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 documentation):
for i in {1..5}; do echo $i; done
Which gives:
1
2
3
4
5
Yet, how can I replace either of the range endpoints with a variable? This doesn't work:
END=5
for i in {1..$END}; do echo $i; done
Which prints:
{1..5}
for i in $(seq 1 $END); do echo $i; doneedit: I prefer seq over the other methods because I can actually remember it ;)
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.