A very stupid bash question: using a variable as an argument

[aborruso]

Hi all,
I have this input csv file

a a,b,c
1,2,a
3,5,t
4,7,o

If I run mlr --csv cut -f "a a","b" input.csv I have

a a,b
1,2
3,5
4,7

If I run in the shell these commands I have instead an empty result.

p="\"a a\",\"b\""
mlr --csv cut -f "$p" input.csv

How do I must use a variable as an argument?

[NikosAlexandris]

Andrea, why not exclude the double quotes (as they are not appearing in your input.csv):

➜  p="a a,b" ;echo "$p"
a a,b

➜  mlr --csv cut -f "$p" input.csv
a a,b
1,2
3,5
4,7

[aborruso]

I feel very very stupid.

The single mlr command is mlr --csv cut -f "a a","b" input.csv. If I run it without double quotes - mlr --csv cut -f a a,b input.csv - it does not work.

For this reason, I have added the escaped double quotes in the variable: p="\"a a\",\"b\"".

But I'm stupid, it's not necessary, bash does everything right and magically for me.

Thank you @NikosAlexandris