Brace expansion with a Bash variable - {0..$foo}

2019-01-01 06:09发布

WEEKS_TO_SAVE=4
mkdir -p weekly.{0..$WEEKS_TO_SAVE}

gives me a folder called weekly.{0..4}

Is there a secret to curly brace expansion while creating folders I'm missing?

5条回答
其实,你不懂
2楼-- · 2019-01-01 06:17

Another way of doing it without eval and calling mkdir only once:

WEEKS_TO_SAVE=4
mkdir -p $(seq -f "weekly.%.0f" 0 $WEEKS_TO_SAVE)
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唯独是你
3楼-- · 2019-01-01 06:19

Brace expansion does not support it. You will have to do it using a loop.

Brace expansion is performed before any other expansions, and any characters special to other expansions are preserved in the result. It is strictly textual. Bash does not apply any syntactic interpretation to the context of the expansion or the text between the braces. To avoid conflicts with parameter expansion, the string ‘${’ is not considered eligible for brace expansion

.

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其实,你不懂
4楼-- · 2019-01-01 06:22

Curly braces don't support variables in BASH, you can do this:

 for (( c=0; c<=WEEKS_TO_SAVE; c++ ))
 do
    mkdir -p weekly.${c}
 done
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宁负流年不负卿
5楼-- · 2019-01-01 06:23

If you happen to have zsh installed on your box, your code as written will work with Z-shell if you use #!/bin/zsh as your interpreter:

Example

$ WEEKS_TO_SAVE=4
$ echo {0..$WEEKS_TO_SAVE}
0 1 2 3 4
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一个人的天荒地老
6楼-- · 2019-01-01 06:36

bash does brace expansion before variable expansion, so you get weekly.{0..4}.
Because the result is predictable and safe(Don't trust user input), you can use eval in your case:

$ WEEKS_TO_SAVE=4
$ eval "mkdir -p weekly.{0..$((WEEKS_TO_SAVE))}"

note:

  1. eval is evil
  2. use eval carefully

Here, $((..)) is used to force the variable to be evaluated as an integer expression.

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