Can I use a variable in a Bash brace expansion?

Question

Below is some sort of pseudo-code for what I'm trying to accomplish:

#!/bin/bash

# I already have the variable below figured out (positive integer):
numlines=$([returns number of lines containing specific characters in a file])

# This is basically what I want to do with it:
for i in {1..$numlines}; do
    # the part below is already figured out as well:        
    do some other stuff
done

I can execute it fine from the command line by inserting the actual number in the `{1..n}' sequence. I just need to know if it's possible to include a variable here and how to go about doing it.

I have tried exporting it
I have tried putting the variable itself in curly braces inside the sequence: {1..${numlines}}
I have tried putting it in double-quotes hoping it would expand: {1.."$numlines"}
I have tried escaping the $: {1..\$numlines}

Do I need to use a set -[something] command in order for this variable to get expanded? I have even tried some forms of using eval...all to no avail.

I just need to know if there is something simple or obscure that I am missing or if this is even possible before I waste anymore time on it.

I could throw together a really, really hackish way of doing it to make it work as needed, but I'd like to avoid that if at all possible and learn the right way to go about doing it.

Related: [How can I use $variable in a shell brace expansion of a sequence?](https://unix.stackexchange.com/q/7738/170373) — ilkkachu, Dec 20 '18 at 11:26

score 13 · Accepted Answer · edited Nov 20 '15 at 02:48

13

Unfortunately, there is no way to use a variable in that expansion (AFAIK), since variable expansion happens after brace expansion.

Fortunately, there's a tool that does the same job.

for i in $(seq 1 $numlines); do
    # stuff
done

seq is from GNU coreutils; no idea how to do it in POSIX.

edited Nov 20 '15 at 02:48

chepner

7,341
1
26
27

answered Nov 19 '15 at 23:27

Tom Hunt

9,808
4
25
43

1

(Plus 1). `seq` is good for GNU systems and, if I recall correctly, the most recent OSX. On other BSD systems, one can use [jot](https://www.freebsd.org/cgi/man.cgi?jot(1)) instead. – John1024 Nov 19 '15 at 23:31
`seq` works perfectly. Thank you so much for your prompt answer. – rubynorails Nov 19 '15 at 23:40
This was not part of my question -- but what is the syntax (if any) for doing this in reverse order, such as `{16..1}`? `$(seq $numlines 1)` did not work. I guess I can always `man seq`, but just wondering if anyone knew off the top of their head. – rubynorails Nov 19 '15 at 23:50
1

Just figured out how to do it in reverse from [this link](http://unix.stackexchange.com/questions/64861/how-to-display-numbers-in-reverse-order-using-seq1) -- `for i in $(seq $numlines -1 1)` – rubynorails Nov 19 '15 at 23:59
`seq ${numlines} -1 0` – DopeGhoti Nov 20 '15 at 00:00

Petr Skocik · Answer 2 · 2015-11-20T02:53:21.817

10

If you must avoid seq, which as Tom Hunt points out seems to be the usual solution to this, then an eval definitely can do it (though, I wouldn't encourage it):

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

You can stay POSIX by avoiding the {} expansion, and simply do math and integer comparisons on $numlines:

while [ ! "$numlines" -eq 0 ]; do
     echo "$numlines"
     : $((numlines-=1))
done

Outside of POSIX, bash and ksh and zsh also have C-style for loops:

for((i=0; i<numlines; i++)); do echo $i; done

edited Nov 20 '15 at 02:53

answered Nov 19 '15 at 23:33

Petr Skocik

28,176
14
81
141

1

I really appreciate this answer as well. While `seq` worked fine for my scenario and seemed to be the simplest solution, it is good to know that there are other (even POSIX) alternatives. Thanks for this. – rubynorails Nov 19 '15 at 23:43
2

There's really no reason to use `eval`; if you have brace expansion, you have the C-style loop. – chepner Nov 20 '15 at 02:46
@PSkocik - If I could choose 2 answers, I would also choose this one. When I came across the fact that I needed to do this in reverse, your `eval` example was the simplest and would have saved me from having to search for an alternate way of doing it using `seq`. The `while` loop is a bit bulky for me. I like to keep things short and sweet, and I never could get the C-style `for` loop to work by giving `i` the value of 0 or 1. It never returned correctly and was always a little off. I'm sure it could be tweaked to work correctly, but these are definitely helpful solutions, nonetheless. – rubynorails Nov 20 '15 at 03:19
1

The `eval` approach is problematic if there is anything non-trivial inside the body of the loop. I imagine it would not be very readable if you needed to nest two such loops. – kasperd Nov 20 '15 at 10:38

score 10 · Answer 3 · answered Nov 20 '15 at 01:36

Sure. If you want a for loop that increments an integer variable, use the form of the for loop that increments an integer variable (or more generally performs arithmetic on the loop variable(s)).

for ((i=1; i<=numlines; i++)); do … done

This construct works in bash (and ksh93 and zsh), but not in plain sh. In plain sh, use a while loop and the test ([ … ]) construct.

i=1
while [ "$i" -le "$numlines" ]; do
  …
  i=$((i+1))
done

Can I use a variable in a Bash brace expansion?

3 Answers3

Linked

Related