How to print character just after a particular character in a string using bash

Question

I have a string like

abc def 0/1 dls dl 9/0 1//o

I want to take 0 and 1 (the characters just after and before the first / in the string) into two variables, say $NUM1 and $NUM2 , and do some further logic on them.

Answer 1

Using just shell parameter expansion :

$ str='abc def 0/1 dls dl 9/0 1//o'
$ num1=${str%%/*}    # Remove first "/" and everything after it
$ echo "$num1"
abc def 0
$ num1=${num1##* }   # Remove last space and everything before it
$ echo "$num1"
0
$ num2=${str#*/}     # Remove first "/" and everything before it
$ echo "$num2"
1 dls dl 9/0 1//o
$ num2=${num2%% *}   # Remove first space and everything after it
$ echo "$num2"
1

# , ## , % and %% have to be used carefully here. A single repetition removes the shortest match of the pattern following, and the double usage removes the longest pattern match.

---

Alternatively, after the first expansion, you just want the first or last character of the string, which can be solved with substring expansion:

$ echo "$num1"
abc def 0
$ echo "${num1:(-1)}"  # Extract last character
0
$ echo "$num2"
1 dls dl 9/0 1//o
$ echo "${num2:0:1}"   # Extract first character
1

Negative indices requires at least Bash 4.2. For older Bash versions, we can use

echo "${num1:${#num1}-1}"

---

And finally, we could use Bash regular expressions, as demonstrated in glenn jackman's answer .

Answer 2

You can use bash regular expressions, which populate the BASH_REMATCH array with the captured subexpressions:

if [[ $a =~ (.)/(.) ]]; then
  num1=${BASH_REMATCH[1]}
  num2=${BASH_REMATCH[2]}
fi
echo $num1 $num2

0 1

Don't get into the habit of using ALL_CAPS_VARNAMES: one day you'll accidentally write PATH=/my/path and then wonder why your script is broken. Leave upper case variables for the shell & OS

Answer 3

You can use sed to parse the line with regex to extract the numbers:

NUM1=$(echo 'abc def 0/1 dls dl 9/0 1//o' | sed 's@[^0-9]*\([0-9]*\)/[0-9]*.*@\1@')
NUM2=$(echo 'abc def 0/1 dls dl 9/0 1//o' | sed 's@[^0-9]*[0-9]*/\([0-9]*\).*@\1@')
echo $NUM1
0
echo $NUM2
1

Answer 4

I assume you are implicitly separating characters based on spaces. So you want the characters from the space to the first '/' and from the first '/' to the next space. I would probably use sed:

echo "abc def 0/1 dls dl 9/0 1//o" | sed -r 's%^[^/]* ([^ ]*)/([^ ]*) .*%\1 \2%'

But note this won't work if there are no spaces preceding the first '/'. You may wish to have another pattern for that case.