Can i use cut to set a variable?

Question

I want to cut a variable and use the result to manipulate a variable at this name

`echo ${!var1} | cut -b$var2`=1

But thats don't work. Do you have any idea to do that ?

The complete code

for (( var1=1; ${!var1}!=""; var1++ )) 
    do for (( var2=1; `echo ${!var1} | cut -b$var2`!=""; var2++ ))
        do `echo ${!var1} | cut -b$var2`=1 
    done
done

http://mywiki.wooledge.org/BashFAQ/006#Assigning_indirect.2Freference_variables — Lightness Races in Orbit, Feb 07 '19 at 16:00
A concrete example would help, but anything involving setting a variable whose name is calculated at run time will need to involve `printf -v`. — chepner, Feb 07 '19 at 16:14
I have `var0 = abcdef` and i want to assign `f = 1`. For that i want to extract the "f" of var0. Here `var2 = 6`. `var1 = var0`, so he will invoke var0 — wxi, Feb 08 '19 at 09:36

tripleee · Answer 1 · 2019-02-07T16:25:54.303

1

Something like this?

ref=${!var1}
printf -v "${ref:$var2:1}" '%s' 1

edited Feb 07 '19 at 16:25

answered Feb 07 '19 at 16:14

tripleee

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With quotes! and I think the PE `${!var1:var2:1}` works (can't test now). – gniourf_gniourf Feb 07 '19 at 16:23
2

And maybe `declare` or `printf -v` instead of `eval` would be slightly less dangerous... – gniourf_gniourf Feb 07 '19 at 16:24

Can i use cut to set a variable?

1 Answers1