Cant input date variable in awk

nish43 · 09-30-2018, 06:52 AM

I have a directory /user/reports under which many files are there, one of them is :

Code:

report.active_user.30092018.77325.csv

I need output as number after date i.e. 77325 from above file name.

I created below command to find a value from file name:

Code:

ls /user/reports | awk -F. '/report.active_user.30092018/ {print $(NF-1)}'

Now, I want current date to be passed in above command as variable and get result:

Code:

ls /user/reports | awk -F. '/report.active_user.$(date +'%d%m%Y')/ {print $(NF-1)}'

But not getting required output.

Tried bash script:

Code:

#!/usr/bin/env bash

_date=`date +%d%m%Y`

 active=$(ls /user/reports | awk -F. '/report.active_user.${_date}/ {print $(NF-1)}')


 echo $active

But still output is blank.

Please help with proper syntax.

BW-userx · 09-30-2018, 08:51 AM

something like this maybe?

Code:

#shows files in directory
userx@slackcurrent:~/testdir$ for i in $(ls) ; do echo $i ; done
report.active_user.30092018.77320.csv
report.active_user.30092018.77321.csv

shows results in getting just the digits to the right of date

userx@slackcurrent:~/testdir$ for i in $(ls) ; do f=${i%.*} ; f=${f##*.} ; echo $f ; done
77320
77321

Turbocapitalist · 09-30-2018, 09:19 AM

Also keep track of the quoting. The single quotes ' ' will provide the contents literally. The double quotes " " will have their contents interpreted. Compare "$a" versus '$a' in the shell.
With double quotes you'll have to escape any dollar signs that you want passed to AWK as literals. Either way enclosing the same quotes inside the same quotes needs escaping too, as in "\"\"" and '\'\''

Maybe pass the date as a variable using -v or --assign

Code:

awk -v d=$(date +"%d%m%Y") 'BEGIN{p="report.active_user."d} $0~p{ print $(NF-1)}' nish43-01.data

grail · 10-01-2018, 02:17 AM

Or you could try and learn how awk deals with dates:

https://www.gnu.org/software/gawk/ma...Time-Functions

Turbocapitalist · 10-01-2018, 02:33 AM

Quote:

Originally Posted by grail

Or you could try and learn how [g]awk deals with dates:

It's not portable:

Quote:

They are gawk extensions; they are not specified in the POSIX standard.

However, that may not matter in some cases.

nish43 · 10-01-2018, 03:26 AM

Quote:

Originally Posted by BW-userx

something like this maybe?

Code:

#shows files in directory
userx@slackcurrent:~/testdir$ for i in $(ls) ; do echo $i ; done
report.active_user.30092018.77320.csv
report.active_user.30092018.77321.csv

shows results in getting just the digits to the right of date

userx@slackcurrent:~/testdir$ for i in $(ls) ; do f=${i%.*} ; f=${f##*.} ; echo $f ; done
77320
77321

This worked .. Thanks everyone for quick response.

KenJackson · 10-02-2018, 10:42 AM

Your original attempt should work just fine if you reposition two single-quotes.

Code:

ls /user/reports | awk -F. '/report.active_user.'$(date +%d%m%Y)'/ {print $(NF-1)}'