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Spring RestTemplate: how to send the payload body as compact form so without \r\n

I am dealing with an API that don't accept multi line json body and accept only 1 json line body (json compact form)
The below payload is valid because it's compact in just 1 line
And the below payload is not passing because it's multiline
I have the same problem in the Java/Spring code where I got this error while posting my object in restemplate.
Is there a way to convert the payload body into 1 single json line?
Code I am using to post the payload via RestTemplate
private HttpHeaders headers() {
HttpHeaders headers = new HttpHeaders();
headers.setAccept(List.of(MediaType.APPLICATION_JSON, MediaType.TEXT_PLAIN));
return headers;
}
post(ParameterizedTypeReference type, REQUEST myObject, URI uri) {
HttpEntity<REQUEST> entity = new HttpEntity<>(myObject, headers());
ResponseEntity<String> res = restTemplate.exchange(uri, HttpMethod.POST, entity , type);
}

The solution that worked for me is to annotate my request class with a custom JsonSerializer
This MyRequestClassSerializer#serialize will be called once restTemplate.exchange(uri, HttpMethod.POST, entity , type); is executed
Hence the payload will be compacted in 1 line by using JsonGenerator#writeRawValue
public class MyRequestClassSerializer extends JsonSerializer<MyRequestClass> {
#Override
public void serialize(MyRequestClass value, JsonGenerator jsonGenerator, SerializerProvider provider) throws IOException {
jsonGenerator.writeStartObject();
ObjectMapper mapper = ((ObjectMapper) jsonGenerator.getCodec());
jsonGenerator.writeFieldName("FieldName");
String stringValue = mapper.writeValueAsString(value);
jsonGenerator.writeRawValue(stringValue);
jsonGenerator.writeEndObject();
}
}
#JsonSerialize(using = MyRequestClassSerializer.class)
public class MyRequestClass{
...
}

Java consuming restful api with httpclient. Having trouble with Jackson json MismatchedInputException

This is my first time using jackson/consuming apis/httpclient. I'm getting this error com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot deserialize value of type java.util.ArrayList<WallHaven> from Object value (token JsonToken.START_OBJECT) . The api I'm trying to consume is https://wallhaven.cc/help/api
try {
HttpClient client = HttpClient.newHttpClient();
HttpRequest request = HttpRequest.newBuilder()
.GET()
.uri(URI.create("https://wallhaven.cc/api/v1/w/pkgkkp"))
.build();
HttpResponse<String> response = client.send(request, HttpResponse.BodyHandlers.ofString());
ObjectMapper mapper = new ObjectMapper();
List<WallHaven> posts = mapper.readValue(response.body(), new TypeReference<List<WallHaven>>() {
});
posts.forEach(System.out::println);
} catch (Exception e) {
e.printStackTrace();
}
The api json format is https://pastebin.com/tbSaVJ1T
Here's my WallHaven class
public class WallHaven {
public Data data;
public WallHaven(Data data) {
this.data = data;
}
public WallHaven() {
}
#Override
public String toString() {
return "WallHaven{" +
"data=" + data.getPath() +
'}';
}
}
Data contains all the other classes/variables

This is happening because you're trying to deserialize a Json Object into a List in java. The error message explains it by saying that the starting character (JsonToken.START_OBJECT) is the start of a json object not a json array, so you can't deserialize it directly into a List, but should deserialize it into an object.
Try changing:
List<WallHaven> posts = mapper.readValue(response.body(), new TypeReference<List<WallHaven>>())
into
WallHaven post = mapper.readValue(response.body(), new TypeReference<WallHaven>())

Retrieve value from returned JSON object in Java

I am a receiving a JSON object and I need to save the values to my DB. But I'm having an issue figuring out how to retrieve the particular values in the JSON object.
In this case, I want to retrieve the values of 'originationNumber' and 'messageBody'
The response object -
{"originationNumber":"***","destinationNumber":"***","messageKeyword":"KEYWORD_***","messageBody":"Answer ","previousPublishedMessageId":"1slamq6mdpucd8q4i7iabf1sikc629ga253tr6o0","inboundMessageId":"88bc02fc-aff3-4277-ac1d-f27b6d3b6abb"}
Method to receive message -
public String getReceivedMessages(Messaging receivedMessage) {
BasicAWSCredentials awsCredentials = new BasicAWSCredentials(awsAccessKey, awsSecretKey);
AmazonSQS sqsClient = AmazonSQSClientBuilder.standard()
.withCredentials(new AWSStaticCredentialsProvider(awsCredentials))
.withRegion(String.valueOf(awsRegion)).build();
StringBuilder sb = new StringBuilder();
String queueUrl = "https://sqs.us-east-1.amazonaws.com/1234567/GetReceivedMessages";
List<Message> messages = sqsClient.receiveMessage(new ReceiveMessageRequest(queueUrl)
.withMaxNumberOfMessages(1).withWaitTimeSeconds(20)).getMessages();
for (Message message : messages) {
sb.append(message.getBody());
sqsClient.deleteMessage(queueUrl, message.getReceiptHandle());
}
// Save messages to DB
String userId = connectionRequestRepository.getUserId();
Date date = new Date();
Timestamp now = new Timestamp(date.getTime());
receivedMessage.setUserId(userId);
receivedMessage.setOriginationNumber("");
receivedMessage.setDestinationNumber("***");
receivedMessage.setMessageBody("");
receivedMessage.setMessageType("RECEIVED");
receivedMessage.setCreatedAt(now);
messagingRepository.save(receivedMessage);
System.out.println(sb); <--- Prints response object to console
return sb.toString();
}

You can use jackson library for that.
Solution 1: You can use ObjectMapper as below:
Message Class to map JSON to Java Object:
public class Message {
private String originationNumber;
private String messageBody;
// public getter and setters methods
}
Create Object From JSON String:
Message message = null;
ObjectMapper objectMapper = new ObjectMapper();
objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
try {
message = objectMapper.readValue(payload, Message.class);
} catch (JsonProcessingException e) {
// Log Or do some action as per need
}
Here message will have those values. DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES is used here to skip exception as there are other fields also in JSON and not present in Java Class (as we don't need them).
Solution 2: Alternatively you can use JsonNode from same Jackson library and read nodes one by one as below:
ObjectMapper mapper = new ObjectMapper();
JsonNode actualObj = mapper.readTree("{\"originationNumber\":\"***\",\"destinationNumber\":\"***\",\"messageKeyword\":\"KEYWORD_***\",\"messageBody\":\"Answer \",\"previousPublishedMessageId\":\"1slamq6mdpucd8q4i7iabf1sikc629ga253tr6o0\",\"inboundMessageId\":\"88bc02fc-aff3-4277-ac1d-f27b6d3b6abb\"}");
String originationNumber = actualObj.get("originationNumber");
String messageBody = actualObj.get("messageBody");
In this approach you won't need to create Message class.

You can convert the json string into a json object
https://www.javatpoint.com/how-to-convert-string-to-json-object-in-java
String string = "{\"originationNumber\":\"***\",\"destinationNumber\":\"***\",\"messageKeyword\":\"KEYWORD_***\",\"messageBody\":\"Answer \",\"previousPublishedMessageId\":\"1slamq6mdpucd8q4i7iabf1sikc629ga253tr6o0\",\"inboundMessageId\":\"88bc02fc-aff3-4277-ac1d-f27b6d3b6abb\"}";
JSONObject json = new JSONObject(string);
System.out.println(json.toString());
String destinationNumber = json.getString("destinationNumber");
System.out.println(destinationNumber);

where ur json like
{"originationNumber":"***",
"destinationNumber":"***",
"messageKeyword":"KEYWORD_***",
"messageBody":"Answer","previousPublishedMessageId":"1slamq6mdpucd8q4i7iabf1sikc629ga253tr6o0",
"inboundMessageId":"88bc02fc-aff3-4277-ac1d-f27b6d3b6abb"
}
it like
"key":Value
i think ur code will be like
receivedMessage.setUserId(userId);
receivedMessage.setOriginationNumber("originationNumber");
receivedMessage.setDestinationNumber("destinationNumber");
receivedMessage.setMessageBody("messageBody");
receivedMessage.setMessageType("RECEIVED");
receivedMessage.setCreatedAt(now);
messagingRepository.save(receivedMessage);

How to get raw JSON body in Spring REST controller?

The API below accept a json string from client, and the map it into a Email object. How can I get request body (email) as a raw String? (I want both raw-string and typed version of email parameter)
PS: This question is NOT a duplicate of: How to access plain json body in Spring rest controller?
#PostMapping(value = "/mailsender")
public ResponseEntity<Void> sendMail(#RequestBody Email email) {
//...
return new ResponseEntity<>(HttpStatus.OK);
}

You can do it in more than one way, listing two
1. **Taking string as the paramater**,
#PostMapping(value = "/mailsender")
public ResponseEntity<Void> sendMail(#RequestBody String email) {
//... the email is the string can be converted to Json using new JSONObject(email) or using jackson.
return new ResponseEntity<>(HttpStatus.OK);
}
2. **Using Jackson**
#PostMapping(value = "/mailsender")
public ResponseEntity<Void> sendMail(#RequestBody Email email) {
//...
ObjectMapper mapper = new ObjectMapper();
String email = mapper.writeValueAsString(email); //this is in string now
return new ResponseEntity<>(HttpStatus.OK);
}

Spring uses Jackson for this in the back, you could use it to serialize it to an string. Like so:
#Autowired private ObjectMapper jacksonMapper;
#PostMapping(value = "/mailsender")
public ResponseEntity<Void> sendMail(#RequestBody Email email) {
//...
log.info("Object as String: " + jacksonMapper.writeValueAsString(email));
return new ResponseEntity<>(HttpStatus.OK);
}

you can create json of type string using GSON library
Gson gson = new Gson();
#PostMapping(value = "/endpoint")
public ResponseEntity<Void> actionController(#RequestBody Car car) {
//...
log.info("Object as String: " + this.gson.toJson(car));
return new ResponseEntity<>(HttpStatus.OK);
}

I did not get all things about this question, but I try to answer as I understand. Well,
if you want to get request body:
as you say How to access plain json body in Spring rest controller? here already writen how to do this. If something wrong, maybe you send wrong json or not suitable type as you wite inside Email class. Maybe your request comes url filter
second way try like this:
private final ObjectMapper mapper = new ObjectMapper();
#PostMapping(value = "/mailsender")
public ResponseEntity<Void> sendMail(HttpServletRequest req) {
// read request body
InputStream body = req.getInputStream();
byte[] result = ByteStreams.toByteArray(body);
String text =new String(result,"UTF-8");
//convert to object
Email email = mapper.readValue(body, Email .class);
return new ResponseEntity<>(HttpStatus.OK);
}
If you want to convert object to json string read this post

Error - Returning JSON along with HTTP status code error JAX-RS

#POST
#Consumes({MediaType.APPLICATION_JSON})
#Produces({MediaType.APPLICATION_JSON})
#Path("/data/services")
public Response DiscoverDevice(BlockDevsPost blockdevice) {
for (DeviceIdentifier device : blockdevice.getDevice()) {
String dev = device.Device();
System.out.println("DEVICE "+ dev);
if (dev == null || dev.equals("")){
return Response.status(Response.Status.BAD_REQUEST).entity("Device cannot be null or empty.").build();
}
}
}
Getting this error when fired POST from REST Client when dev is null. I am not able to get JSON and this error is thrown:
Unexpected character (D) at position 0. Device Identifier cannot be null or empty.
Where D in Device Identifier marked as Red which means it is not returning JSON as response.

Your client is expecting to get JSON but you have set a plain string in the Response entity and application/json as content-type. You need to return a valid JSON. For example
return Response
.status(Response.Status.BAD_REQUEST)
.entity("{\"error\":\"Device cannot be null or empty.\"}")
.build();
You can also build the json response string using your preferred mapper (you will need to add a dependency). This is an example using Jackson
Jackson using API
ObjectMapper mapper = new ObjectMapper();
ObjectNode objectNode = mapper.createObjectNode();
objectNode.put("error", "Device cannot be null or empty.");
String json = mapper.writeValueAsString(objectNode);
Jackson using POJO
class ErrorBean{
private String error;
//getters and setters
}
ObjectMapper mapper = new ObjectMapper();
ErrorBeanerrorBean = new ErrorBean();
errorBean.setError ("Device cannot be null or empty.");
String json = mapper.writeValueAsString(errorBean);
You can also return POJO from your service method and let the JAX-RS implementation to convert them to JSON (this means change the response type). See https://www.mkyong.com/webservices/jax-rs/json-example-with-jersey-jackson/