What should be @JsonFormat(pattern="") in Spring-boot for given example

Question

Using Spring-boot RESTful service,in that receiving JSON file which contain DateTime in following format "2017-08-16T16:08:25.000Z" and JsonFormat tries are e.g

@JsonFormat(pattern="yyyy-MM-dd HH:mm") 
@JsonFormat(JsonFormat.Shape.STRING,pattern="yyyy-MM-dd@HH:mm:ss.SSSZ")

but none of format match accepting. jar used as shown below:

import com.fasterxml.jackson.annotation.JsonFormat;

so what should be JSON format pattern to accept such json data -Date with Time

Answer 1

You need to read the javadocs carefully. Quoting from the javadoc of @JsonFormat

• Date: Shape can be JsonFormat.Shape.STRING or JsonFormat.Shape.NUMBER; pattern may contain SimpleDateFormat-compatible pattern definition.

and quoting from the javadoc of SimpleDateFormat:

Within date and time pattern strings, unquoted letters from 'A' to 'Z' and from 'a' to 'z' are interpreted as pattern letters [...].
Text can be quoted using single quotes (') to avoid interpretation.

Hence you need to use 'T' instead of T in the pattern for @JsonFormat:

pattern="yyyy-MM-dd'T'HH:mm:ss.SSSZ"