.getMessageConverters().add(
new ByteArrayHttpMessageConverter());
String fileName = "path + file name";
// FileUtils is from Apache Commons IO
// import org.apache.commons.io.FileUtils;
byte[] requestBody = FileUtils.readFileToByteArray(new File(fileName));
HttpEntity<byte[]> requestEntity = new HttpEntity<byte[]>(requestBody , requestHeaders);
// Create the HTTP PUT request,
ResponseEntity<byte[]> response =
restTemplate.exchange("URL ...." , HttpMethod.PUT ,
requestEntity , byte[].class);