How to produce a range with step n in bash? (generate a sequence of numbers with increments)

问题

The way to iterate over a range in bash is

for i in {0..10}; do echo $i; done

What would be the syntax for iterating over the sequence with a step? Say, I would like to get only even number in the above example.

回答1:

I'd do

for i in `seq 0 2 10`; do echo $i; done

(though of course seq 0 2 10 will produce the same output on its own).

Note that seq allows floating-point numbers (e.g., seq .5 .25 3.5) but bash's brace expansion only allows integers.

回答2:

Bash 4's brace expansion has a step feature:

for {0..10..2}; do
  ..
done

No matter if Bash 2/3 (C-style for loop, see answers above) or Bash 4, I would prefer anything over the 'seq' command.

回答3:

Pure Bash, without an extra process:

for (( COUNTER=0; COUNTER<=10; COUNTER+=2 )); do
    echo $COUNTER
done

回答4:

#!/bin/bash
for i in $(seq 1 2 10)
do
   echo "skip by 2 value $i"
done

回答5:

> seq 4
1
2
3
4

> seq 2 5
2
3
4
5

> seq 4 2 12
4
6
8
10
12

> seq -w 4 2 12
04
06
08
10
12

> seq -s, 4 2 12
4,6,8,10,12

来源：https://stackoverflow.com/questions/966020/how-to-produce-a-range-with-step-n-in-bash-generate-a-sequence-of-numbers-with