don't have to chain so many tools. Just one awk command does the job
COMPANY_NAME=$(awk -F"=" '/company_name/{gsub(/;$/,"",$2) ;print $2}' file.txt)
Some refinements to answer above. To remove more than one char you add multiple question marks. For example, to remove last two chars from variable $SRC_IP_MSG, you can use:
SRC_IP_MSG=${SRC_IP_MSG%??}
Don't abuse cat
s. Did you know that grep
can read files, too?
The canonical approach would be this:
grep "company_name" file.txt | cut -d '=' -f 2 | sed -e 's/;$//'
the smarter approach would use a single perl
or awk
statement, which can do filter and different transformations at once. For example something like this:
COMPANY_NAME=$( perl -ne '/company_name=(.*);/ && print $1' file.txt )
I am not finding that sed 's/;$//'
works. It doesn't trim anything, though I'm wondering whether it's because the character I'm trying to trim off happens to be a "$". What does work for me is sed 's/.\{1\}$//'
.
you can strip the beginnings and ends of a string by N characters using this bash construct, as someone said already
$ fred=abcdefg.rpm
$ echo ${fred:1:-4}
bcdefg
HOWEVER, this is not supported in older versions of bash.. as I discovered just now writing a script for a Red hat EL6 install process. This is the sole reason for posting here. A hacky way to achieve this is to use sed with extended regex like this:
$ fred=abcdefg.rpm
$ echo $fred | sed -re 's/^.(.*)....$/\1/g'
bcdefg
In Bash using only one external utility:
IFS='= ' read -r discard COMPANY_NAME <<< $(grep "company_name" file.txt)
COMPANY_NAME=${COMPANY_NAME/%?}
Using sed
, if you don't know what the last character actually is:
$ grep company_name file.txt | cut -d '=' -f2 | sed 's/.$//'
"Abc Inc"
cat file.txt | grep "company_name" | cut -d '=' -f 2 | cut -d ';' -f 1
foo="hello world"
echo ${foo%?}
hello worl
This will remove the last character contained in your COMPANY_NAME var regardless if it is or not a semicolon:
echo "$COMPANY_NAME" | rev | cut -c 2- | rev
I'd use head --bytes -1
, or head -c-1
for short.
COMPANY_NAME=`cat file.txt | grep "company_name" | cut -d '=' -f 2 | head --bytes -1`
head
outputs only the beginning of a stream or file. Typically it counts lines, but it can be made to count characters/bytes instead. head --bytes 10
will output the first ten characters, but head --bytes -10
will output everything except the last ten.
NB: you may have issues if the final character is multi-byte, but a semi-colon isn't
I'd recommend this solution over sed
or cut
because
head
was designed to do, thus less command-line options and an easier-to-read commandI believe the cleanest way to strip a single character from a string with bash is:
echo ${COMPANY_NAME:: -1}
but I haven't been able to embed the grep piece within the curly braces, so your particular task becomes a two-liner:
COMPANY_NAME=$(grep "company_name" file.txt); COMPANY_NAME=${COMPANY_NAME:: -1}
This will strip any character, semicolon or not, but can get rid of the semicolon specifically, too. To remove ALL semicolons, wherever they may fall:
echo ${COMPANY_NAME/;/}
To remove only a semicolon at the end:
echo ${COMPANY_NAME%;}
Or, to remove multiple semicolons from the end:
echo ${COMPANY_NAME%%;}
For great detail and more on this approach, The Linux Documentation Project covers a lot of ground at http://tldp.org/LDP/abs/html/string-manipulation.html
Assuming the quotation marks are actually part of the output, couldn't you just use the -o switch to return everything between the quote marks?
COMPANY_NAME="\"ABC Inc\";" | echo $COMPANY_NAME | grep -o "\"*.*\""
Source: Stackoverflow.com