Why does mktime return -1 in Awk?

Question

I have file with below data

# date format in file is (YYYY MM DD HH MM SS)
 file_create_date       file_execute_date       file_name     
"2017 10 10 12 24 30" , "2017 10 11 09 23 00" , LTE--cNum--T201710110923--W101114450--CHGLSMR01LEMS1--sam--4.0.csv.gz

i want output like this

 file_create_date       file_execute_date       difference          file_name     
"2017 10 10 12 20 30" , "2017 10 11 13 31 59" ,  "1 Day 01:11:29, "LTE--cNum--T201710110923--W101114450--CHGLSMR01LEMS1--sam--4.0.csv.gz"

my script is :

awk -F',' '{print $1 "," , $2  , (mktime($2)-mktime($1))/86400 " Day",$3}' Filename

But mktime() returns -1.

Could you please use code tags for your samples as it is not at all clear from this. — RavinderSingh13, Nov 02 '17 at 11:05

Ed Morton · Answer 1 · 2017-11-02T11:23:19.730

You're passing mktime a string containing double quotes and blank chars at the start and end. Remove those.

$ echo '"2017 10 10 12 24 30" , "2017 10 11 09 23 00"'
"2017 10 10 12 24 30" , "2017 10 11 09 23 00"

$ echo '"2017 10 10 12 24 30" , "2017 10 11 09 23 00"' | awk -F',' '{print mktime($1)}'
-1

$ echo '"2017 10 10 12 24 30" , "2017 10 11 09 23 00"' | awk -F',' '{gsub(/\s*"\s*/,""); print mktime($1)}'
1507656270

Why does mktime return -1 in Awk?

1 Answers1