Unix Script to remove last seven characters from a variable - unix

Question

Need to remove the last seven characters from a variable.
For example if my variable string is
COLUMN_NAME||','||
then it should output COLUMN_NAME
I have tried the below but last pipe symbol only getting removed
var=$(lastline%|)
var=$(lastline%|*)
Result : COLUMN_NAME||','|

Answer 1

To remove the last 7 characters:
$ var="COLUMN_NAME||','||"
$ echo "${var%???????}"
COLUMN_NAME
To remove everything after the first pipe:
$ echo "${var%%|*}"
COLUMN_NAME
See https://www.gnu.org/software/bash/manual/bashref.html#Shell-Parameter-Expansion
and https://www.gnu.org/software/bash/manual/bashref.html#Pattern-Matching

Answer 2

The old school way
echo "COLUMN_NAME||','||" | rev|cut -c 8-|rev
So you are just reversing the string, deleting first 7 characters and again reversing the string.
Use the $variable with echo to do the same.
You can also use awk like below which would be faster.
awk '{print substr($0, 1, length($0)-7)}'
Example:
$ export variable1="COLUMN_NAME||','||"
$ echo $variable1|rev|cut -c 8-|rev
COLUMN_NAME
$ echo $variable1|awk '{print substr($0, 1, length($0)-7)}'
COLUMN_NAME

Answer 3

You need to use two % to strip the longest match:
$ r="COLUMN_NAME||','||"
$ echo ${r%%|*}
COLUMN_NAME
As BashFAQ says in Removing part of a string:
% means "remove the shortest possible match from the end of the
variable's contents".
%% means "remove the longest possible match from the end of the
variable's contents".

Answer 4

You could find the length of your string/s and index from that point:
str1="1234567890foobar";
strlen=${#str1};
str2=${str1:0:$strlen-7};
echo $str2;