How to convert an input parameter to integer value in a for loop in bash?

Question

in my bash script I try to use a number as an input variable for a for loop I run the script as

./script.sh InputFolder/ Number_of_iterations

the script should work inside the given folder and run a for loop as many times as the Number_of_iterations variable is set to. But somehow I can't set the variable as an integer.

this is an example of my loop in the script:

for i in {1..$(($2))}
do
 echo "Welcome $i times"
done

I have tried already the double brackets $(($...)) option as well as the double quotations "...", but the output I keep getting is

Welcome {1..5} times

which make me think, this is not an integer. I would appreciate any help in reading the input parameter as an integer into the script.

If you insist on the {1..$ construct you can use something similar to this: for i in $(eval echo "{1..$2}"); do — Mikael Kjær, Jan 22 '18 at 10:29

score 11 · Accepted Answer · edited Feb 19 '22 at 06:45

11

You can do this two ways:

With ksh93-compatible shells (ksh93, zsh, bash):

for (( i=1;i<=$2;i++ ))
do
 echo "Welcome $i times"
done

Here we set i to 1 and loop, incrementing it until it is less than or equal to $2, outputting:

Welcome 1 times
Welcome 2 times

With POSIX shells on GNU systems:

for i in $(seq "$2")
do
 echo "Welcome $i times"
done

The seq command (GNU specific) will output numbers from 1 to the number specified in $2 on separate lines. Assuming you've not modified $IFS (which by default contains the line delimiter character), the command substitution will split that into as many elements for for to loop on.

edited Feb 19 '22 at 06:45

Joseph Cass

3

answered Jan 22 '18 at 10:22

Raman Sailopal

1,432

thanks, this has solved my problem. I have used the second option. – Assa Yeroslaviz Jan 22 '18 at 10:59
first one is not working in Ubuntu18.06 – Samuel Jan 06 '21 at 07:40

score 3 · Answer 2 · 2018-01-22T10:30:00.447

3

In Bash this construct that you're trying to use: {1..$(($2))} or {1..$2} is the brace expansion. The reason why it doesn't work for you is that in the order of shell expansions the brace expansion is the first one. So when this expansion takes place $2 is still only $2, not 5 as you'd like. The brace expansion doesn't work then. It simply produces 1..$2, and then when it's time to expand $2, the shell does so, and you're left with this one word:

{1..5}

The braces are still there because they have not been consumed by the brace expansion.

As for solutions, see Raman Sailopal's answer.

edited Jan 22 '18 at 10:30

answered Jan 22 '18 at 10:27

This depends on the shell. In ksh93, {1..$2} would work without issues. – Kusalananda Jan 22 '18 at 10:29
... and that's ok since brace expansions are a non-standard extension, so shells can choose to do them in any way they see fit. – Kusalananda Jan 22 '18 at 10:33
@Kusalananda Is the order of expansions the same in all shells? – Jan 22 '18 at 10:34
With regards to the standard syntax and grammar, I would think so. But I have nothing backing that up. – Kusalananda Jan 22 '18 at 10:35
@Kusalananda, also in zsh (where the syntax comes from) and yash. bash is the only shell (with support for {x..y}) where that doesn't work AFAIK. – Stéphane Chazelas Jan 22 '18 at 10:39
@Kusalananda Is it a reasonable question to ask? – Jan 22 '18 at 10:39
@tomasz, it's been asked many times before. It's a FAQ here – Stéphane Chazelas Jan 22 '18 at 10:39
@StéphaneChazelas Also shells like pdksh that has brace expansion but only supports alternation and not sequences. But that's nit-picking. – Kusalananda Jan 22 '18 at 10:43

How to convert an input parameter to integer value in a for loop in bash?

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