CODEWITHSUNDEEP
bashglob
sergiers
sergiers

Reputation: 31

How to print a list of regex matches using variables and asterisk

I have this list of files:

perspectium-02-16-2019-1.log.gz  perspectium-07-27-2019-3.log.gz  perspectium-10-18-2019-1.log.gz
perspectium-07-27-2019-1.log.gz  perspectium-10-16-2019-1.log.gz  perspectium-10-18-2019-2.log.gz
perspectium-07-27-2019-2.log.gz  perspectium-10-17-2019-1.log.gz  perspectium-10-18-2019-3.log.gz

I have calculated one date:

d=$(date -d '-1 day' '+%m-%d-%Y')

I want to delete those files that matches the condition:

perspectium-$d-*.log.gz

But if I make an echo of this, it will print the asterisk literally, so It wont match any file:

perspectium-10-17-2019-*.log.gz

Can you please give me some clue? Thanks in advice.

Upvotes: 0

Views: 56

Answers (1)

arheops
arheops

Reputation: 15259

You can use following:

output=`ls perspectium-$d-*.log.gz`
echo $output

or

for file in perspectium-$d-*.log.gz
do
echo $file
done;

or just do without quotes:

echo perspectium-$d-*.log.gz

But as already said, care about possible issues with incorrect matching.

Here is output for you case:

[root@pro-sip tmp]# d=10-18-2019
[root@pro-sip tmp]# ls perspectium-$d-*.log.gz
perspectium-10-18-2019-1.log.gz  perspectium-10-18-2019-2.log.gz
[root@pro-sip tmp]# for i in  perspectium-$d-*.log.gz  
> do
> echo $i
> done;
perspectium-10-18-2019-1.log.gz
perspectium-10-18-2019-2.log.gz
[root@pro-sip tmp]# echo perspectium-$d-*.log.gz
perspectium-10-18-2019-1.log.gz perspectium-10-18-2019-2.log.gz

Upvotes: 1

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