bash for i in {1..$1} doesn't work as expected

问题:

Here is myscript.sh

#!/bin/bash for i in {1..$1}; do     echo $1 $i; done 

If I run myscript.sh 3 the output is

3 {1..3} 

instead of

3 1 3 2 3 3 

Clearly $3 contains the right value, so why doesn't for i in {1..$1} behave the same as if I had written for i in {1..3} directly?

回答1:

You should use a C-style for loop to accomplish this:

for ((i=1; i

This avoids external commands and nasty eval statements.

回答2:

Because brace expansion occurs before expansion of variables. http://www.gnu.org/software/bash/manual/bashref.html#Brace-Expansion.

If you want to use braces, you could so something grim like this:

for i in `eval echo {1..$1}`; do     echo $1 $i; done 

Summary: Bash is vile.

回答3:

You can use seq command:

for i in `seq 1 $1` 

Or you can use the C-style for...loop:

for((i=1;i

回答4:

I know you've mentioned bash in the heading, but I would add that 'for i in {$1..$2}' works as intended in zsh. If your system has zsh installed, you can just change your shebang to zsh.

Using zsh with the example 'for i in {$1..$2}' also has the added benefit that $1 can be less than $2 and it still works, something that would require quite a bit of messing about if you wanted that kind of flexibility with a C-style for loop.

回答5:

Here is a way to expand variables inside braces without eval:

end=3 declare -a 'range=({'"1..$end"'})' 

We now have a nice array of numbers:

for i in ${range[@]};do echo $i;done 1 2 3 

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