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How to use jackson in Java if a json can represent 2 classes?

I'm using Jackson to deserialize objects in Java, e.g. here's one of hello world examples:
Foo foo = new ObjectMapper().readValue(jsonFoo, Foo.class);
There's a problem with my json string, though: it may be either
{
"error":{
"code":"404",
"message":"Not Found"
}
}
or
[{fooFields}, {fooFields}]
So usually for the latter case, I would use:
Foo[] foos = new ObjectMapper().readValue(jsonFoo, []Foo.class);
or
Error error = new ObjectMapper().readValue(jsonFoo, Error.class);
How can I try to parse jsonFoo to either Error or [Foo] and return another custom Error object if there's an exception?
Currently, I only managed to come up with a nested try-catch construction which doesn't look really nice.
try {
Foo[] foos = new ObjectMapper().readValue(jsonFoo, []Foo.class);
return new Bar(foos);
catch {
try {
Error error = new ObjectMapper().readValue(jsonFoo, Error.class);
return new Bar(error);
catch (IOException e)... // new Bar(InvalidInputError);
}
return new Bar(InvalidInputError);
Is there a way to rewrite it in a more concise way?

I usually use JsonNode.
public static void main(String[] args) throws IOException {
ObjectMapper objectMapper = new ObjectMapper();
String jsonStr1 = "[{\"key\":\"value\"}]";
JsonNode jsonNode1 = objectMapper.readTree(jsonStr1);
boolean array = jsonNode1.isArray();
String jsonStr2 = "{\"key\":\"value\"}";
JsonNode jsonNode2 = objectMapper.readTree(jsonStr2);
boolean object = jsonNode2.isObject();
}
More information

How to enforce a date value input in a REST API?

The user needs to make a POST to /api/date with something like March 13, 2019 or 08/19/2020. As long as it's a date, it should be accepted.
I have something like this (Using Dropwizard framework)
#POST
public void post(String date)
{
validateDate(date);
//continue
}
private void validateDate(String date)
{
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("yyyy-MM-dd");
try
{
LocalDateTime.parse(date, formatter);
}
catch (DateTimeParseException e)
{
//not a date
}
}
I'm not sure if I'm in the right approach, there must be a better way to validate strings as dates.

You can accept multiple formats for a date time using the optional syntax ([<your format>])*. eg
DateTimeFormatter formatter = DateTimeFormatter.ofPattern(
"[yyyy-MM-dd][dd-MM-yyyy][MMMM dd, yyyy]");

EDIT: It is not really clear if you want to know how to validate dates correctly or how to handle invalid inputs to your REST API. My answer shows the latter.
You should use a return value for your post method. You can return javax.ws.rs.core.Response, with that you can control the HTTP code and response object you want to return.
On success, you would normally return the created object with a 200 success code.
On failure, you would return an error code (like 400 Bad request) with a detailed error message ("Date must be in the format yyyy-MM-dd").
To create the response, you can use the ResponseBuilder.
Example:
Response.ok( yourObject ).build(); //success
Response.status( Status.BAD_REQUEST ).entity( yourErrorMessageObject ).build(); // failure
So I would change the code to this:
#POST
public Response post(String date)
{
if(!isDateValid(date)){
return Response.status( Status.BAD_REQUEST ).entity( buildErrorMessage()).build();
}
//continue
Response.ok().build(); // returns nothing on success (like void)
}
private boolean isDateValid(String date)
{
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("yyyy-MM-dd");
try
{
LocalDateTime.parse(date, formatter);
return true;
}
catch (DateTimeParseException e)
{
//not a date
return false;
}
}

DynamoDB - how to get Primary Key (which is random id) from database to make endpoind class?

I've made method that I use to edit Item from database.
This is how my method looks:
public Product editProduct(PrimaryKey primaryKey, Product content) {
UpdateItemSpec updateItemSpec = new UpdateItemSpec().withPrimaryKey(primaryKey).withValueMap(createValueMap(content));
UpdateItemOutcome itemOutcome = databaseController.getTable(PRODUCT_TABLE).updateItem(updateItemSpec);
return convertToProduct(itemOutcome);
}
private Map<String, Object> createValueMap(Product content) {
Map<String, Object> result = new HashMap<>();
result.put("name", content.getName());
result.put("calories", content.getCalories());
result.put("fat", content.getFat());
result.put("carbo", content.getCarbo());
result.put("protein", content.getProtein());
result.put("productKinds", content.getProductKinds());
result.put("author", content.getAuthor());
result.put("media", content.getMedia());
result.put("approved", content.getApproved());
return result;
}
private Product convertToProduct(UpdateItemOutcome itemOutcome) {
Product product = new Product();
product.setName(itemOutcome.getItem().get("name").toString());
product.setCalories(itemOutcome.getItem().getInt("calories"));
product.setFat(itemOutcome.getItem().getDouble("fat"));
product.setCarbo(itemOutcome.getItem().getDouble("carbo"));
product.setProtein(itemOutcome.getItem().getDouble("protein"));
product.setProductKinds(itemOutcome.getItem().getList("productKinds"));
ObjectMapper objectMapper = new ObjectMapper();
try {
Author productAuthor = objectMapper.readValue(itemOutcome.getItem().getString("author"), Author.class);
product.setAuthor(productAuthor);
} catch (IOException e) {
e.printStackTrace();
}
try {
Media productMedia = objectMapper.readValue(itemOutcome.getItem().getString("media"), Media.class);
product.setMedia(productMedia);
} catch (IOException e) {
e.printStackTrace();
}
return product;
}
Now I want to create endpoint class for this method but I have problem, I need to get primarykey as parameter (it's looks like this for example: 2567763a-d21e-4146-8d61-9d52c2561fc0) and I don't know how to do this.
At the moment my class looks like that:
public class EditProductLambda implements RequestHandler<Map<String, Object>, ApiGatewayResponse> {
private LambdaLogger logger;
#Override
public ApiGatewayResponse handleRequest(Map<String, Object> input, Context context) {
logger = context.getLogger();
logger.log(input.toString());
try{
Product product = RequestUtil.parseRequest(input, Product.class);
//PrimaryKey primaryKey = XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
KitchenService kitchenService = new KitchenService(new DatabaseController(context, Regions.EU_CENTRAL_1), logger);
Product editedProduct = kitchenService.editProduct(primaryKey, product);
return ResponseUtil.generateResponse(HttpStatus.SC_CREATED, editedProduct);
} catch (IllegalArgumentException e){
return ResponseUtil.generateResponse(HttpStatus.SC_BAD_REQUEST, e.getMessage());
}
}
Can someone give me some advice how to do that? Or maybe my method is done wrong?

So first you have to create a trigger to Lambda function and ideal prefer here would be an API gateway. You can pass your data as query string or as a request body to API gateway.
You can use body mapping template in the integration request section of API gateway and get request body/query string. Construct a new json at body mapping template, which will have data from request body/query string. As we are adding body mapping template your business logic will get the json we have constructed at body mapping template.
Inside body mapping template to get query string please do ,
$input.params('querystringkey')
For example inside body mapping template (If using query string),
#set($inputRoot = $input.path('$'))
{
"primaryKey" : "$input.params('$.primaryKey')"
}
if passing data as body then,
#set($inputRoot = $input.path('$'))
{
"primaryKey" : "$input.path('$.primaryKey')"
}
Please read https://aws.amazon.com/blogs/compute/tag/mapping-templates/ for more details on body mapping template

Can't convert from string to Joda LocalTime with DefaultFormattingConversionService

Am unable to convert string to Joda LocalTime with DefaultFormattingConversionService.
If I pass time as "12:00" it says time is too short, but if I pass it as "12:00:00", it says it is malformed.
import org.joda.time.LocalTime;
import org.springframework.format.support.DefaultFormattingConversionService;
public class SpringLocalTimeFormatterTry {
public static void main(String[] args) {
DefaultFormattingConversionService service = new DefaultFormattingConversionService();
try {
System.out.println(service.convert("12:00", LocalTime.class));
}
catch (Exception e) {
System.out.println(e.getMessage());
}
try {
System.out.println(service.convert("12:00:00", LocalTime.class));
}
catch (Exception e) {
System.out.println(e.getMessage());
}
}
}
How to use it correctly or fix?

The vanilla settings of DefaultFormattingConversionService use platform default locale, which, I assume from the error, are the same as mine, ie. English. That means, that for time you need to add the AM/PM indicator. This works for me:
System.out.println(service.convert("10:12 am", LocalTime.class));
>> 10:12:00.000
To handle your desired time format, you can add an extra converter:
service.addConverter(new Converter<String, LocalTime>() {
#Override
public LocalTime convert(String source) {
return LocalTime.parse(source);
}
});
Then, both examples pass:
System.out.println(service.convert("12:00", LocalTime.class));
>> 12:00:00.000
System.out.println(service.convert("12:00:00", LocalTime.class));
>> 12:00:00.000
You can skip registering the default converters by creating the service with
new DefaultFormattingConversionService(false);
Finally, I assume in the production code you are actually getting the ConversionService from the ConversionServiceFactoryBean, so you can configure that as follows:
#Bean
public ConversionServiceFactoryBean conversionService() {
ConversionServiceFactoryBean conversionServiceFactoryBean = new ConversionServiceFactoryBean();
Set<Converter<?, ?>> myConverters = new HashSet<>();
myConverters.add(new Converter<String, LocalTime>() {
#Override
public LocalTime convert(String source) {
return LocalTime.parse(source);
}
});
conversionServiceFactoryBean.setConverters(myConverters);
return conversionServiceFactoryBean;
}

Try this:
DateTimeFormatter dtf = DateTimeFormat.forPattern("HH:mm:ss");
LocalTime localTime = dtf.parseLocalTime("12:00:00");
System.out.println("Time"+localTime);

How to store Date field as ISODate() using jackson in MongoDb

I am trying to persist a java object having java.util.Date field in mongo collection using fasterxml jackson.
The problem is the default nature of objectMapper is to store Date as NumberLong type.
For e.g , a createdTime field of java.util.Date type gets stored as below:
"createdTime" : NumberLong("1427728445176")
I want to store it in ISODate format which is available in mongo Shell.
Now, i know there is way to format object mapper to store Date in a String dateformat.
But I am ONLY looking for ISODate() format.
For e.g
"createdTime" : ISODate("2015-01-20T16:39:42.132Z")
Is there a way to do that ?
Please advise gurus .
Thanks in advance for help.

What you need is the Jackson Joda Module. If you import that into your classpath, you can do the following on your mapper to write it as your desired Timestamp:
ObjectMapper mapper = new ObjectMapper();
mapper.registerModule(new JodaModule());
mapper.configure(SerializationFeature.WRITE_DATES_AS_TIMESTAMPS, true);
mapper.writeValueAsString(date);
You can replace date in the code sample above with your POJO as necessary.
Edit:
It looks like what you really want is a custom serializer. That would look something like this:
public class IsoDateSerializer extends JsonSerializer<DateTime> {
#Override
public void serialize(DateTime value, JsonGenerator jgen, SerializerProvider provider) {
String isoDate = ISODateTimeFormat.dateTime().print(value);
jgen.writeRaw("ISODATE(\"" + isoDate + "\")");
}
Then you'll either register it on the mapper for all DateTime types
mapper.addSerializer(DateTime.class, new IsoDateSerializer());
or specify it on the function using annotations
#JsonSerializer(using = IsoDateSerializer.class)
public DateTime createdTime;

I was able to serialize the date string to ISODate format. I wrote a customer date serializer like below.
public void serialize(Date date, JsonGenerator jgen, SerializerProvider provider) throws IOException {
String dateValue = getISODateString(date);
String text = "{ \"$date\" : \""+ dateValue +"\"}";
jgen.writeRawValue(text);
}
Based on request from user #mmx73, I am adding code for customer Date DeSeriaizer.
public class IsoDateDeSerializer extends JsonDeserializer<Date> {
#Override
public Date deserialize(JsonParser jsonParser, DeserializationContext deserializationContext)
throws IOException, JsonProcessingException {
ObjectCodec oc = jsonParser.getCodec();
JsonNode node = oc.readTree(jsonParser);
String dateValue = node.get("$date").asText();
//DateFormat df = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss'Z'");
DateFormat df = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSS'Z'");
Date date = null;
try {
date = df.parse(dateValue);
} catch (ParseException e) {
e.printStackTrace();
}
return date;
}
}

None of these answers accomplished what I wanted. I was having trouble because when I serialized the JSON string to MongoDB, it was stored as a String. A nicely formatted string, but a string none the less.
I use the com.fasterxml.jackson.databind.ObjectMapper to convert my objects to/from JSON and I wanted to continue using this class. I have the following method:
public enum JsonIntent {NONE, MONGODB};
public static ObjectMapper getMapper(final JsonIntent intent) {
ObjectMapper mapper = new ObjectMapper();
// Setting to true saves the date as NumberLong("1463597707000")
// Setting to false saves the data as "2016-05-18T19:30:52.000+0000"
mapper.configure(SerializationFeature.WRITE_DATES_AS_TIMESTAMPS, false);
mapper.registerModule(new JodaModule());
if (intent == JsonIntent.MONGODB) {
// If you want a date stored in MONGO as a date, then you must store it in a way that MONGO
// is able to deal with it.
SimpleModule testModule = new SimpleModule("MyModule", new Version(1, 0, 0, null, null, null));
testModule.addSerializer(Date.class, new StdSerializer<Date>(Date.class) {
private static final long serialVersionUID = 1L;
#Override
public void serialize(Date value, JsonGenerator jgen, SerializerProvider provider) throws IOException {
try {
if (value == null) {
jgen.writeNull();
} else {
jgen.writeStartObject();
jgen.writeFieldName("$date");
String isoDate = ISODateTimeFormat.dateTime().print(new DateTime(value));
jgen.writeString(isoDate);
jgen.writeEndObject();
}
} catch (Exception ex) {
Logger.getLogger(JsonUtil.class.getName()).log(Level.SEVERE, "Couldn't format timestamp " + value + ", writing 'null'", ex);
jgen.writeNull();
}
}
});
testModule.addDeserializer(Date.class, new StdDeserializer<Date>(Date.class) {
private static final long serialVersionUID = 1L;
#Override
public Date deserialize(JsonParser jp, DeserializationContext dc) throws IOException, JsonProcessingException {
JsonNode tree = jp.readValueAsTree();
SimpleDateFormat dateFormat = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ssZ");
try {
return ISODateTimeFormat.dateTime().parseDateTime(tree.get("$date").textValue()).toDate();
} catch (Throwable t) {
throw new IOException(t.getMessage(), t);
}
}
});
mapper.registerModule(testModule);
}
return mapper;
}
Now, I can run the following test code:
BObjectMapper mapper = getMapper(JsonUtil.JsonIntent.NONE);
Date d1 = new Date();
String v = mapper.writeValueAsString(d1);
System.out.println("Joda Mapping: " + v);
Date d2 = mapper.readValue(v, Date.class);
System.out.println("Decoded Joda: " + d2);
mapper = getMapper(JsonUtil.JsonIntent.MONGODB);
v = mapper.writeValueAsString(d1);
System.out.println("Mongo Mapping: " + v);
d2 = mapper.readValue(v, Date.class);
System.out.println("Decoded Mongo: " + d2);
The output is as follows:
Joda Mapping: "2016-06-13T14:58:11.937+0000"
Decoded Joda: Mon Jun 13 10:58:11 EDT 2016
Mongo Mapping: {"$date":"2016-06-13T10:58:11.937-04:00"}
Decoded Mongo: Mon Jun 13 10:58:11 EDT 2016
Note that the JSON that will be sent to MONGODB defines the value containing a field named "$date". This tells MongoDB that this is a date object it seems.
When I look at Mongo, I see the following:
"importDate" : ISODate("2016-05-18T18:55:07Z")
Now, I can access the field as a date rather than as a string.
To add an encoded JSON string to Mongo, my code is as follows:
MongoDatabase db = getDatabase();
Document d = Document.parse(json);
db.getCollection(bucket).insertOne(d);
In this case, "json" is the encoded JSON string. Because it is coming from a JSON string, it has no way of knowing the types unless it infers this, which is why we needed the "$date" portion. The "bucket" is just a string indicating which table to use.
As a side note, I found out that if I pull a BSON object from Mongo and convert it to a JSON string by calling doc.toJson() (where doc is of type org.bison.Document as returned from a query), the date object is stored with a long value rather than a formatted text string. I did not check to see if I could push data into mongo after formatting in this way, but, you can modify the deserializer shown above to support this as follows:
testModule.addDeserializer(Date.class, new StdDeserializer<Date>(Date.class) {
private static final long serialVersionUID = 1L;
#Override
public Date deserialize(JsonParser jp, DeserializationContext dc) throws IOException, JsonProcessingException {
JsonNode tree = jp.readValueAsTree();
SimpleDateFormat dateFormat = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ssZ");
try {
// Mongo will return something that looks more like:
// {$date:<long integer for milliseconds>}
// so handle that as well.
JsonNode dateNode = tree.get("$date");
if (dateNode != null) {
String textValue = dateNode.textValue();
if (!Util.IsNullOrEmpty(textValue)) {
return ISODateTimeFormat.dateTime().parseDateTime(textValue).toDate();
}
return Util.MillisToDate(dateNode.asLong());
}
return null;
} catch (Throwable t) {
Util.LogIt("Exception: " + t.getMessage());
throw new IOException(t.getMessage(), t);
}
}
});
You can convert milliseconds to a Date or DateTime as follows:
/**
* Convert milliseconds to a date time. If zero or negative, just return
* null.
*
* #param milliseconds
* #return
*/
public static Date MillisToDate(final long milliseconds) {
if (milliseconds < 1) {
return null;
}
Calendar calendar = Calendar.getInstance();
calendar.setTimeInMillis(milliseconds);
return calendar.getTime();
}
public static DateTime MillisToDateTime(final long milliseconds) {
if (milliseconds < 1) {
return null;
}
return new DateTime(milliseconds);
}

In case you get some message like
com.fasterxml.jackson.core.JsonGenerationException: Can not write a field name, expecting a value.
Make sure you use writeRawValue instead within the accepted answer. That ends the field correctly, otherwise the next field to be serialize may throw this error.

You can solve this issue by reading/writing bson instead of json. Here is a test class:
package com.nagra.jongo.mapper;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.module.SimpleModule;
import de.undercouch.bson4jackson.BsonFactory;
import de.undercouch.bson4jackson.deserializers.BsonDateDeserializer;
import de.undercouch.bson4jackson.serializers.BsonDateSerializer;
import org.junit.Assert;
import org.junit.Before;
import org.junit.Test;
import java.io.IOException;
import java.util.Date;
/**
* Uses Bson4Jackson 2.9.0
*
* <!-- https://mvnrepository.com/artifact/de.undercouch/bson4jackson -->
* <dependency>
* <groupId>de.undercouch</groupId>
* <artifactId>bson4jackson</artifactId>
* <version>2.9.2</version>
* </dependency>
*/
public class ObjectMapperTest {
private ObjectMapper mapper = new ObjectMapper(new BsonFactory());
private static class WrappedDate {
private Date Date = new Date(0);
public WrappedDate() {
}
public Date getDate() {
return Date;
}
public void setDate(Date Date) {
this.Date = Date;
}
}
#Before
public void setUp() {
SimpleModule module = new SimpleModule();
module.addSerializer(Date.class, new BsonDateSerializer());
module.addDeserializer(Date.class, new BsonDateDeserializer());
mapper.registerModule(module);
}
#Test
public void testDate() throws IOException {
WrappedDate date = new WrappedDate();
byte[] b = mapper.writeValueAsBytes(date);
WrappedDate i = mapper.readValue(b, WrappedDate.class);
Assert.assertEquals(date.getDate(), i.getDate());
System.out.println(i.getDate());
}}