Bash array with variable

Question

Any idea how I can declare a bash array with a variable embedded?

For example, i is a integer that increments in a for loop. I want to continue to increment i and append it to the end of the array being declared like so:

declare -a DB$iFIELDS
DB$iFIELDS[$j]=blah blah blah

So you want to create a new array every time in loop OR append to an existing array? — anubhava
– anubhava, Commented Apr 30, 2014 at 19:20
This is just creating a new (incremented) array each time. There is a separate loop that fills the array. — Atomiklan
– Atomiklan, Commented Apr 30, 2014 at 19:36
Use printf: printf -v DB${i}FIELDS[$j] '%s' "blah blah blah". By the way, Bash is not exactly supposed to be used this way... — gniourf_gniourf
– gniourf_gniourf, Commented Apr 30, 2014 at 19:48
@gniourf_gniourf is right. If your data structures are getting this complicated (and really, it doesn't take much), it's time to pick another language. — chepner
– chepner, Commented Apr 30, 2014 at 19:50

chepner · Accepted Answer · 2014-04-30 19:56:34Z

1

You can use declare to make the assignment as well, since to some extent the [] is as much a part of a variable name as it is an indexing operator.

$ i=3
$ declare -a DB${i}FIELDS
# ...
$ j=6
$ declare "DB${i}FIELDS[$j]=blah blah blah"
$ set | grep "DB.*FIELDS"
DB3FIELDS=([6]="blah blah blah")

answered Apr 30, 2014 at 19:56

chepner

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1 Comment

opinion_no9 Over a year ago

yes, declare works fine with indexed (-a) and associated (-A) arrays.

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