How to manipulate a variable name in Bash?

Question

I have a variable name in Bash as:

var=(any_word)_SuSE_11_design_guides

I want to manipulate the name such that I can have:

x=$(echo "$var" | some operation)
echo $x
SuSE_11_design_guides

Basically I want to remove every character behind first _ when detected. How to achieve this?

what you're saying makes NO sense, if you are "removing" everything behind first _ then echo $x would have whatever's (any_word) and NOT SuSE_11_design_guides. Because if you wanted that then you'd simply set x=SuSE_11_design_guides and you're done. Instead do you mean that you trying to get (any_word) in x? — Ahmed Masud
– Ahmed Masud, Commented Feb 4, 2017 at 4:45

Inian · Accepted Answer · 2017-02-04 04:51:32Z

1

I think you meant retain Everything behind first _ , which can be achieved using parameter-expansion.

string="(anyword)_SuSE_11_design_guides"
printf "%s\n" "${string#*_}"
SuSE_11_design_guides

answered Feb 4, 2017 at 4:46

Inian

87k15 gold badges166 silver badges183 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Jameson · Accepted Answer · 2017-02-04 04:47:29Z

1

It's unclear whether you want to print SuSE_11_design_guides, or whatever came before. So, I'll answer both.

# Print the content before the first `_`
echo $var | awk -F_ '{ print $1 }'

If you want to print the trailing part:

echo $var | sed -E 's/.*(SuSE_11_design_guides).*/\1/'

answered Feb 4, 2017 at 4:47

Jameson

6,7296 gold badges37 silver badges58 bronze badges

clearlight · Accepted Answer · 2017-02-04 13:34:01Z

0

$ var=(any_word)_SuSE_11_design_guides
$ echo ${var##${var%%SuSE_11_design_guides}}
SuSE_11_design_guides

The above echo statement is comprised of two nested bash strings function. Breaking down what this does, it is the equivalent of:

let a=${var%%SuSE_11_design_guides}

The above line strips "SuSE_11_design_guides" off the tail of $var returning the prefix

${var##$a}

The above strips whatever is in $a (which happens to be the prefix as per the previous step), from the beginning of $var, leaving the suffix.

I combined them in the original example just because I could. It definitely makes it harder to read, but I was going for shock value :-)

answered Feb 4, 2017 at 5:30

clearlight

12.7k11 gold badges61 silver badges82 bronze badges