topic Is there a better way to convert the age output into a more readable format (i.e. days)? in Getting Data In

Is there a better way to convert the age output into a more readable format (i.e. days)?

kppradhan — Fri, 27 Mar 2020 17:03:39 GMT

Hello,

I'm attempting to track AWS related password events in my Splunk.

I am sifting through my index and receiving the data I need -- however I am having an issue with converting the "age" from a unix based notation.

I am using the following to determine the age of passwords:

| eval age =_time

My output is as follows:

PasswordLastUsed                                                                            age
018448995162    user    2020-02-14T20:49:08+00:00   1585319203
018448995162    user    2020-02-13T16:59:30+00:00   1585319203

Is there a better way to convert the age output into a more readable format (i.e. days)?

Thanks,

Kiran

Re: Is there a better way to convert the age output into a more readable format (i.e. days)?

richgalloway — Fri, 27 Mar 2020 19:16:04 GMT

The _time field is the date and time when the event occurred. It's probably not the best choice for password age.

That said, use the strftime function to make epochs readable.

... | eval age = strftime(_time, "%Y-%m-%d %H:%M:S")

You can use the convert command, instead.

... | convert timeformat="%Y-%m-%d %H:%M:S" ctime(_time) as age

Re: Is there a better way to convert the age output into a more readable format (i.e. days)?

woodcock — Tue, 31 Mar 2020 06:52:12 GMT

Like this:

... | eval age = now() - _time
| fieldformat age = tostring(age, "duration")