How to generate for loop number sequence by using variable names in bash?

Question

Could anybody explain how can I use variable names in bash for loop to generate a sequence of numbers?

for year in {1990..1997}
do
  echo ${year}
done

Results:

1990 1991 1992 1993 1994 1995 1996 1997

However

year_start=1990
year_end=1997
for year in {${year_start}..${year_end}}
do
  echo ${year}
done

Results:

{1990..1997}

How can I make the second case work? otherwise I have to write tedious while loops. Thanks very much.

Bill · Accepted Answer · 2013-05-11T14:02:13.717

You can use something like this

for (( year = ${year_start}; year <=${year_end}; year++ ))
do
  echo ${year}
done

seq command is outdated and best avoided.

And if you want real bad to use the following syntax,

for year in {${year_start}..${year_end}}

please modify it as follows:

for year in $(eval echo "{$year_start..$year_end}")

Personally, I prefer using for (( year = ${year_start}; year <=${year_end}; year++ )).

score 4 · Answer 2 · edited May 11 '13 at 13:35

4

Try following:

start=1990; end=1997; for year in $(seq $start $end); do echo $year; done

edited May 11 '13 at 13:35

chepner

answered May 11 '13 at 13:28

Sithsu

@chepner Would like to know the reason for not using backticks. Could they cause other problems? – Sithsu May 11 '13 at 13:42
1

`seq` command is outdated, and not recommended for `bash v3.x+` – Bill May 11 '13 at 13:49
@Sithsu They're hard to read, and hard to nest. There's no real reason to use them in place of `$( ... )`. – chepner May 11 '13 at 14:20

2 Answers2