Jackson deserialize ISO8601 formatted date-time into Java8 Instant

问题

I'm trying to deserialize an ISO8601 formatted date into Java8 Instant using Jackson. I registered JavaTimeModule with the ObjectMapper, turned off WRITE_DATES_AS_TIMESTAMPS.

However, if one tries to deserialize 2016-03-28T19:00:00.000+01:00 it will not work, because it seems that JavaTimeModule will only deserialize date-times formatted with UTC timezone offset (e.g. 2016-03-28T18:00:00.000Z). I then tried using @JsonFormat annotation like this:

@JsonFormat(shape=JsonFormat.Shape.STRING, pattern="yyyy-MM-dd'T'HH:mm:ss.SSSZ", timezone = "UTC")

And like this:

@JsonFormat(shape=JsonFormat.Shape.STRING, pattern="yyyy-MM-dd'T'HH:mm:ss.SSSZ", timezone = JsonFormat.DEFAULT_TIMEZONE)

However, neither of these work and I get an exception:

com.fasterxml.jackson.databind.JsonMappingException: Unsupported field: YearOfEra (through reference chain: org.example.Article["date"])

Which implies that timezone parameter is ignored and date time formatter doesn't know how to format an Instant without a timezone.

Is there a way to deserialize a ISO8601 string that's not in UTC time zone offset to Java 8 Instant using Jackson and JavaTimeModule without writing custom deserializer?

回答1:

You need to set the explicit time zone via XXX in your modell class:

@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd'T'HH:mm:ss.SSSXXX")

(see: https://docs.oracle.com/javase/7/docs/api/java/text/SimpleDateFormat.html)

回答2:

If you want to serialize Date objects into ISO-8601, you don't need to specify a pattern at all - ISO-8601 is the default pattern. It is kind of mentioned in the JsonFormat Java doc:

Common uses include choosing between alternate representations -- for example, whether Date is to be serialized as number (Java timestamp) or String (such as ISO-8601 compatible time value) -- as well as configuring exact details with pattern() property.

[emphasasis mine] you should understand from the above text that specifying shape = STRING would mean an ISO-8601 format but you can choose something else using the pattern property.

In my experience, this always turns out a UTC date format (with the time zone rendered as +0000), which could be the default time zone in my VM (even though my operating system clock is not set to UTC).

回答3:

In Jackson 2.9.8 (current one as I'm writing this) it's better to use Instant instead of Date.

You have to add a dependency:

<dependency>
    <groupId>com.fasterxml.jackson.datatype</groupId>
    <artifactId>jackson-datatype-jsr310</artifactId>
    <version>2.9.8</version>
</dependency>

Also, register the module and configure SerializationFeature.WRITE_DATES_AS_TIMESTAMPS to false.

new ObjectMapper()
                .findAndRegisterModules()
                .configure(SerializationFeature.WRITE_DATES_AS_TIMESTAMPS, false);

More information about Jackson for Java8 here: https://github.com/FasterXML/jackson-modules-java8

回答4:

The format "Z" does not work with "+01:00" as this is a different pattern. JsonFormat is using SimpleDateFormat patterns. "Z" in upper case only represents strict RFC 822. You have to use syntax like: "+0100", without colon.

See: ISO 8601:2004, SimpleDateFormat patterns

来源：https://stackoverflow.com/questions/36252556/jackson-deserialize-iso8601-formatted-date-time-into-java8-instant

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