How do I iterate over a range of numbers defined by variables in Bash?

Question

How do I iterate over a range of numbers in Bash when the range is given by a variable?

I know I can do this (called \"sequence expression\" in the Bash documentatio

Answer 1

The POSIX way

If you care about portability, use the example from the POSIX standard:

i=2
end=5
while [ $i -le $end ]; do
    echo $i
    i=$(($i+1))
done

Output:

2
3
4
5

Things which are not POSIX:

• (( )) without dollar, although it is a common extension as mentioned by POSIX itself.
• [[. [ is enough here. See also: What is the difference between single and double square brackets in Bash?
• for ((;;))
• seq (GNU Coreutils)
• {start..end}, and that cannot work with variables as mentioned by the Bash manual.
• let i=i+1: POSIX 7 2. Shell Command Language does not contain the word let, and it fails on bash --posix 4.3.42

• the dollar at i=$i+1 might be required, but I'm not sure. POSIX 7 2.6.4 Arithmetic Expansion says:

  If the shell variable x contains a value that forms a valid integer constant, optionally including a leading plus or minus sign, then the arithmetic expansions "$((x))" and "$(($x))" shall return the same value.

  but reading it literally that does not imply that $((x+1)) expands since x+1 is not a variable.

Answer 2

The seq method is the simplest, but Bash has built-in arithmetic evaluation.

END=5
for ((i=1;i<=END;i++)); do
    echo $i
done
# ==> outputs 1 2 3 4 5 on separate lines

The for ((expr1;expr2;expr3)); construct works just like for (expr1;expr2;expr3) in C and similar languages, and like other ((expr)) cases, Bash treats them as arithmetic.

Answer 3

If you're doing shell commands and you (like I) have a fetish for pipelining, this one is good:

seq 1 $END | xargs -I {} echo {}

Answer 4

Another layer of indirection:

for i in $(eval echo {1..$END}); do
    ∶

Answer 5

If you need it prefix than you might like this

 for ((i=7;i<=12;i++)); do echo `printf "%2.0d\n" $i |sed "s/ /0/"`;done

that will yield

07
08
09
10
11
12

Answer 6

I know this question is about bash, but - just for the record - ksh93 is smarter and implements it as expected:

$ ksh -c 'i=5; for x in {1..$i}; do echo "$x"; done'
1
2
3
4
5
$ ksh -c 'echo $KSH_VERSION'
Version JM 93u+ 2012-02-29

$ bash -c 'i=5; for x in {1..$i}; do echo "$x"; done'
{1..5}