Can't input date variable in bash

Question

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

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:

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:

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

But not getting required output.

Tried bash script:

#!/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.

See: Difference between single and double quotes in bash

Cyrus
– Cyrus

2018-09-30 09:01:02 +00:00
Commented Sep 30, 2018 at 9:01 — Cyrus
– Cyrus, Commented Sep 30, 2018 at 9:01

bbichero · Accepted Answer · 2018-09-30 10:40:51Z

1

As @cyrus said you must use double quotes in your variable assignment because simple quote are use only for string and not for containing variables.

Bas use case

number=10
string='I m sentence with or wihtout var $number'
echo $string

Correct use case

number=10
string_with_number="I m sentence with var $number"
echo $string_with_number

You can use simple quote but not englobe all the string

number=10
string_with_number='I m sentence with var '$number
echo $string_with_number

answered Sep 30, 2018 at 10:40

bbichero

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1 Comment

glenn jackman Over a year ago

If you're taking the variable out of single quotes, put the variable in double quotes: string_with_number='I m sentence with var '"$number"

glenn jackman · Accepted Answer · 2018-09-30 13:03:16Z

0

Don't parse ls

You don't need awk for this: you can manage with the shell's capabilities

for file in report.active_user."$(date "+%d%m%Y")"*; do
    tmp=${file%.*}      # remove the extension
    number=${tmp##*.}   # remove the prefix up to and including the last dot
    echo "$number"
done

See https://www.gnu.org/software/bash/manual/bashref.html#Shell-Parameter-Expansion

answered Sep 30, 2018 at 13:03

glenn jackman

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