print upto second last character in unix

shell unix

If you are using BASH then it is fairly straight forward to remove last character:

s="string1,string2,"echo "${s%?}"

? matches any single character and %? removes any character from right hand side.

That will output:

string1,string2

Otherwise you can use this sed to remove last character:

echo "$s" | sed 's/.$//'string1,string2

shell unix

You can do it with bash "parameter substitution":

string=12345new=${string:0:$((${#string}-1))}echo $new1234

where I am saying:

new=${string:a:b}

where:

a=0 (meaning starting from the first character)

and:

b=${#string} i.e. the length of the string minus 1, performed in an arithmetic context, i.e. inside `$((...))`

shell unix

str="something"echo $str | cut -c1-$((${#str}-1))

will give result as

somethin

If you have two different variables, then you can try this also.

str="something"strlen=9echo $str | cut -c1-$((strlen-1))

cut -c1-8 will print from first character to eighth.

CodeHunter