Evaluate expression inside bash for loop [duplicate]

问题

If I do this, I get the result as expected.

for i in {125..129}; do echo $i; done
125
126
127
128
129

But when I do this? I get something weired.

for i in {$((1+(25-1)*500))..$((25*500))}; do echo $i; done

{12001..12500}

I wish to pass a variable inside the loop variables like $((1+($j-1)*500))

回答1:

Bash's brace expansion has limitations. What you want is seq:

for i in $( seq $((1+(25-1)*500)) $((25*500)) ); do echo $i; done

The above will loop over all numbers from 12001 to 12500.

Discussion

seq is similar to bash's braces:

$ echo {2..4}
2 3 4
$ echo $(seq 2 4)
2 3 4

The key advantage of seq is that its arguments can include not just arithmetic expressions, as shown above, but also shell variables:

$ x=4; echo $(seq $((x-2)) $x)
2 3 4

By contrast, the brace notation will accept neither.

seq is a GNU utility and is available on all linux systems as well as recent versions of OSX. Older BSD systems can use a similar utility called jot.

回答2:

Brace expansion is the very first expansion that occurs, before parameter, variable, and arithmetic expansion. Brace expansion with .. only occurs if the values before and after the .. are integers or single characters. Since the arithmetic expansion in your example has not yet occurred, they aren't single characters or integers, so no brace expansion occurs.

You can force reexpansion to occur after arithmetic expansion with eval:

for i in $(eval echo {$((1+(25-1)*500))..$((25*500))}); do echo $i;

回答3:

You query is very similar to : shell script "for" loop syntax

brace expansion, {x..y} is performed before other expansions, so you cannot use that for variable length sequences.

Instead try

for i in seq $((1+(25-1)*500)) $((25*500)); do echo $i; done

回答4:

It is just echoing the text which is exactly as it says:

{12001..12500}

That is "{"+12001+"..."+12500+"}"

回答5:

Don't do this with a for loop. The {..} notation is not that flexible:

i=$((1+(25-1)*500)); while test $i -le $((25*500)); do echo $((i++)); done

回答6:

Try this

for (( i= $((1+(25-1)*500)); i<=$((25*500)); i++ )); do echo $i; done

or this

for i in $(seq $(( 1+(25-1)*500 )) $(( 25*500 )) ); do echo $i; done

来源：https://stackoverflow.com/questions/28623266/evaluate-expression-inside-bash-for-loop

标签

bash

shell

unix