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\}$//'.
Related
I've a file that contains info that I'm retrieving such way
Command
cat 2018_02_15_09_01_08_result.tsv | grep -o [A-Z]\\*[0-9]*:[0-9]* | sort | uniq | sed -e 's/^/HLA-/' |tr '\n' ',' | sed '$ s/.$//'
Output
HLA-A*30:02,HLA-B*18:01,HLA-C*05:01
But I'm trying to save this in variable, the asterisk and a letter disappears, I've tried several ways, adding/removing commas etc and I'm yet not able to print it properly.
hla=`cat 2018_02_15_09_01_08_result.tsv | grep -o [A-Z]\\*[0-9]*:[0-9]* | sort | uniq | sed -e 's/^/HLA-/' |tr '\n' ',' | sed '$ s/.$//'`
echo $hla
HLA-05:01,HLA-18:01,HLA-30:02
echo "$hla"
HLA-05:01,HLA-18:01,HLA-30:02
There are multiple errors here, most of which will be aptly diagnosed by http://shellcheck.net/ without any human intervention.
You really should single-quote your regular expressions unless you specifically require the shell to perform wildcard expansion and whitespace tokenization on the regex before executing the command.
The obsolescent `command` in backticks introduces some unfortunate additional shell handling on the string inside the backticks. The solution since the 1990s is to prefer the $(command) syntax for command substitution, which does not exhibit this problem.
The cat is useless; grep knows full well how to read a file.
Try this refactored code:
hla=$(grep -o '[A-Z]*[0-9]*:[0-9]*' 2018_02_15_09_01_08_result.tsv |
sort -u | sed -e 's/^/HLA-/' |tr '\n' ',' | sed '$ s/.$//')
echo "$hla"
The double quotes around the variable interpolation in the echo are necessary and useful; notice also the line wraps for legibility and the use of sort -u in preference over sort | uniq (and generally try to reduce the number of processes -- once I understand what the sed | tr | sed does I can probably propose a simplification for that, too). Perhaps the simplest fix would be to refactor all of this into a single Awk script, but without access to the input, it's hard to tell you in more detail what that might look like.
(Also, are you really sure you need to capture the value to a variable? Often variable=value; echo "$variable" is just an obscure and inefficient way to say echo "value". And variable=$(command); echo "$variable" is better written simply command and capturing the command's standard output just so you can print it to standard output is a pure waste of cycles, unless you are planning to do something more with that variable's value.)
I've solved it by saving the output of the command with a redirection:
cat 2018_02_15_09_01_08_result.tsv |
grep -o [A-Z]\\*[0-9]*:[0-9]* |
sort | uniq |
sed -e 's/^/HLA-/' |tr '\n' ',' | sed '$ s/.$//' > out_file
hla=`cat out_file`
echo $hla
which gets me the expected HLA-A*30:02,HLA-B*18:01,HLA-C*05:01. Not the ideal solution, but it works.
I've got the string 10.11.12. I'd like to get everything up to but not including the second dot, in other words in this case I want to return 10.11.
I'm using bash but can't figure it out. I was hoping sed might help but I've spent ages googling and can't figure it out. e.g. this didn't work:
$ echo 10.11.12 | sed "s/\d*\.\d*/\0/p"
10.11.12
10.11.12
help!
See BashFAQ #100 ("How can I do string manipulations in bash?"). Like many common operations, this can be done with a parameter expansion.
s=10.11.12
result="${s%.*}" # Remove everything after the last .
echo "$result"
Of course, you could go directly to:
echo "${s%.*}"
In perl:
echo 10.11.12 | perl -pe 's/(.*)\..*/$1/'
Gives what you expect
You could do the following without sed.
echo 10.12.13 | rev | cut -d "." -f2- | rev
I'm having a bit of an issue cutting the output up from egrep. I have output like:
From: First Last
From: First Last
From: First Last
I want to cut out the "From: " (essentially leaving the "First Last").
I tried
cut -d ":" -f 7
but the output is just a bunch of blank lines.
I would appreciate any help.
Here's the full code that I am trying to use if it helps:
egrep '^From:' $file | cut -d ":" -f 7
NOTE: I've already tested the egrep portion of the code and it works as expected.
The cut command lines in your question specify colon-separated fields and that you want the output to consist only of field 7; since there is no 7th field in your input, the result you're getting isn't what you intend.
Since the "From:" prefix appears to be identical across all lines, you can simply cut from the 7th character onward:
egrep '^From:' $file | cut -c7-
and get the result you intend.
you were really close.
I think you only need to replace ":" with " " as separator and add "-" after the "7": like this:
cut -d " " -f 2-
I tested and works pretty well.
The -f argument is for what fields. Since there is only one : in the line, there's only two fields. So changing -f 7 to -f 2- will give you want you want. Albeit with a leading space.
You can combine the egrep and cut parts into one command with sed:
sed -n 's/^From: //gp' $file
sed -n turns off printing by default, and then I am using p in the sed command explicitly to print the lines I want.
You can use sed:
sed 's/^From: *//'
OR awk:
awk -F ': *' '$1=="From"{print $2}'
OR grep -oP
grep -oP '^From: *\K.*'
Here is a Bash one-liner:
grep ^From file.txt | while read -a cols; do echo ${cols[#]:1}; done
See: Handling positional parameters at wiki.bash-hackers.org
cut itself is a very handy tool in bash
cut -d (delimiter character) -f (fields that you want as output)
a single field is given directly as -f 3 ,
range of fields can be selected as -f 5-9
so in your this particular case code would be
egrep '^From:' $file | cut -d\ -f 2-3
the delimiter is space here and can be escaped using a \
-f 1 corresponds to " From " and 2-3 corresponds to " First Last "
I am writing shell script for embedded Linux in a small industrial box. I have a variable containing the text pid: 1234 and I want to strip first X characters from the line, so only 1234 stays. I have more variables I need to "clean", so I need to cut away X first characters and ${string:5} doesn't work for some reason in my system.
The only thing the box seems to have is sed.
I am trying to make the following to work:
result=$(echo "$pid" | sed 's/^.\{4\}//g')
Any ideas?
The following should work:
var="pid: 1234"
var=${var:5}
Are you sure bash is the shell executing your script?
Even the POSIX-compliant
var=${var#?????}
would be preferable to using an external process, although this requires you to hard-code the 5 in the form of a fixed-length pattern.
Here's a concise method to cut the first X characters using cut(1). This example removes the first 4 characters by cutting a substring starting with 5th character.
echo "$pid" | cut -c 5-
Use the -r option ("use extended regular expressions in the script") to sed in order to use the {n} syntax:
$ echo 'pid: 1234'| sed -r 's/^.{5}//'
1234
Cut first two characters from string:
$ string="1234567890"; echo "${string:2}"
34567890
pipe it through awk '{print substr($0,42)}' where 42 is one more than the number of characters to drop. For example:
$ echo abcde| awk '{print substr($0,2)}'
bcde
$
Chances are, you'll have cut as well. If so:
[me#home]$ echo "pid: 1234" | cut -d" " -f2
1234
Well, there have been solutions here with sed, awk, cut and using bash syntax. I just want to throw in another POSIX conform variant:
$ echo "pid: 1234" | tail -c +6
1234
-c tells tail at which byte offset to start, counting from the end of the input data, yet if the the number starts with a + sign, it is from the beginning of the input data to the end.
Another way, using cut instead of sed.
result=`echo $pid | cut -c 5-`
I found the answer in pure sed supplied by this question (admittedly, posted after this question was posted). This does exactly what you asked, solely in sed:
result=\`echo "$pid" | sed '/./ { s/pid:\ //g; }'\``
The dot in sed '/./) is whatever you want to match. Your question is exactly what I was attempting to, except in my case I wanted to match a specific line in a file and then uncomment it. In my case it was:
# Uncomment a line (edit the file in-place):
sed -i '/#\ COMMENTED_LINE_TO_MATCH/ { s/#\ //g; }' /path/to/target/file
The -i after sed is to edit the file in place (remove this switch if you want to test your matching expression prior to editing the file).
(I posted this because I wanted to do this entirely with sed as this question asked and none of the previous answered solved that problem.)
Rather than removing n characters from the start, perhaps you could just extract the digits directly. Like so...
$ echo "pid: 1234" | grep -Po "\d+"
This may be a more robust solution, and seems more intuitive.
This will do the job too:
echo "$pid"|awk '{print $2}'
I am a newbie in Bash and I am doing some string manipulation.
I have the following file among other files in my directory:
jdk-6u20-solaris-i586.sh
I am doing the following to get jdk-6u20 in my script:
myvar=`ls -la | awk '{print $9}' | egrep "i586" | cut -c1-8`
echo $myvar
but now I want to convert jdk-6u20 to jdk1.6.0_20. I can't seem to figure out how to do it.
It must be as generic as possible. For example if I had jdk-6u25, I should be able to convert it at the same way to jdk1.6.0_25 so on and so forth
Any suggestions?
Depending on exactly how generic you want it, and how standard your inputs will be, you can probably use AWK to do everything. By using FS="regexp" to specify field separators, you can break down the original string by whatever tokens make the most sense, and put them back together in whatever order using printf.
For example, assuming both dashes and the letter 'u' are only used to separate fields:
myvar="jdk-6u20-solaris-i586.sh"
echo $myvar | awk 'BEGIN {FS="[-u]"}; {printf "%s1.%s.0_%s",$1,$2,$3}'
Flavour according to taste.
Using only Bash:
for file in jdk*i586*
do
file="${file%*-solaris*}"
file="${file/-/1.}"
file="${file/u/.0_}"
do_something_with "$file"
done
i think that sed is the command for you
You can try this snippet:
for fname in *; do
newname=`echo "$fname" | sed 's,^jdk-\([0-9]\)u\([0-9][0-9]*\)-.*$,jdk1.\1.0_\2,'`
if [ "$fname" != "$newname" ]; then
echo "old $fname, new $newname"
fi
done
awk 'if(match($9,"i586")){gsub("jdk-6u20","jdk1.6.0_20");print $9;}'
The if(match()) supersedes the egrep bit if you want to use it. You could use substr($9,1,8) instead of cut as well.
garph0 has a good idea with sed; you could do
myvar=`ls jdk*i586.sh | sed 's/jdk-\([0-9]\)u\([0-9]\+\).\+$/jdk1.\1.0_\2/'`
You're needing the awk in there is an artifact of the -l switch on ls. For pattern substitution on lines of text, sed is the long-time champion:
ls | sed -n '/^jdk/s/jdk-\([0-9][0-9]*\)u\([0-9][0-9]*\)$/jdk1.\1.0_\2/p'
This was written in "old-school" sed which should have greater portability across platforms. The expression says:
don't print lines unless they match -n
on lines beginning with 'jdk' do:
on a line that contains only "jdk-IntegerAuIntegerB"
change it to "jdk.1.IntegerA.0_IntegerB"
and print it
Your sample becomes even simpler as:
myvar=`echo *solaris-i586.sh | sed 's/-solaris-i586\.sh//'`