如何在 bash 中生成步骤 n 的范围?(生成带增量的数字序列) [英] How to produce a range with step n in bash? (generate a sequence of numbers with increments)

问题描述

在 bash 中迭代一个范围的方法是

The way to iterate over a range in bash is

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

使用步骤迭代序列的语法是什么?比如说，我想在上面的例子中只得到偶数.

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.

推荐答案

我愿意

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

(当然 seq 0 2 10 会自行产生相同的输出).

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

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.

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

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