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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
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 have a string that is something like "info_A!__B????????C_*". I wan to remove the special characters from it but keep underscores and letters. I tried with [:word:] (ASCII letters and _) character set, but it says "invalid character set". any idea how to handle this ? Thanks.
text="info_!_????????_*"
if [ -z `echo $text | tr -dc "[:word:]"` ]
......
Using bash parameter expansion:
$ var='info_A!__B????????C_*'
$ echo "${var//[^[:alnum:]_]/}"
info_A__BC_
A sed one-liner would be
sed 's/[^[:alnum:]_]//g' <<< 'info_!????????*'
gives you
info_
An awk one-liner would be
awk '{gsub(/[^[:alnum:]_]/,"",$0)} 1' <<< 'info_!??A_??????*pi9ngo^%$_mingo745'
gives you
info_A_pi9ngo_mingo745
If you don't wish to have numbers in the output then change :alnum: to :alpha:.
My tr doesn't understand [:word:]. I had to do like this:
$ x=$(echo 'info_A!__B????????C_*' | tr -cd '[:alnum:]_')
$ echo $x
info_A__BC_
Not sure if its robust way but it worked for your sample text.
sed one-liner:
echo "SamPlE_#tExT%, really ?" | sed -e 's/[^a-z^A-Z|^_]//g'
SamPlE_tExTreally
I am trying to only change the first letter of a string to lowercase using a Shell script. Ideally a simple way to go from CamelCase to lowerCamelCase.
GOAL:
$DIR="SomeString"
# missing step
$echo $DIR
someString
I have found some great resources for doing this to the entire string but not just altering the first letter and leaving the remaining string untouched.
If your shell is recent enough, you can use the following parameter expansion:
DIR="SomeString" # Note the missing dollar sign.
echo ${DIR,}
Alternative solution (will work on old bash too)
DIR="SomeString"
echo $(echo ${DIR:0:1} | tr "[A-Z]" "[a-z]")${DIR:1}
prints
someString
for assing to variable
DIR2="$(echo ${DIR:0:1} | tr "[A-Z]" "[a-z]")${DIR:1}"
echo $DIR2
prints
someString
alternative perl
DIR3=$(echo SomeString | perl -ple 's/(.)/\l$1/')
DIR3=$(echo SomeString | perl -nle 'print lcfirst')
DIR3=$(echo "$DIR" | perl -ple 's/.*/lcfirst/e'
some terrible solutions;
DIR4=$(echo "$DIR" | sed 's/^\(.\).*/\1/' | tr "[A-Z]" "[a-z]")$(echo "$DIR" | sed 's/^.//')
DIR5=$(echo "$DIR" | cut -c1 | tr '[[:upper:]]' '[[:lower:]]')$(echo "$DIR" | cut -c2-)
All the above is tested with OSX's /bin/bash.
With sed:
var="SomeString"
echo $var | sed 's/^./\L&/'
^ means the start of the line
\L is the command to make the match in lowercase
& is the whole match
Perl solution:
DIR=SomeString
perl -le 'print lcfirst shift' "$DIR"
Since awk hasn't yet been mentioned, here's another way you could do it (requires GNU awk):
dir="SomeString"
new_dir=$(awk 'BEGIN{FS=OFS=""}{$1=tolower($1)}1' <<<"$dir")
This sets the input and output field separators to an empty string, so each character is a field. The tolower function does what you think it does. 1 at the end prints the line. If your shell doesn't support <<< you can do echo "$dir" | awk ... instead.
If you are looking for a POSIX compliant solution then have a look at typeset.
var='SomeString'
typeset -lL1 b="$var"
echo "${b}${var#?}"
Output:
someString
The typeset command creates a special variable that is lowercase, left aligned and one char long. ${var#?} trims the first occurrence of pattern from the start of $var and ? matches a single
character.
Does anybody knows a way to convert "40900000" to "409-00-000" with single command, sed or awk.
I already tried couple of ways with sed but no luck at all. I need to do this in a bulk, there is around 40k line and some of this lines are not proper, so they need to be fixed.
Thanks in advance
Using GNU sed, I would do it like this:
sed -r 's/([0-9]{3})([0-9]{2})([0-9]{3})/\1-\2-\3/' filename
# or, equivalently
sed -E 's/([0-9]{3})([0-9]{2})([0-9]{3})/\1-\2-\3/' filename
The -r or -E enables extended regex mode, which avoids the need to escape all the parentheses
\1 is the first capture group (the bits in between the ( ))
[0-9] means the range zero to nine
{3} means three of the preceeding character or range
edit: Thanks for all the comments.
On other systems that lack the -r switch, or its alias -E, you have to escape the ( ) and { } above. That leaves you with:
sed 's/\([0-9]\{3\}\)\([0-9]\{2\}\)\([0-9]\{3\}\)/\1-\2-\3/' filename
At the expense of repetition, you can avoid some of the escapes by simply repeating the [0-9]:
sed 's/\([0-9][0-9][0-9]\)\([0-9][0-9]\)\([0-9][0-9][0-9]\)/\1-\2-\3/' filename
For the record, Perl is equally capable of doing this sort of thing:
perl -pwe 's/(\d{3})(\d{2})(\d{3})/$1-$2-$3/' filename
-p means print
-w means enable warnings
-e means execute one line
\d is the "digit" character class (zero to nine)
No need to run external commands, bash or ksh can do it themselves.
$ a=12345678
$ [ ${#a} = 8 ] && { b=${a:0:3}-${a:3:2}-${a:5};a=$b;}
$ echo $a
123-45-678
$ a=abc-de-fgh
$ [ ${#a} = 8 ] && { b=${a:0:3}-${a:3:2}-${a:5};a=$b;}
$ echo $a
abc-de-fgh
You can use sed, like this:
sed 's/\([0-9][0-9][0-9]\)\([0-9][0-9]\)\([0-9][0-9][0-9]\)/\1-\2-\3/'
or more succinctly, with extended regex syntax:
sed -E 's/([0-9]{3})([0-9]{2})([0-9]{3})/\1-\2-\3/'
For golfing:
$ echo "40900000" | awk '$1=$1' FIELDWIDTHS='3 2 3' OFS='-'
409-00-000
With sed:
sed 's/\(...\)\(..\)\(...\)/\1-\2-\3/'
The dot matches character, and the surrounding with \( and \) makes it a group. The \1 references the first group.
Just for the fun of it, an awk
echo "40900000" | awk '{a=$0+0} length(a)==8 {$0=substr(a,1,3)"-"substr(a,4,2)"-"substr(a,6)}1'
409-00-000
This test if there are 8 digits.
A more complex version (need gnu awk due to gensub):
echo "40900000" | awk --re-interval '{print gensub(/([0-9]{3})([0-9]{2})([0-9]{3})/,"\\1-\\2-\\3","g")}'
409-00-000
echo "409-00-000" | awk --re-interval '{print gensub(/([0-9]{3})([0-9]{2})([0-9]{3})/,"\\1-\\2-\\3","g")}'
409-00-000
Turnarround from STDIN:
echo "40900000" | grep -E "[0-9]{8}" | cut -c "1-3,4-5,6-8" --output-delimiter=-
from file:
grep -E "[0-9]{8}" filename | cut -c "1-3,4-5,6-8" --output-delimiter=-
But I prefect Tom Fenech's solution.