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Question

How to handle Julian time in parsers

  • October 27, 2025
  • 2 replies
  • 57 views

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Hey Everyone,

Have anyone handled Julian time format in parsers? I has “day of the year” and I am not sure how to handle this?

Julian Time format - 202528209:12:13

Description of the time: 2025(year)282(day of the year)09:12:13(time)

 

Let me know if anyone has handled this previously in parsers? or it is possible to handle or not?

 

Thank you,

2 replies

Eoved
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  • Bronze 1
  • October 28, 2025

This is possible.
You need to use the date filter plugin. First, grok the string 202528209:12:13 into a temporary field, then use the date filter to parse it with the specific format pattern yyyyDDDHH:mm:ss

filter {
grok {
match => { "message" => "^%{DATA:log_timestamp} %{GREEDYDATA:rest_of_message}" }
}
if [log_timestamp] {
date {
match => [ "log_timestamp", "yyyyDDDHH:mm:ss" ]
target => "event.idm.read_only_udm.metadata.event_timestamp"
on_error => "date_parse_failure"
}
}
}

You will find the date function listed under the section "Transform data using other functions”:
Parser syntax reference  |  Google Security Operations  |  Google Cloud Documentation


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  • Author
  • New Member
  • October 28, 2025

Hi,

 

I have already tried this yyyyDDDHH:mm:ss pattern but it is not working. “DDD” is not working for “day of year”.