Bash remove first and last characters from a string

Question

I have a string like that:

|abcdefg|

And i want to get a new string called in someway (like string2) with the original string without the two | at the start and at the end of it

so that i will have this

abcdefg

is that possible in bash?

Answer 1

You can do

string="|abcdefg|"
string2=${string#"|"}
string2=${string2%"|"}
echo $string2

Or if your string length is constant, you can do

string="|abcdefg|"
string2=${string:1:7}
echo $string2

Also, this should work

echo "|abcdefg|" | cut -d "|" -f 2

Also this

echo "|abcdefg|" | sed 's/^|\(.*\)|$/\1/'

Answer 2

Here's a solution that is independent of the length of the string (bash):

string="|abcdefg|"
echo "${string:1:${#string}-2}"

Answer 3

And another one:

string="|abcdefg|"
echo "${string//|/}"

Answer 4

You can also use sed to remove the | not just referencing the symbol itself but using positional references as in:

$ echo "|abcdefg|" | sed 's:^.\(.*\).$:\1:'
abcdefg

Where ':' are the delimiters (you can replace them with / or any character not in the query, any sign following the s will do it) Here ^ (caret) means at the beginning of the input string and $ (dollar) means at the end. The . (point) that it's after the caret and the one that it's before the dollar sign represents a single character. So in other words we are deleting the first and last characters. Take in mind this will delete any characters even if | it's not present in the string.

EX:

$ echo "abcdefg" | sed 's:^.\(.*\).$:\1:'
bcdef

Answer 5

Going off a few posts listed here it seems the simplest way to do it is:

string="|abcdefg|"
echo ${string:1:-1}

edit: works on ubuntu with bash 4.2; does not work on centOS with bash 4.1