Related
Below is the snippet of a shell script from a larger script. It removes the quotes from the string that is held by a variable. I am doing it using sed, but is it efficient? If not, then what is the efficient way?
#!/bin/sh
opt="\"html\\test\\\""
temp=`echo $opt | sed 's/.\(.*\)/\1/' | sed 's/\(.*\)./\1/'`
echo $temp
Use tr to delete ":
echo "$opt" | tr -d '"'
NOTE: This does not fully answer the question, removes all double quotes, not just leading and trailing. See other answers below.
There's a simpler and more efficient way, using the native shell prefix/suffix removal feature:
temp="${opt%\"}"
temp="${temp#\"}"
echo "$temp"
${opt%\"} will remove the suffix " (escaped with a backslash to prevent shell interpretation).
${temp#\"} will remove the prefix " (escaped with a backslash to prevent shell interpretation).
Another advantage is that it will remove surrounding quotes only if there are surrounding quotes.
BTW, your solution always removes the first and last character, whatever they may be (of course, I'm sure you know your data, but it's always better to be sure of what you're removing).
Using sed:
echo "$opt" | sed -e 's/^"//' -e 's/"$//'
(Improved version, as indicated by jfgagne, getting rid of echo)
sed -e 's/^"//' -e 's/"$//' <<<"$opt"
So it replaces a leading " with nothing, and a trailing " with nothing too. In the same invocation (there isn't any need to pipe and start another sed. Using -e you can have multiple text processing).
If you're using jq and trying to remove the quotes from the result, the other answers will work, but there's a better way. By using the -r option, you can output the result with no quotes.
$ echo '{"foo": "bar"}' | jq '.foo'
"bar"
$ echo '{"foo": "bar"}' | jq -r '.foo'
bar
There is a straightforward way using xargs:
> echo '"quoted"' | xargs
quoted
xargs uses echo as the default command if no command is provided and strips quotes from the input, see e.g. here. Note, however, that this will work only if the string does not contain additional quotes. In that case it will either fail (uneven number of quotes) or remove all of them.
If you came here for aws cli --query, try this. --output text
You can do it with only one call to sed:
$ echo "\"html\\test\\\"" | sed 's/^"\(.*\)"$/\1/'
html\test\
The shortest way around - try:
echo $opt | sed "s/\"//g"
It actually removes all "s (double quotes) from opt (are there really going to be any more double quotes other than in the beginning and the end though? So it's actually the same thing, and much more brief ;-))
The easiest solution in Bash:
$ s='"abc"'
$ echo $s
"abc"
$ echo "${s:1:-1}"
abc
This is called substring expansion (see Gnu Bash Manual and search for ${parameter:offset:length}). In this example it takes the substring from s starting at position 1 and ending at the second last position. This is due to the fact that if length is a negative value it is interpreted as a backwards running offset from the end of parameter.
Update
A simple and elegant answer from Stripping single and double quotes in a string using bash / standard Linux commands only:
BAR=$(eval echo $BAR) strips quotes from BAR.
=============================================================
Based on hueybois's answer, I came up with this function after much trial and error:
function stripStartAndEndQuotes {
cmd="temp=\${$1%\\\"}"
eval echo $cmd
temp="${temp#\"}"
eval echo "$1=$temp"
}
If you don't want anything printed out, you can pipe the evals to /dev/null 2>&1.
Usage:
$ BAR="FOO BAR"
$ echo BAR
"FOO BAR"
$ stripStartAndEndQuotes "BAR"
$ echo BAR
FOO BAR
This is the most discrete way without using sed:
x='"fish"'
printf " quotes: %s\nno quotes: %s\n" "$x" "${x//\"/}"
Or
echo $x
echo ${x//\"/}
Output:
quotes: "fish"
no quotes: fish
I got this from a source.
Linux=`cat /etc/os-release | grep "ID" | head -1 | awk -F= '{ print $2 }'`
echo $Linux
Output:
"amzn"
Simplest ways to remove double quotes from variables are
Linux=`echo "$Linux" | tr -d '"'`
Linux=$(eval echo $Linux)
Linux=`echo ${Linux//\"/}`
Linux=`echo $Linux | xargs`
All provides the Output without double quotes:
echo $Linux
amzn
I know this is a very old question, but here is another sed variation, which may be useful to someone. Unlike some of the others, it only replaces double quotes at the start or end...
echo "$opt" | sed -r 's/^"|"$//g'
If you need to match single or double quotes, and only strings that are properly quoted. You can use this slightly more complex regex...
echo $opt | sed -E "s|^(['\"])(.*)\1$|\2|g"
This uses backrefences to ensure the quote at the end is the same as at the start.
In Bash, you could use the following one-liner:
[[ "${var}" == \"*\" || "${var}" == \'*\' ]] && var="${var:1:-1}"
This will remove surrounding quotes (both single and double) from the string stored in var while keeping quote characters inside the string intact. Also, this won't do anything if there's only a single leading quote or only a single trailing quote or if there are mixed quote characters at start/end.
Wrapped in a function:
#!/usr/bin/env bash
# Strip surrounding quotes from string [$1: variable name]
function strip_quotes() {
local -n var="$1"
[[ "${var}" == \"*\" || "${var}" == \'*\' ]] && var="${var:1:-1}"
}
str="'hello world'"
echo "Before: ${str}"
strip_quotes str
echo "After: ${str}"
My version
strip_quotes() {
while [[ $# -gt 0 ]]; do
local value=${!1}
local len=${#value}
[[ ${value:0:1} == \" && ${value:$len-1:1} == \" ]] && declare -g $1="${value:1:$len-2}"
shift
done
}
The function accepts variable name(s) and strips quotes in place. It only strips a matching pair of leading and trailing quotes. It doesn't check if the trailing quote is escaped (preceded by \ which is not itself escaped).
In my experience, general-purpose string utility functions like this (I have a library of them) are most efficient when manipulating the strings directly, not using any pattern matching and especially not creating any sub-shells, or calling any external tools such as sed, awk or grep.
var1="\"test \\ \" end \""
var2=test
var3=\"test
var4=test\"
echo before:
for i in var{1,2,3,4}; do
echo $i="${!i}"
done
strip_quotes var{1,2,3,4}
echo
echo after:
for i in var{1,2,3,4}; do
echo $i="${!i}"
done
I use this regular expression, which avoids removing quotes from strings that are not properly quoted, here the different outputs are shown depending on the inputs, only one with begin-end quote was affected:
echo '"only first' | sed 's/^"\(.*\)"$/\1/'
Output: >"only first<
echo 'only last"' | sed 's/^"\(.*\)"$/\1/'
Output: >"only last"<
echo '"both"' | sed 's/^"\(.*\)"$/\1/'
Output: >both<
echo '"space after" ' | sed 's/^"\(.*\)"$/\1/'
Output: >"space after" <
echo ' "space before"' | sed 's/^"\(.*\)"$/\1/'
Output: > "space before"<
STR='"0.0.0"' ## OR STR="\"0.0.0\""
echo "${STR//\"/}"
## Output: 0.0.0
There is another way to do it. Like:
echo ${opt:1:-1}
If you try to remove quotes because the Makefile keeps them, try this:
$(subst $\",,$(YOUR_VARIABLE))
Based on another answer: https://stackoverflow.com/a/10430975/10452175
I'm trying to format output in a way that inserts newline characters after each 'line', with lines denoted by double quotes (""). The quotes themselves are temporary and to be stripped in a later step.
Input:
"a",1,"aa""b",2,"bb"
Output:
a,1,aa
b,2,bb
I've tried:
sed 's/""/\n/'
sed 's/""/\/g'
tr '""' '\n'
But tr seems to replace every quote character and sed seems to insert \n as text instead of a newline. What can I do to make this work?
echo '"a",1,"aa""b",2,"bb"' |awk -v RS='""' '{$1=$1} {gsub(/"/,"")}1'
a,1,aa
b,2,bb
or using sed:
echo '"a",1,"aa""b",2,"bb"' |sed -e 's/""/\n/' -e 's/"//g' # OR sed -e 's/""/\n/;s/"//g'
a,1,aa
b,2,bb
awk solution: Here the default record separator is changed from new line to "". So awk will consider the EOL when it hits "".
sed solution: Here first "" are converted into new line and second replacement is to remove " from each line.
neech#nicolaw.uk:~ $ cat file.txt
"a",1,"aa""b",2,"bb"
neech#nicolaw.uk:~ $ sed 's/""/\n/' file.txt | tr -d '"'
a,1,aa
b,2,bb
You seem to be dealing with POSIX sed, which does not have support for the \n notation. Insert an actual new-line into the pattern, either:
sed 's/""/\
/'
Or:
sed 's/""/\'$'\n''/'
E.g.:
sed 's/""/\
/' | tr -d \"
Output:
a,1,aa
b,2,bb
As suggested by George Vasiliou if you have perl you could use:
> echo '"a",1,"aa""b",2,"bb"' | perl -pe 's/""/"\n"/g;s/"//g'
This avoids the non portable sed problem.
Or for a crappy hack version.
Replace the "" with another character and then use tr (since tr should work with \n) to replace it with \n instead then remove the single " after.
So you can get the "" replaced with newline like this:
sed 's/""/#/g' | tr '#' '\n'
Then the rest follows:
> echo '"a",1,"aa""b",2,"bb"'| sed 's/""/#/g' | tr '#' '\n' | sed 's/\"//g'
Below is the snippet of a shell script from a larger script. It removes the quotes from the string that is held by a variable. I am doing it using sed, but is it efficient? If not, then what is the efficient way?
#!/bin/sh
opt="\"html\\test\\\""
temp=`echo $opt | sed 's/.\(.*\)/\1/' | sed 's/\(.*\)./\1/'`
echo $temp
Use tr to delete ":
echo "$opt" | tr -d '"'
NOTE: This does not fully answer the question, removes all double quotes, not just leading and trailing. See other answers below.
There's a simpler and more efficient way, using the native shell prefix/suffix removal feature:
temp="${opt%\"}"
temp="${temp#\"}"
echo "$temp"
${opt%\"} will remove the suffix " (escaped with a backslash to prevent shell interpretation).
${temp#\"} will remove the prefix " (escaped with a backslash to prevent shell interpretation).
Another advantage is that it will remove surrounding quotes only if there are surrounding quotes.
BTW, your solution always removes the first and last character, whatever they may be (of course, I'm sure you know your data, but it's always better to be sure of what you're removing).
Using sed:
echo "$opt" | sed -e 's/^"//' -e 's/"$//'
(Improved version, as indicated by jfgagne, getting rid of echo)
sed -e 's/^"//' -e 's/"$//' <<<"$opt"
So it replaces a leading " with nothing, and a trailing " with nothing too. In the same invocation (there isn't any need to pipe and start another sed. Using -e you can have multiple text processing).
If you're using jq and trying to remove the quotes from the result, the other answers will work, but there's a better way. By using the -r option, you can output the result with no quotes.
$ echo '{"foo": "bar"}' | jq '.foo'
"bar"
$ echo '{"foo": "bar"}' | jq -r '.foo'
bar
There is a straightforward way using xargs:
> echo '"quoted"' | xargs
quoted
xargs uses echo as the default command if no command is provided and strips quotes from the input, see e.g. here. Note, however, that this will work only if the string does not contain additional quotes. In that case it will either fail (uneven number of quotes) or remove all of them.
If you came here for aws cli --query, try this. --output text
You can do it with only one call to sed:
$ echo "\"html\\test\\\"" | sed 's/^"\(.*\)"$/\1/'
html\test\
The shortest way around - try:
echo $opt | sed "s/\"//g"
It actually removes all "s (double quotes) from opt (are there really going to be any more double quotes other than in the beginning and the end though? So it's actually the same thing, and much more brief ;-))
The easiest solution in Bash:
$ s='"abc"'
$ echo $s
"abc"
$ echo "${s:1:-1}"
abc
This is called substring expansion (see Gnu Bash Manual and search for ${parameter:offset:length}). In this example it takes the substring from s starting at position 1 and ending at the second last position. This is due to the fact that if length is a negative value it is interpreted as a backwards running offset from the end of parameter.
Update
A simple and elegant answer from Stripping single and double quotes in a string using bash / standard Linux commands only:
BAR=$(eval echo $BAR) strips quotes from BAR.
=============================================================
Based on hueybois's answer, I came up with this function after much trial and error:
function stripStartAndEndQuotes {
cmd="temp=\${$1%\\\"}"
eval echo $cmd
temp="${temp#\"}"
eval echo "$1=$temp"
}
If you don't want anything printed out, you can pipe the evals to /dev/null 2>&1.
Usage:
$ BAR="FOO BAR"
$ echo BAR
"FOO BAR"
$ stripStartAndEndQuotes "BAR"
$ echo BAR
FOO BAR
This is the most discrete way without using sed:
x='"fish"'
printf " quotes: %s\nno quotes: %s\n" "$x" "${x//\"/}"
Or
echo $x
echo ${x//\"/}
Output:
quotes: "fish"
no quotes: fish
I got this from a source.
Linux=`cat /etc/os-release | grep "ID" | head -1 | awk -F= '{ print $2 }'`
echo $Linux
Output:
"amzn"
Simplest ways to remove double quotes from variables are
Linux=`echo "$Linux" | tr -d '"'`
Linux=$(eval echo $Linux)
Linux=`echo ${Linux//\"/}`
Linux=`echo $Linux | xargs`
All provides the Output without double quotes:
echo $Linux
amzn
I know this is a very old question, but here is another sed variation, which may be useful to someone. Unlike some of the others, it only replaces double quotes at the start or end...
echo "$opt" | sed -r 's/^"|"$//g'
If you need to match single or double quotes, and only strings that are properly quoted. You can use this slightly more complex regex...
echo $opt | sed -E "s|^(['\"])(.*)\1$|\2|g"
This uses backrefences to ensure the quote at the end is the same as at the start.
In Bash, you could use the following one-liner:
[[ "${var}" == \"*\" || "${var}" == \'*\' ]] && var="${var:1:-1}"
This will remove surrounding quotes (both single and double) from the string stored in var while keeping quote characters inside the string intact. Also, this won't do anything if there's only a single leading quote or only a single trailing quote or if there are mixed quote characters at start/end.
Wrapped in a function:
#!/usr/bin/env bash
# Strip surrounding quotes from string [$1: variable name]
function strip_quotes() {
local -n var="$1"
[[ "${var}" == \"*\" || "${var}" == \'*\' ]] && var="${var:1:-1}"
}
str="'hello world'"
echo "Before: ${str}"
strip_quotes str
echo "After: ${str}"
My version
strip_quotes() {
while [[ $# -gt 0 ]]; do
local value=${!1}
local len=${#value}
[[ ${value:0:1} == \" && ${value:$len-1:1} == \" ]] && declare -g $1="${value:1:$len-2}"
shift
done
}
The function accepts variable name(s) and strips quotes in place. It only strips a matching pair of leading and trailing quotes. It doesn't check if the trailing quote is escaped (preceded by \ which is not itself escaped).
In my experience, general-purpose string utility functions like this (I have a library of them) are most efficient when manipulating the strings directly, not using any pattern matching and especially not creating any sub-shells, or calling any external tools such as sed, awk or grep.
var1="\"test \\ \" end \""
var2=test
var3=\"test
var4=test\"
echo before:
for i in var{1,2,3,4}; do
echo $i="${!i}"
done
strip_quotes var{1,2,3,4}
echo
echo after:
for i in var{1,2,3,4}; do
echo $i="${!i}"
done
I use this regular expression, which avoids removing quotes from strings that are not properly quoted, here the different outputs are shown depending on the inputs, only one with begin-end quote was affected:
echo '"only first' | sed 's/^"\(.*\)"$/\1/'
Output: >"only first<
echo 'only last"' | sed 's/^"\(.*\)"$/\1/'
Output: >"only last"<
echo '"both"' | sed 's/^"\(.*\)"$/\1/'
Output: >both<
echo '"space after" ' | sed 's/^"\(.*\)"$/\1/'
Output: >"space after" <
echo ' "space before"' | sed 's/^"\(.*\)"$/\1/'
Output: > "space before"<
STR='"0.0.0"' ## OR STR="\"0.0.0\""
echo "${STR//\"/}"
## Output: 0.0.0
There is another way to do it. Like:
echo ${opt:1:-1}
If you try to remove quotes because the Makefile keeps them, try this:
$(subst $\",,$(YOUR_VARIABLE))
Based on another answer: https://stackoverflow.com/a/10430975/10452175
How to remove extra spaces in variable HEAD?
HEAD=" how to remove extra spaces "
Result:
how to remove extra spaces
Try this:
echo "$HEAD" | tr -s " "
or maybe you want to save it in a variable:
NEWHEAD=$(echo "$HEAD" | tr -s " ")
Update
To remove leading and trailing whitespaces, do this:
NEWHEAD=$(echo "$HEAD" | tr -s " ")
NEWHEAD=${NEWHEAD%% }
NEWHEAD=${NEWHEAD## }
Using awk:
$ echo "$HEAD" | awk '$1=$1'
how to remove extra spaces
Take advantage of the word-splitting effects of not quoting your variable
$ HEAD=" how to remove extra spaces "
$ set -- $HEAD
$ HEAD=$*
$ echo ">>>$HEAD<<<"
>>>how to remove extra spaces<<<
If you don't want to use the positional paramaters, use an array
ary=($HEAD)
HEAD=${ary[#]}
echo "$HEAD"
One dangerous side-effect of not quoting is that filename expansion will be in play. So turn it off first, and re-enable it after:
$ set -f
$ set -- $HEAD
$ set +f
This horse isn't quite dead yet: Let's keep beating it!*
Read into array
Other people have mentioned read, but since using unquoted expansion may cause undesirable expansions all answers using it can be regarded as more or less the same. You could do
set -f
read HEAD <<< $HEAD
set +f
or you could do
read -rd '' -a HEAD <<< "$HEAD" # Assuming the default IFS
HEAD="${HEAD[*]}"
Extended Globbing with Parameter Expansion
$ shopt -s extglob
$ HEAD="${HEAD//+( )/ }" HEAD="${HEAD# }" HEAD="${HEAD% }"
$ printf '"%s"\n' "$HEAD"
"how to remove extra spaces"
*No horses were actually harmed – this was merely a metaphor for getting six+ diverse answers to a simple question.
Here's how I would do it with sed:
string=' how to remove extra spaces '
echo "$string" | sed -e 's/ */ /g' -e 's/^ *\(.*\) *$/\1/'
=> how to remove extra spaces # (no spaces at beginning or end)
The first sed expression replaces any groups of more than 1 space with a single space, and the second expression removes any trailing or leading spaces.
echo -e " abc \t def "|column -t|tr -s " "
column -t will:
remove the spaces at the beginning and at the end of the line
convert tabs to spaces
tr -s " " will squeeze multiple spaces to single space
BTW, to see the whole output you can use cat - -A: shows you all spacial characters including tabs and EOL:
echo -e " abc \t def "|cat - -A
output: abc ^I def $
echo -e " abc \t def "|column -t|tr -s " "|cat - -A
output:
abc def$
Whitespace can take the form of both spaces and tabs. Although they are non-printing characters and unseen to us, sed and other tools see them as different forms of whitespace and only operate on what you ask for. ie, if you tell sed to delete x number of spaces, it will do this, but the expression will not match tabs. The inverse is true- supply a tab to sed and it will not match spaces, even if the number of them is equal to those in a tab.
A more extensible solution that will work for removing either/both additional space in the form of spaces and tabs (I've tested mixing both in your specimen variable) is:
echo $HEAD | sed 's/^[[:blank:]]*//g'
or we can tighten-up #Frontear 's excellent suggestion of using xargs without the tr:
echo $HEAD | xargs
However, note that xargs would also remove newlines. So if you were to cat a file and pipe it to xargs, all the extra space- including newlines- are removed and everything put on the same line ;-).
Both of the foregoing achieved your desired result in my testing.
Try this one:
echo ' how to remove extra spaces ' | sed 's/^ *//g' | sed 's/$ *//g' | sed 's/ */ /g'
or
HEAD=" how to remove extra spaces "
HEAD=$(echo "$HEAD" | sed 's/^ *//g' | sed 's/$ *//g' | sed 's/ */ /g')
I would make use of tr to remove the extra spaces, and xargs to trim the back and front.
TEXT=" This is some text "
echo $(echo $TEXT | tr -s " " | xargs)
# [...]$ This is some text
echo variable without quotes does what you want:
HEAD=" how to remove extra spaces "
echo $HEAD
# or assign to new variable
NEW_HEAD=$(echo $HEAD)
echo $NEW_HEAD
output: how to remove extra spaces
I would like to replace blank spaces/white spaces in a string with commas.
STR1=This is a string
to
STR1=This,is,a,string
Without using external tools:
echo ${STR1// /,}
Demo:
$ STR1="This is a string"
$ echo ${STR1// /,}
This,is,a,string
See bash: Manipulating strings.
Just use sed:
echo $STR1 | sed 's/ /,/g'
or pure BASH way::
echo ${STR1// /,}
kent$ echo "STR1=This is a string"|awk -v OFS="," '$1=$1'
STR1=This,is,a,string
Note:
if there are continued blanks, they would be replaced with a single comma. as example above shows.
This might work for you:
echo 'STR1=This is a string' | sed 'y/ /,/'
STR1=This,is,a,string
or:
echo 'STR1=This is a string' | tr ' ' ','
STR1=This,is,a,string
How about
STR1="This is a string"
StrFix="$( echo "$STR1" | sed 's/[[:space:]]/,/g')"
echo "$StrFix"
**output**
This,is,a,string
If you have multiple adjacent spaces in your string and what to reduce them to just 1 comma, then change the sed to
STR1="This is a string"
StrFix="$( echo "$STR1" | sed 's/[[:space:]][[:space:]]*/,/g')"
echo "$StrFix"
**output**
This,is,a,string
I'm using a non-standard sed, and so have used ``[[:space:]][[:space:]]*to indicate one or more "white-space" characters (including tabs, VT, maybe a few others). In a modern sed, I would expect[[:space:]]+` to work as well.
STR1=`echo $STR1 | sed 's/ /,/g'`