Bash - command substitution with variables

Question

I am still new to bash, so I was hoping someone more experience than me would be able to help me. I have the following bit of code:

DAY1=$(date +"%d/%b/%Y")
DAY2=$(date --date="-1 days" +"%d/%b/%Y")
...
DAY14=$(date --date="-13 days" +"%d/%b/%Y")
COUNT=0

What I would like to accomplish is to refer to the substituted commands as something like

$DAY($COUNT+1)

to produce variables from $DAY1 to $DAY14. That bit of code obviously doesn't work, but I needed some ideas on how to make this happen. I use a loop that increments the COUNT variable like so:

for i in $(seq 1 14); do
  let COUNT+=1
done

How about using an array?

EJK
– EJK

2013-12-09 01:24:47 +00:00
Commented Dec 9, 2013 at 1:24 — EJK
– EJK, Commented Dec 9, 2013 at 1:24

chepner · Accepted Answer · 2013-12-09 15:54:00Z

1

Don't use eval; it's a bad habit to get into. If you don't want to use an array, use indirect parameter expansion.

DAY1=$(date +"%d/%b/%Y")
DAY2=$(date --date="-1 days" +"%d/%b/%Y")
...
DAY14=$(date --date="-13 days" +"%d/%b/%Y")

COUNT=7
var="DAY$COUNT"
echo "${!var}"   # Displays the value of $COUNT7

answered Dec 9, 2013 at 15:54

chepner

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Comments

EJK · Accepted Answer · 2013-12-09 01:29:56Z

1

You could construct the variable expression and use eval to execute the command. For example:

for i in {1..14}; do
  let COUNT+=1

  command=echo DAY${COUNT}
  eval $command

done

Another approach, that might make more sense would be to use bash arrays.

answered Dec 9, 2013 at 1:29

EJK

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1 Comment

marat75 Over a year ago

I need to pass the variables into grep, so an array would probably make more sense here.

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