Splunk Search

How do I properly convert to UNIX time using strptime with this specific example?

Communicator

The new myTimefield is blank for some reason -- anyone know why? Consider the below code I'm using:

|makeresults
|eval originalTime = "01/05/19"
|eval myTime = strptime(originalTime, "%m/%d/%Y")
|table originalTime myTime

Oddly enough, it DOES work if I use |eval originalTime = "11:55" with eval myTime = strptime(originalTime, "%H:%M"). Why won't my original query work? Because of this, I'm unable to convert time to UNIX time in my CSVs.

0 Karma
1 Solution

SplunkTrust
SplunkTrust

Your string format time is wrong. Try this

|makeresults
 |eval originalTime = "01/05/19"
 |eval myTime = strptime(originalTime, "%m/%d/%y")
 |table originalTime myTime

Your upper case %Y is for year with century (i.e. 2019) . You specified without century (i.e. 19) which is %y

View solution in original post

Builder

It should be lowercase y in the format...

| makeresults 
| eval originalTime = "01/05/19" 
| eval myTime=strptime(originalTime,"%m/%d/%y")
| table originalTime myTime

With the strptime function, you must specify the time format of the string X so that the function can convert the string time into the correct UNIX time.
http://docs.splunk.com/Documentation/Splunk/7.2.1/SearchReference/DateandTimeFunctions#strptime.28X....

0 Karma

SplunkTrust
SplunkTrust

Your string format time is wrong. Try this

|makeresults
 |eval originalTime = "01/05/19"
 |eval myTime = strptime(originalTime, "%m/%d/%y")
 |table originalTime myTime

Your upper case %Y is for year with century (i.e. 2019) . You specified without century (i.e. 19) which is %y

View solution in original post

Communicator

Ahh you're right, I missed that! Thanks!