How do I remove the first and the last quotes?
echo "\"test\"" | sed 's/"//' | sed 's/"$//'
The above is working as expected, But I guess there must be a better way.
You can combine the sed calls into one:
echo "\"test\"" | sed 's/"//;s/"$//'
The command you posted will remove the first quote even if it's not at the beginning of the line. If you want to make sure that it's only done if it is at the beginning, then you can anchor it like this:
echo "\"test\"" | sed 's/^"//;s/"$//'
Some versions of sed don't like multiple commands separated by semicolons. For them you can do this (it also works in the ones that accept semicolons):
echo "\"test\"" | sed -e 's/^"//' -e 's/"$//'
Maybe you prefer something like this:
echo '"test"' | sed 's/^"\(.*\)"$/\1/'
if you are sure there are no other quotes besides the first and last, just use /g modifier
$ echo "\"test\"" | sed 's/"//g'
test
If you have Ruby(1.9+)
$ echo $s
blah"te"st"test
$ echo $s | ruby -e 's=gets.split("\"");print "#{s[0]}#{s[1..-2].join("\"")+s[-1]}"'
blahte"sttest
Note the 2nd example the first and last quotes which may not be exactly at the first and last positions.
example with more quotes
$ s='bl"ah"te"st"tes"t'
$ echo $s | ruby -e 's=gets.split("\"");print "#{s[0]}#{s[1..-2].join("\"")+s[-1]}"'
blah"te"st"test
Related
I have a simple .csv file with lines that holds 't' values. Here is the example:
2ABC;t;t;t;tortuga;fault;t;t;bored
I want to replace them to '1' using sed.
If I make sed "s/;t;/;1;/g" I get the next result:
2ABC;1;t;1;tortuga;fault;1;t;bored
As you can see, consecutive ';t;' have been replaced through one. Yes, I can replace all ';t;' by sed -e "s/;t;/;1;/g" -e "s/;t;/;1;/g" but this is boring.
How can I make the replacement by one sed command?
If there is something to replace, branch to replace again.
sed ': again; /;t;/{ s//;1;/; b again }'
Overall, parsing cvs with sed is crude. Consider awk.
awk -F';' -v OFS=';' '{ for(i=1;i<=NF;++i) if ($i=="t") $i=1 } 1'
Lookarounds is helpful in such cases:
$ s='t;2ABC;t;t;t;tortuga;fault;t;t;bored;t'
$ echo "$s" | perl -lpe 's/(?<![^;])t(?![^;])/1/g'
1;2ABC;1;1;1;tortuga;fault;1;1;bored;1
echo '2ABC;t;t;t;tortuga;fault;t;t;bored' |
— gawk-specific solution
gawk -be '(ORS = RT)^!(NF = NF)' FS='^t$' OFS=1 RS=';'
— cross-awk-solution
{m,g,n}awk 'gsub(FS, OFS, $!(NF = NF))^_' FS=';t;' OFS=';1;' RS=
2ABC;1;1;1;tortuga;fault;1;1;bored
In my script, have a possible version number: 15.03.2 set to variable $STRING. These numbers always change. I want to strip it down to: 15.03 (or whatever it will be next time).
How do I remove everything after the second . using sed?
Something like:
$(echo "$STRING" | sed "s/\.^$\.//")
(I don't know what ^, $ and others do, but they look related, so I just guessed.)
I think the better tool here is cut
echo '15.03.2' | cut -d . -f -2
This might work for you (GNU sed):
sed 's/\.[^.]*//2g' file
Remove the second or more occurrence of a period followed by zero or non-period character(s).
$ echo '15.03.2' | sed 's/\([^.]*\.[^.]*\)\..*/\1/'
15.03
More generally to skip N periods:
$ echo '15.03.2.3.4.5' | sed -E 's/(([^.]*\.){2}[^.]*)\..*/\1/'
15.03.2
$ echo '15.03.2.3.4.5' | sed -E 's/(([^.]*\.){3}[^.]*)\..*/\1/'
15.03.2.3
$ echo '15.03.2.3.4.5' | sed -E 's/(([^.]*\.){4}[^.]*)\..*/\1/'
15.03.2.3.4
How do you escape line beginning and line end in bracket expressions in sed?
For example, let's say I want to replace both comma, line beginning, and line end in each line with pipe:
echo "a,b,c" | sed 's/,/|/g'
# a|b|c
echo "a,b,c" | sed 's/^/|/g'
# |a,b,c
echo "a,b,c" | sed 's/$/|/g'
# a,b,c|
echo "a,b,c" | sed 's/[,^$]/|/g'
# a|b|c
I would expect the last command to produce |a|b|c|. I also tried escaping the line beginning and line end via backslash, with no change.
With GNU sed with extended regular expressions, you can do:
$ echo "a,b,c" | /opt/gnu/bin/sed -E 's/^|,|$/|/g'
|a|b|c|
$
The -E option enables the extended regular expressions, as does -r, but -E is also used by other sed variants for the same purpose, unlike -r.
However, for reasons which elude me, the BSD (macOS) variant of sed produces:
$ echo "a,b,c" | sed -E 's/^|,|$/|/g'
|a|b|c
$
I can't think why.
If this variability is unacceptable, go with the three-substitution solution:
$ echo "a,b,c" | sed -e "s/^/|/" -e "s/$/|/" -e "s/,/|/g"
|a|b|c|
$
which should work with any variant of sed. However, note that echo "" | sed …3 subs… produces || whereas the -E variant produces |. I'm not sure if there's an easy fix for that.
You tried this, but it didn't do what you wanted:
$ echo "a,b,c" | sed 's/[,^$]/|/g'
a|b|c
$
This is what should be expected. Inside character classes, most special characters lose their special-ness. There is nothing special about $ (or , but it isn't a metacharacter anyway) in a character class; ^ is only special at the start of the class and it negates the character class. That means that what follows shows the correct, expected behaviour from this permutation of the contents of your character class:
$ echo "a,b\$\$b,c" | sed 's/[^,$]/|/g'
|,|$$|,|
$
It mapped all the non-comma, non-dollar characters to pipes. I should be using single quotes around the echo; then the backslashes wouldn't be necessary. I just followed the question's code quietly.
Following sed may help you in same.
echo "a,b,c" | sed 's/^/|/;s/,/|/g;s/$/|/'
Output will be as follows.
|a|b|c|
How to replace '.' with '_' within the part of the line before the '=' char in the input below
Need single sed command to do all three
echo "few.num.dots=/home/user/.hidden/folder.dot" | sed 's/\./_/g'
required output => few_num_dots=/home/user/.hidden/folder.dot
echo "var=nodot" | sed 's/\./_/g'
required output => var=nodot
echo "var.one=onedot.notthis" | sed 's/\./_/g'
required output => var_one=onedot.notthis
You can use a conditional branching using the t command. It does a loop until the substitution command fails, and the command replaces any . character followed by an equal sign:
echo "few.num.dots=/home/user/.hidden/folder.dot" |
sed ':a; s/\.\([^=]*=\)/_\1/; ta'
It yields:
few_num_dots=/home/user/.hidden/folder.dot
perl?
echo "few.num.dots=/home/user/.hidden/folder.dot" |
perl -pe 's/^[^=]+/ ($x=$&) =~ tr{.}{_}; $x /e'
few_num_dots=/home/user/.hidden/folder.dot
awk?
awk -F= -v OFS='=' '{gsub(/\./,"_",$1)} 1'
You can do it this way as well,
echo "few.num.dots=/home/user/.hidden/folder.dot" |
sed -e '1,/./s/\./_/' -e '1,/./s/\./_/'
few_num_dots=/home/user/.hidden/folder.dot
First -e replaces the first occurrence of the pattern ., next -e replaces the next one...
Using awk
$ echo "few.num.dots=/home/user/.hidden/folder.dot" |awk '/=/{gsub(/\./,"_",$1)}1' FS="=" OFS="="
few_num_dots=/home/user/.hidden/folder.dot
$ echo "var.one=onedot.notthis" |awk '/=/{gsub(/\./,"_",$1)}1' FS="=" OFS="="
var_one=onedot.notthis
This might work for you (GNU sed):
sed 's/=/\n&/;h;y/./_/;G;s/\n.*\n.*\n//' file
Insert a marker to divide the line, copy the line, translate the characters, append the original line and using the marker reconstitute the line.
In a bash script, files with spaces show up as "File\ with\ spaces.txt" and I want to substitute those slashed-spaces with either _ or +.
How can I tell sed to do that? I had no success using;
$1=~/File\ with\ spaces.txt
ext=$1
web=$(echo "$ext" | sed 's/\ /+/')
I'm open to suggestions if there's a better way than through sed.
[EDIT]: Foo Bah's solution works well, but it substitutes only the first space because the text following it is treated as arguments rather than part of the $1. Any way around this?
sed 's/\\\\ /+/';
\\\\ evaluates to a \\ at the shell level, and then into a literal \ within sed.
Sed recognises \ as space just fine:
bee#i20 ~ $ echo file\ 123 | sed 's/\ /+/'
file+123
Your bash script syntax is all wrong, though.
Not sure what you were trying to do with the script, but here is an example of replacing spaces with +:
ext=~/File\ with\ spaces.txt
web=`echo "$ext" | sed 's/\ /+/g'`
echo $web
Upd:
Oh, and you need the g flag to replace all occurences of space, not only the first one. Fixed above.
you want to escape the slash:
web=$(echo "$ext" | sed 's/\\ /_/g')
single quotes are your friend
the following should be used with single quoted args for $1 and $2
#!/bin/bash
ESCAPE='\\'
if [ $# -ne 2 ];then
echo "$0 <TO_ESCAPE> <IN_STRING>"
echo args should be in single quotes!!
exit 1
fi
TO_ESCAPE="${1}"
IN_STRING="${2}"
if [ ${TO_ESCAPE} = '\' ];then
TO_ESCAPE='\\'
fi
echo "${IN_STRING}" | sed "s/${TO_ESCAPE}/${ESCAPE}${TO_ESCAPE}/g"