How to convert string to integer in Bash?

Question

I have found many answers about this on StackOverflow but I can't apply them to my code.

I used this command to get last day of current month:

LASTDAY=`cal $(date +"%m %Y") | grep . | fmt -1 | tail -1`

then I use this code:

for i in {1..${LASTDAY}}
do
    # code for processing here!
done

But always get this warning: line 12: [: {1..31}: integer expression expected

and i is {1..31} but I expected i is a number in range [1,31]

I have tried this:

LASTDAY=$((LASTDAY+0))

LASTDAY=$( echo "$LASTDAY - 0" | bc )

LASTDAY=$(printf "%d" "$LASTDAY")

but it can't solve this problem. What's wrong in my code? and how to fix it?

Thanks in advanced.

I'm not a bash head, but I think the range expression like {1..31} maybe a new feature to bash. What version are you using? bash --version, should output your version. Add that bit of information to your question above. Good luck. — shellter
– shellter, Commented Oct 1, 2013 at 3:30
You are right. However, I have run this expression many times but it is failed in the case I asked above. FYI, my bash version is: 3.2.25(1)-release (x86_64-redhat-linux-gnu) — PhatHV
– PhatHV, Commented Oct 1, 2013 at 3:44
Brace expansion has been in bash for a long time. However, it occurs prior to parameter expansion; since the braces do not contain two integer literals, the expression is treated literally as the string {1..${LASTDAY}}. Then parameter expansion is applied to produce {1..31}. That string is rejected by the code that receives $i, since that string is not an integer. — chepner
– chepner, Commented Oct 1, 2013 at 13:01

CS Pei · Accepted Answer · 2013-10-01 03:25:23Z

9

Use the following instead of for i in {1..$Lastday}

for i in $(seq 1 ${LASTDAY}) ; do  echo $i done

This will work.

answered Oct 1, 2013 at 3:25

CS Pei

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1 Comment

gavenkoa Over a year ago

Be careful about range length - some shells have limit on 64 KiB line length. In bash use may use usual for (( expr1 ; expr2 ; expr3 )) ; do list ; done

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