bash 如何在bash中使用步骤n生成范围？（生成带有增量的数字序列）
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How to produce a range with step n in bash? (generate a sequence of numbers with increments)
提问by SilentGhost
The way to iterate over a range in bash is
在 bash 中迭代范围的方法是
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.
用一个步骤迭代序列的语法是什么？说，我想在上面的例子中只得到偶数。
回答by chaos
I'd do
我会做
for i in `seq 0 2 10`; do echo $i; done
(though of course seq 0 2 10will produce the same output on its own).
（虽然当然seq 0 2 10会自己产生相同的输出）。
Note that seqallows floating-point numbers (e.g., seq .5 .25 3.5) but bash's brace expansion only allows integers.
请注意，seq允许浮点数（例如，seq .5 .25 3.5），但 bash 的大括号扩展只允许整数。
回答by TheBonsai
Bash 4's brace expansion has a step feature:
Bash 4的大括号扩展有一个 step 特性：
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.
无论是 Bash 2/3（C 风格的 for 循环，参见上面的答案）还是 Bash 4，我都喜欢使用 'seq' 命令。
回答by Fritz G. Mehner
Pure Bash, without an extra process:
纯 Bash，没有额外的过程：
for (( COUNTER=0; COUNTER<=10; COUNTER+=2 )); do
echo $COUNTER
done
回答by z -
#!/bin/bash
for i in $(seq 1 2 10)
do
echo "skip by 2 value $i"
done
回答by Mir-Ismaili
> 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
