Here is what is the file beast.txt:
savage
beast
tank
beauty
I am looking to get it to look like:
tbwbhf
cfbtu
ubol
cfbvuz
This what I have but when I run this it changes the letters to "[a-z]+1".
cat beast.txt | sed 's/[a-z]/[a-z]+1/g' > savage.txt
Do I have to use other special characters to in the sed command or should this be done with a loop? Thank you for your help.
This is not a job for sed:
$ echo 'savage beast tank beauty' | tr a-z b-za
tbwbhf cfbtu ubol cfbvuz
If you really want to use sed, you can do:
$ b2z=bcdefghijklmnopqrstuvwxyz
$ echo 'savage beast tank beauty' | sed -e "y/a$b2z/${b2z}a/"
tbwbhf cfbtu ubol cfbvuz
I used this '$ echo 'savage beast tank beauty' | tr a-z b-za' and it worked.
I am now having trouble reversing what I just did.
Related
I am working on shell script and new to it. I want to extract the string between double $$ characters, for example:
input:
$$extractabc$$
output
extractabc
I used grep and sed but not working out. Any suggestions are welcome!
You could do
awk -F"$" '{print $3}' file.txt
assuming the file contained input:$$extractabc$$ output:extractabc. awk splits your data into pieces using $ as a delimiter. First item will be input:, next will be empty, next will be extractabc.
You could use sed like so to get the same info.
sed -e 's/.*$$\(.*\)$$.*/\1/' file.txt
sed looks for information between $$s and outputs that. The goal is to type something like this .*$$(.*)$$.*. It's greedy but just stay with me.
looks for .* - i.e. any character zero or more times before $$
then the string should have $$
after $$ there'll be any character zero or more times
then the string should have another $$
and some more characters to follow
between the 2 $$ is (.*). String found between $$s is given a placeholder \1
sed finds such information and publishes it
Using grep PCRE (where available) and look-around:
$ echo '$$extractabc$$' | grep -oP "(?<=\\$\\$).*(?=\\$\\$)"
extractabc
echo '$$extractabc$$' | awk '{gsub(/\$\$/,"")}1'
extractabc
Here is an other variation:
echo "$$extractabc$$" | awk -F"$$" 'NF==3 {print $2}'
It does test of there are two set of $$ and only then prints whats between $$
Does also work for input like blabla$$some_data$$moreblabla
How about remove all the $ in the input?
$ echo '$$extractabc$$' | sed 's/\$//g'
extractabc
Same with tr
$ echo '$$extractabc$$' | tr -d '$'
extractabc
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
In bash i want to replace
H+O2=O+OH
with
H+O_2=O+OH
I tried countless things with awk and sed but nothing worked so far.
The Problem is that i dont know how to save the string and then just add the _ instead of replacing the O or the 2.
Thanks!
You can use sed:
$> s='H+O2=O+OH'
$> sed -E 's/([A-Z])([0-9])/\1_\2/g' <<< "$s"
H+O_2=O+OH
If Perl is an option, you can use a technique very similar to Anubhava's:
echo 'H+O2=O+OH' | perl -pe 's/([A-Z])([0-9])/\1_\2/g'
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 $2}'
COMPANY_NAME=`cat file.txt | grep "company_name" | cut -d '=' -f 2`
outputs something like this
"Abc Inc";
What I want to do is I want to remove the trailing ";" as well. How can i do that? I am a beginner to bash. Any thoughts or suggestions would be helpful.
This will remove the last character contained in your COMPANY_NAME var regardless if it is or not a semicolon:
echo "$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\}$//'.