I need to cut letter X out of a word:
For example: I need to cut the first letter out of Star Wars, the fourth out of munich,...
1 star wars
4 munich
5 casino royale
7 the fast and the furious
52 a fish called wanda
to
tar wars
munch
casio royale
the fat and the furious
a fish called wanda
I already tried it with cut, but it didn't work.
This was my command:
sed 's/^\([0-9]*\) \(.*\)/ echo \2 | cut -c \1/'
So it gave me this output:
echo star wars | cut -c 5
echo munich | cut -c 5
echo casino royale | cut -c 5
echo the fast and the furious | cut -c 5
echo a fish called wanda | cut -c 52
And than if I send it to bash. I only get the X th letter of the word.
I need to do the exercise with sed and other commands. But I can't use awk or perl.
Thanks
You can use just bash and its parameter expansion:
while read n s ; do
echo "${s:0:n-1}${s:n}"
done < input.txt
If you need one line, just remove the newlines and add a semicolon:
while read n s ; do echo "${s:0:n-1}${s:n}" ; done < input.txt
If you really need to use sed and cut, it's also doable, but a bit less readable:
cat -n input.txt \
| sed 's/\t\([0-9]\+\).*/s=\\(.\\{\1\\}\\).=\\1=/' \
| sed -f- <(sed 's/[0-9]*//' input.txt) \
| cut -c2-
Explanation:
number the lines
turn each line into a sed command that searches for the given number of characters and removes the one following them
run the generated sed command on the original file with the numbers removed
remove the extra leading space
This might work for you (GNU sed):
sed 's/^\([0-9]*\) \(.*\)/echo '\''\2'\''|sed '\''s\/.\/\/\1'\''/e' file
This uses the e flag of the s command to evaluate the RHS and runs a second sed invocation using the backreferences from the LHS. Perhaps easier on the eye is this:
sed -r 's/^([0-9]*) (.*)/echo "\2"|sed "s#.##\1"/e' file
you can use sed in this way:
sed -e '1s/\([a-z]\)\{1\}//' -e 's/^[0-9]\+\s\+\(.*\)/\1/g' file.txt
The first sed regular expresion works on the first line and replace the first character and the second regular expresion works with rest of text, 1 column: one number or more, 2 column; one space or more, after this I put the remaining test in one match \(.*\) and replaced all with this match.
Related
I have a file with names:
Smith, John.
Brown, Aaron K.
And want to get:
Smith, J
Brown, A K
or better:
SmithJ
BrownAK
Can this task be solved in bash?
You can solve it with different tools and different methods. I will show two solutions using sed and one without.
Solution 1
You want to use some command on part of the line.
You can remove all non-uppercase characters from a string with echo "${string}" | tr -cd "[:upper:]".
With sed s/../../e the resulting line from the substitition is given to the shell.
Combining these give you:
sed -r 's/([^,]*)(.*)/echo "\1\$(echo "\2" | tr -cd "[:upper:]")"/e' file
Solution 2
Less creative but easier to write is temporarily splitting each line in two lines, and execute the substition on the even lines. Put the lines together and your finished.
sed -e 's/,/\n/' file | sed '0~2s/[^A-Z]//g' | paste -d '' - -
Solution 3
With the tr from the first and the paste from the second solution you can avoid sed.
Be aware that the tr characterset must include a newline.
paste -d '' <(cut -d, -f1 file) <(cut -d, -f2 file | tr -cd ':[A-Z]:\n')
IMHO the second solution looks best. The first one is slow on large files.
I want to remove some n lines from the end of a file. Can this be done using sed?
For example, to remove lines from 2 to 4, I can use
$ sed '2,4d' file
But I don't know the line numbers. I can delete the last line using
$sed $d file
but I want to know the way to remove n lines from the end. Please let me know how to do that using sed or some other method.
I don't know about sed, but it can be done with head:
head -n -2 myfile.txt
If hardcoding n is an option, you can use sequential calls to sed. For instance, to delete the last three lines, delete the last one line thrice:
sed '$d' file | sed '$d' | sed '$d'
From the sed one-liners:
# delete the last 10 lines of a file
sed -e :a -e '$d;N;2,10ba' -e 'P;D' # method 1
sed -n -e :a -e '1,10!{P;N;D;};N;ba' # method 2
Seems to be what you are looking for.
A funny & simple sed and tac solution :
n=4
tac file.txt | sed "1,$n{d}" | tac
NOTE
double quotes " are needed for the shell to evaluate the $n variable in sed command. In single quotes, no interpolate will be performed.
tac is a cat reversed, see man 1 tac
the {} in sed are there to separate $n & d (if not, the shell try to interpolate non existent $nd variable)
Use sed, but let the shell do the math, with the goal being to use the d command by giving a range (to remove the last 23 lines):
sed -i "$(($(wc -l < file)-22)),\$d" file
To remove the last 3 lines, from inside out:
$(wc -l < file)
Gives the number of lines of the file: say 2196
We want to remove the last 23 lines, so for left side or range:
$((2196-22))
Gives: 2174
Thus the original sed after shell interpretation is:
sed -i '2174,$d' file
With -i doing inplace edit, file is now 2173 lines!
If you want to save it into a new file, the code is:
sed -i '2174,$d' file > outputfile
You could use head for this.
Use
$ head --lines=-N file > new_file
where N is the number of lines you want to remove from the file.
The contents of the original file minus the last N lines are now in new_file
Just for completeness I would like to add my solution.
I ended up doing this with the standard ed:
ed -s sometextfile <<< $'-2,$d\nwq'
This deletes the last 2 lines using in-place editing (although it does use a temporary file in /tmp !!)
To truncate very large files truly in-place we have truncate command.
It doesn't know about lines, but tail + wc can convert lines to bytes:
file=bigone.log
lines=3
truncate -s -$(tail -$lines $file | wc -c) $file
There is an obvious race condition if the file is written at the same time.
In this case it may be better to use head - it counts bytes from the beginning of file (mind disk IO), so we will always truncate on line boundary (possibly more lines than expected if file is actively written):
truncate -s $(head -n -$lines $file | wc -c) $file
Handy one-liner if you fail login attempt putting password in place of username:
truncate -s $(head -n -5 /var/log/secure | wc -c) /var/log/secure
This might work for you (GNU sed):
sed ':a;$!N;1,4ba;P;$d;D' file
Most of the above answers seem to require GNU commands/extensions:
$ head -n -2 myfile.txt
-2: Badly formed number
For a slightly more portible solution:
perl -ne 'push(#fifo,$_);print shift(#fifo) if #fifo > 10;'
OR
perl -ne 'push(#buf,$_);END{print #buf[0 ... $#buf-10]}'
OR
awk '{buf[NR-1]=$0;}END{ for ( i=0; i < (NR-10); i++){ print buf[i];} }'
Where "10" is "n".
With the answers here you'd have already learnt that sed is not the best tool for this application.
However I do think there is a way to do this in using sed; the idea is to append N lines to hold space untill you are able read without hitting EOF. When EOF is hit, print the contents of hold space and quit.
sed -e '$!{N;N;N;N;N;N;H;}' -e x
The sed command above will omit last 5 lines.
It can be done in 3 steps:
a) Count the number of lines in the file you want to edit:
n=`cat myfile |wc -l`
b) Subtract from that number the number of lines to delete:
x=$((n-3))
c) Tell sed to delete from that line number ($x) to the end:
sed "$x,\$d" myfile
You can get the total count of lines with wc -l <file> and use
head -n <total lines - lines to remove> <file>
Try the following command:
n = line number
tail -r file_name | sed '1,nd' | tail -r
This will remove the last 3 lines from file:
for i in $(seq 1 3); do sed -i '$d' file; done;
I prefer this solution;
head -$(gcalctool -s $(cat file | wc -l)-N) file
where N is the number of lines to remove.
sed -n ':pre
1,4 {N;b pre
}
:cycle
$!{P;N;D;b cycle
}' YourFile
posix version
To delete last 4 lines:
$ nl -b a file | sort -k1,1nr | sed '1, 4 d' | sort -k1,1n | sed 's/^ *[0-9]*\t//'
I came up with this, where n is the number of lines you want to delete:
count=`wc -l file`
lines=`expr "$count" - n`
head -n "$lines" file > temp.txt
mv temp.txt file
rm -f temp.txt
It's a little roundabout, but I think it's easy to follow.
Count up the number of lines in the main file
Subtract the number of lines you want to remove from the count
Print out the number of lines you want to keep and store in a temp file
Replace the main file with the temp file
Remove the temp file
For deleting the last N lines of a file, you can use the same concept of
$ sed '2,4d' file
You can use a combo with tail command to reverse the file: if N is 5
$ tail -r file | sed '1,5d' file | tail -r > file
And this way runs also where head -n -5 file command doesn't run (like on a mac!).
#!/bin/sh
echo 'Enter the file name : '
read filename
echo 'Enter the number of lines from the end that needs to be deleted :'
read n
#Subtracting from the line number to get the nth line
m=`expr $n - 1`
# Calculate length of the file
len=`cat $filename|wc -l`
#Calculate the lines that must remain
lennew=`expr $len - $m`
sed "$lennew,$ d" $filename
A solution similar to https://stackoverflow.com/a/24298204/1221137 but with editing in place and not hardcoded number of lines:
n=4
seq $n | xargs -i sed -i -e '$d' my_file
In docker, this worked for me:
head --lines=-N file_path > file_path
Say you have several lines:
$ cat <<EOF > 20lines.txt
> 1
> 2
> 3
[snip]
> 18
> 19
> 20
> EOF
Then you can grab:
# leave last 15 out
$ head -n5 20lines.txt
1
2
3
4
5
# skip first 14
$ tail -n +15 20lines.txt
15
16
17
18
19
20
POSIX compliant solution using ex / vi, in the vein of #Michel's solution above.
#Michel's ed example uses "not-POSIX" Here-Strings.
Increment the $-1 to remove n lines to the EOF ($), or just feed the lines you want to (d)elete. You could use ex to count line numbers or do any other Unix stuff.
Given the file:
cat > sometextfile <<EOF
one
two
three
four
five
EOF
Executing:
ex -s sometextfile <<'EOF'
$-1,$d
%p
wq!
EOF
Returns:
one
two
three
This uses POSIX Here-Docs so it is really easy to modify - especially using set -o vi with a POSIX /bin/sh.
While on the subject, the "ex personality" of "vim" should be fine, but YMMV.
This will remove the last 12 lines
sed -n -e :a -e '1,10!{P;N;D;};N;ba'
I need to cut letter X out of a word:
For example: I need to cut the first letter out of Star Wars, the fourth out of munich,...
1 star wars
4 munich
5 casino royale
7 the fast and the furious
52 a fish called wanda
to
tar wars
munch
casio royale
the fat and the furious
a fish called wanda
I already tried it with cut, but it didn't work.
This was my command:
sed 's/^\([0-9]*\) \(.*\)/ echo \2 | cut -c \1/'
So it gave me this output:
echo star wars | cut -c 5
echo munich | cut -c 5
echo casino royale | cut -c 5
echo the fast and the furious | cut -c 5
echo a fish called wanda | cut -c 52
And than if I send it to bash. I only get the X th letter of the word.
I need to do the exercise with sed and other commands. But I can't use awk or perl.
Thanks
You can use just bash and its parameter expansion:
while read n s ; do
echo "${s:0:n-1}${s:n}"
done < input.txt
If you need one line, just remove the newlines and add a semicolon:
while read n s ; do echo "${s:0:n-1}${s:n}" ; done < input.txt
If you really need to use sed and cut, it's also doable, but a bit less readable:
cat -n input.txt \
| sed 's/\t\([0-9]\+\).*/s=\\(.\\{\1\\}\\).=\\1=/' \
| sed -f- <(sed 's/[0-9]*//' input.txt) \
| cut -c2-
Explanation:
number the lines
turn each line into a sed command that searches for the given number of characters and removes the one following them
run the generated sed command on the original file with the numbers removed
remove the extra leading space
This might work for you (GNU sed):
sed 's/^\([0-9]*\) \(.*\)/echo '\''\2'\''|sed '\''s\/.\/\/\1'\''/e' file
This uses the e flag of the s command to evaluate the RHS and runs a second sed invocation using the backreferences from the LHS. Perhaps easier on the eye is this:
sed -r 's/^([0-9]*) (.*)/echo "\2"|sed "s#.##\1"/e' file
you can use sed in this way:
sed -e '1s/\([a-z]\)\{1\}//' -e 's/^[0-9]\+\s\+\(.*\)/\1/g' file.txt
The first sed regular expresion works on the first line and replace the first character and the second regular expresion works with rest of text, 1 column: one number or more, 2 column; one space or more, after this I put the remaining test in one match \(.*\) and replaced all with this match.
I have a file containing a list of replacement pairs (about 100 of them) which are used by sed to replace strings in files.
The pairs go like:
old|new
tobereplaced|replacement
(stuffiwant).*(too)|\1\2
and my current code is:
cat replacement_list | while read i
do
old=$(echo "$i" | awk -F'|' '{print $1}') #due to the need for extended regex
new=$(echo "$i" | awk -F'|' '{print $2}')
sed -r "s/`echo "$old"`/`echo "$new"`/g" -i file
done
I cannot help but think that there is a more optimal way of performing the replacements. I tried turning the loop around to run through lines of the file first but that turned out to be much more expensive.
Are there any other ways of speeding up this script?
EDIT
Thanks for all the quick responses. Let me try out the various suggestions before choosing an answer.
One thing to clear up: I also need subexpressions/groups functionality. For example, one replacement I might need is:
([0-9])U|\10 #the extra brackets and escapes were required for my original code
Some details on the improvements (to be updated):
Method: processing time
Original script: 0.85s
cut instead of awk: 0.71s
anubhava's method: 0.18s
chthonicdaemon's method: 0.01s
You can use sed to produce correctly -formatted sed input:
sed -e 's/^/s|/; s/$/|g/' replacement_list | sed -r -f - file
I recently benchmarked various string replacement methods, among them a custom program, sed -e, perl -lnpe and an probably not that widely known MySQL command line utility, replace. replace being optimized for string replacements was almost an order of magnitude faster than sed. The results looked something like this (slowest first):
custom program > sed > LANG=C sed > perl > LANG=C perl > replace
If you want performance, use replace. To have it available on your system, you'll need to install some MySQL distribution, though.
From replace.c:
Replace strings in textfile
This program replaces strings in files or from stdin to stdout. It accepts a list of from-string/to-string pairs and replaces each occurrence of a from-string with the corresponding to-string. The first occurrence of a found string is matched. If there is more than one possibility for the string to replace, longer matches are preferred before shorter matches.
...
The programs make a DFA-state-machine of the strings and the speed isn't dependent on the count of replace-strings (only of the number of replaces). A line is assumed ending with \n or \0. There are no limit exept memory on length of strings.
More on sed. You can utilize multiple cores with sed, by splitting your replacements into #cpus groups and then pipe them through sed commands, something like this:
$ sed -e 's/A/B/g; ...' file.txt | \
sed -e 's/B/C/g; ...' | \
sed -e 's/C/D/g; ...' | \
sed -e 's/D/E/g; ...' > out
Also, if you use sed or perl and your system has an UTF-8 setup, then it also boosts performance to place a LANG=C in front of the commands:
$ LANG=C sed ...
You can cut down unnecessary awk invocations and use BASH to break name-value pairs:
while IFS='|' read -r old new; do
# echo "$old :: $new"
sed -i "s~$old~$new~g" file
done < replacement_list
IFS='|' will give enable read to populate name-value in 2 different shell variables old and new.
This is assuming ~ is not present in your name-value pairs. If that is not the case then feel free to use an alternate sed delimiter.
Here is what I would try:
store your sed search-replace pair in a Bash array like ;
build your sed command based on this array using parameter expansion
run command.
patterns=(
old new
tobereplaced replacement
)
pattern_count=${#patterns[*]} # number of pattern
sedArgs=() # will hold the list of sed arguments
for (( i=0 ; i<$pattern_count ; i=i+2 )); do # don't need to loop on the replacement…
search=${patterns[i]};
replace=${patterns[i+1]}; # … here we got the replacement part
sedArgs+=" -e s/$search/$replace/g"
done
sed ${sedArgs[#]} file
This result in this command:
sed -e s/old/new/g -e s/tobereplaced/replacement/g file
You can try this.
pattern=''
cat replacement_list | while read i
do
old=$(echo "$i" | awk -F'|' '{print $1}') #due to the need for extended regex
new=$(echo "$i" | awk -F'|' '{print $2}')
pattern=${pattern}"s/${old}/${new}/g;"
done
sed -r ${pattern} -i file
This will run the sed command only once on the file with all the replacements. You may also want to replace awk with cut. cut may be more optimized then awk, though I am not sure about that.
old=`echo $i | cut -d"|" -f1`
new=`echo $i | cut -d"|" -f2`
You might want to do the whole thing in awk:
awk -F\| 'NR==FNR{old[++n]=$1;new[n]=$2;next}{for(i=1;i<=n;++i)gsub(old[i],new[i])}1' replacement_list file
Build up a list of old and new words from the first file. The next ensures that the rest of the script isn't run on the first file. For the second file, loop through the list of replacements and perform them each one by one. The 1 at the end means that the line is printed.
{ cat replacement_list;echo "-End-"; cat YourFile; } | sed -n '1,/-End-/ s/$/³/;1h;1!H;$ {g
t again
:again
/^-End-³\n/ {s///;b done
}
s/^\([^|]*\)|\([^³]*\)³\(\n\)\(.*\)\1/\1|\2³\3\4\2/
t again
s/^[^³]*³\n//
t again
:done
p
}'
More for fun to code via sed. Try maybe for a time perfomance because this start only 1 sed that is recursif.
for posix sed (so --posix with GNU sed)
explaination
copy replacement list in front of file content with a delimiter (for line with ³ and for list with -End-) for an easier sed handling (hard to use \n in class character in posix sed.
place all line in buffer (add the delimiter of line for replacement list and -End- before)
if this is -End-³, remove the line and go to final print
replace each first pattern (group 1) found in text by second patttern (group 2)
if found, restart (t again)
remove first line
restart process (t again). T is needed because b does not reset the test and next t is always true.
Thanks to #miku above;
I have a 100MB file with a list of 80k replacement-strings.
I tried various combinations of sed's sequentially or parallel, but didn't see throughputs getting shorter than about a 20-hour runtime.
Instead I put my list into a sequence of scripts like "cat in | replace aold anew bold bnew cold cnew ... > out ; rm in ; mv out in".
I randomly picked 1000 replacements per file, so it all went like this:
# first, split my replace-list into manageable chunks (89 files in this case)
split -a 4 -l 1000 80kReplacePairs rep_
# next, make a 'replace' script out of each chunk
for F in rep_* ; do \
echo "create and make executable a scriptfile" ; \
echo '#!/bin/sh' > run_$F.sh ; chmod +x run_$F.sh ; \
echo "for each chunk-file line, strip line-ends," ; \
echo "then with sed, turn '{long list}' into 'cat in | {long list}' > out" ; \
cat $F | tr '\n' ' ' | sed 's/^/cat in | replace /;s/$/ > out/' >> run_$F.sh ;
echo "and append commands to switch in and out files, for next script" ; \
echo -e " && \\\\ \nrm in && mv out in\n" >> run_$F.sh ; \
done
# put all the replace-scripts in sequence into a main script
ls ./run_rep_aa* > allrun.sh
# make it executable
chmod +x allrun.sh
# run it
nohup ./allrun.sh &
.. which ran in under 5 mins, a lot less than 20 hours !
Looking back, I could have used more pairs per script, by finding how many lines would make up the limit.
xargs --show-limits </dev/null 2>&1 | grep --color=always "actually use:"
Maximum length of command we could actually use: 2090490
So just under 2MB; how many pairs would that be for my script ?
head -c 2090490 80kReplacePairs | wc -l
76923
So it seems I could have used 2 * 40000-line chunks
to expand on chthonicdaemon's solution
live demo
#! /bin/sh
# build regex from text file
REGEX_FILE=some-patch.regex.diff
# test
# set these with "export key=val"
SOME_VAR_NAME=hello
ANOTHER_VAR_NAME=world
escape_b() {
echo "$1" | sed 's,/,\\/,g'
}
regex="$(
(echo; cat "$REGEX_FILE"; echo) \
| perl -p -0 -e '
s/\n#[^\n]*/\n/g;
s/\(\(SOME_VAR_NAME\)\)/'"$(escape_b "$SOME_VAR_NAME")"'/g;
s/\(\(ANOTHER_VAR_NAME\)\)/'"$(escape_b "$ANOTHER_VAR_NAME")"'/g;
s/([^\n])\//\1\\\//g;
s/\n-([^\n]+)\n\+([^\n]*)(?:\n\/([^\n]+))?\n/s\/\1\/\2\/\3;\n/g;
'
)"
echo "regex:"; echo "$regex" # debug
exec perl -00 -p -i -e "$regex" "$#"
prefixing lines with -+/ allows empty "plus" values, and protects leading whitespace from buggy text editors
sample input: some-patch.regex.diff
# file format is similar to diff/patch
# this is a comment
# replace all "a/a" with "b/b"
-a/a
+b/b
/g
-a1|a2
+b1|b2
/sg
# this is another comment
-(a1).*(a2)
+b\1b\2b
-a\na\na
+b
-a1-((SOME_VAR_NAME))-a2
+b1-((ANOTHER_VAR_NAME))-b2
sample output
s/a\/a/b\/b/g;
s/a1|a2/b1|b2/;;
s/(a1).*(a2)/b\1b\2b/;
s/a\na\na/b/;
s/a1-hello-a2/b1-world-b2/;
this regex format is compatible with sed and perl
since miku mentioned mysql replace:
replacing fixed strings with regex is non-trivial,
since you must escape all regex chars,
but you also must handle backslash escapes ...
naive escaper:
echo '\(\n' | perl -p -e 's/([.+*?()\[\]])/\\\1/g'
\\(\n
COMPANY_NAME=`cat file.txt | grep "company_name" | cut -d '=' -f 2`
outputs something like this
"Abc Inc";
What I want to do is I want to remove the trailing ";" as well. How can i do that? I am a beginner to bash. Any thoughts or suggestions would be helpful.
This will remove the last character contained in your COMPANY_NAME var regardless if it is or not a semicolon:
echo "$COMPANY_NAME" | rev | cut -c 2- | rev
I'd use sed 's/;$//'. eg:
COMPANY_NAME=`cat file.txt | grep "company_name" | cut -d '=' -f 2 | sed 's/;$//'`
foo="hello world"
echo ${foo%?}
hello worl
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
It's exactly what head was designed to do, thus less command-line options and an easier-to-read command
It saves you having to think about regular expressions, which are cool/powerful but often overkill
It saves your machine having to think about regular expressions, so will be imperceptibly faster
I 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
Using sed, if you don't know what the last character actually is:
$ grep company_name file.txt | cut -d '=' -f2 | sed 's/.$//'
"Abc Inc"
Don't abuse cats. Did you know that grep can read files, too?
The canonical approach would be this:
grep "company_name" file.txt | cut -d '=' -f 2 | sed -e 's/;$//'
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 $1' file.txt )
don't have to chain so many tools. Just one awk command does the job
COMPANY_NAME=$(awk -F"=" '/company_name/{gsub(/;$/,"",$2) ;print $2}' file.txt)
In Bash using only one external utility:
IFS='= ' read -r discard COMPANY_NAME <<< $(grep "company_name" file.txt)
COMPANY_NAME=${COMPANY_NAME/%?}
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 "\"*.*\""
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
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%??}
cat file.txt | grep "company_name" | cut -d '=' -f 2 | cut -d ';' -f 1
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\}$//'.