I tried to loop comma separated values with space, but not able to get the exact value since it has space in the string.
I tried in different ways, but i not able to get desired results.
Can anyone help me on this
#!/bin/ksh
values="('A','sample text','Mark')"
for i in `echo $values | sed 's/[)(]//g' | sed 's/,/ /g'`
do
echo $i
done
My expected output is:
A
sample text
Mark
First, change values to an array. Then iterating over it is a simple matter.
values=(A "sample text" Mark)
for i in "${values[#]}"; do
echo "$i"
done
This is the same as Chepner's answer, only kludgier, (variable substitution), and more dangerous, (the eval...), the better to use the OP's exact $values assignment:
values="('A','sample text','Mark')"
eval values=${values//,/ }
for i in "${values[#]}"; do
echo "$i"
done
It works in ksh, but really, if at all possible try to use Chepner's simpler and safer $values assignment.
Simply trim the quotes
#!/bin/ksh
values="('A','sample text','Mark')"
echo $values | tr -d "()'\"" | tr ',' '\n'
output:
A
sample text
Mark
You should use the single quotes for splitting the string (and quote "$values").
When your sed supports \n for replacement into a line, you can do without a loop:
echo "${values}" | sed "s/[)(]//g;s/','/\n/g;s/'//g"
# or
sed "s/[)(]//g;s/','/\n/g;s/'//g" <<< "${values}"
When the values in your string are without a comma and parentheses, you can use
grep -Eo "[^',()]*" <<< "${values}"
Better is looking for fields between 2 single quotes and remove those single quotes.
grep -Eo "'[^']*'" <<< "${values}" | tr -d "'"
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
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 $x
info_A__BC_
Not sure if its robust way but it worked for your sample text.
sed one-liner:
echo "SamPlE_#tExT%, really ?" | sed -e 's/[^a-z^A-Z|^_]//g'
SamPlE_tExTreally
I used many times [``] to capture output of command to a variable. but with following code i am not getting right output.
#!/bin/bash
export XLINE='($ZWP_SCRIP_NAME),$ZWP_LT_RSI_TRIGGER)R),$ZWP_RTIMER'
echo 'Original XLINE'
echo $XLINE
echo '------------------'
echo 'Extract all word with $ZWP'
#works fine
echo $XLINE | sed -e 's/\$/\n/g' | sed -e 's/.*\(ZWP[_A-Z]*\).*/\1/g' | grep ZWP
echo '------------------'
echo 'Assign all word with $ZWP to XVAR'
#XVAR doesn't get all the values
export XVAR=`echo $XLINE | sed -e 's/\$/\n/g' | sed -e 's/.*\(ZWP[_A-Z]*\).*/\1/g' | grep ZWP` #fails
echo "$XVAR"
and i get:
Original XLINE
($ZWP_SCRIP_NAME),$ZWP_LT_RSI_TRIGGER)R),$ZWP_RTIMER
------------------
Extract all word with $ZWP
ZWP_SCRIP_NAME
ZWP_LT_RSI_TRIGGER
ZWP_RTIMER
------------------
Assign all word with $ZWP to XVAR
ZWP_RTIMER
why XVAR doesn't get all the values?
however if i use $() to capture the out instead of ``, it works fine. but why `` is not working?
Having GNU grep you can use this command:
XVAR=$(grep -oP '\$\KZWP[A-Z_]+' <<< "$XLINE")
If you pass -P grep is using Perl compatible regular expressions. The key here is the \K escape sequence. Basically the regex matches $ZWP followed by one or more uppercase characters or underscores. The \K after the $ removes the $ itself from the match, while its presence is still required to match the whole pattern. Call it poor man's lookbehind if you want, I like it! :)
Btw, grep -o outputs every match on a single line instead of just printing the lines which match the pattern.
If you don't have GNU grep or you care about portability you can use awk, like this:
XVAR=$(awk -F'$' '{sub(/[^A-Z_].*/, "", $2); print $2}' RS=',' <<< "$XLINE")
First, the smallest change that makes your code "work":
echo "$XLINE" | tr '$' '\n' | sed -e 's/.*\(ZWP[_A-Z]*\).*/\1/g' | grep ZWP_
The use of tr replaces a sed expression that didn't actually do what you thought it did -- try looking at its output to see.
One sane alternative would be to rely on GNU grep's -o option. If you can't do that...
zwpvars=( ) # create a shell array
zwp_assignment_re='[$](ZWP_[[:alnum:]_]+)(.*)' # ...and a regex
content="$XLINE"
while [[ $content =~ $zwp_assignment_re ]]; do
zwpvars+=( "${BASH_REMATCH[1]}" ) # found a reference
content=${BASH_REMATCH[2]} # stuff the remaining content aside
done
printf 'Found variable: %s\n' "${zwpvars[#]}"
Suppose I have a Unix shell variable as below
variable=abc,def,ghij
I want to extract all the values (abc, def and ghij) using a for loop and pass each value into a procedure.
The script should allow extracting arbitrary number of comma-separated values from $variable.
Not messing with IFS
Not calling external command
variable=abc,def,ghij
for i in ${variable//,/ }
do
# call your procedure/other scripts here below
echo "$i"
done
Using bash string manipulation http://www.tldp.org/LDP/abs/html/string-manipulation.html
You can use the following script to dynamically traverse through your variable, no matter how many fields it has as long as it is only comma separated.
variable=abc,def,ghij
for i in $(echo $variable | sed "s/,/ /g")
do
# call your procedure/other scripts here below
echo "$i"
done
Instead of the echo "$i" call above, between the do and done inside the for loop, you can invoke your procedure proc "$i".
Update: The above snippet works if the value of variable does not contain spaces. If you have such a requirement, please use one of the solutions that can change IFS and then parse your variable.
If you set a different field separator, you can directly use a for loop:
IFS=","
for v in $variable
do
# things with "$v" ...
done
You can also store the values in an array and then loop through it as indicated in How do I split a string on a delimiter in Bash?:
IFS=, read -ra values <<< "$variable"
for v in "${values[#]}"
do
# things with "$v"
done
Test
$ variable="abc,def,ghij"
$ IFS=","
$ for v in $variable
> do
> echo "var is $v"
> done
var is abc
var is def
var is ghij
You can find a broader approach in this solution to How to iterate through a comma-separated list and execute a command for each entry.
Examples on the second approach:
$ IFS=, read -ra vals <<< "abc,def,ghij"
$ printf "%s\n" "${vals[#]}"
abc
def
ghij
$ for v in "${vals[#]}"; do echo "$v --"; done
abc --
def --
ghij --
I think syntactically this is cleaner and also passes shell-check linting
variable=abc,def,ghij
for i in ${variable//,/ }
do
# call your procedure/other scripts here below
echo "$i"
done
#/bin/bash
TESTSTR="abc,def,ghij"
for i in $(echo $TESTSTR | tr ',' '\n')
do
echo $i
done
I prefer to use tr instead of sed, becouse sed have problems with special chars like \r \n in some cases.
other solution is to set IFS to certain separator
Another solution not using IFS and still preserving the spaces:
$ var="a bc,def,ghij"
$ while read line; do echo line="$line"; done < <(echo "$var" | tr ',' '\n')
line=a bc
line=def
line=ghij
Here is an alternative tr based solution that doesn't use echo, expressed as a one-liner.
for v in $(tr ',' '\n' <<< "$var") ; do something_with "$v" ; done
It feels tidier without echo but that is just my personal preference.
The following solution:
doesn't need to mess with IFS
doesn't need helper variables (like i in a for-loop)
should be easily extensible to work for multiple separators (with a bracket expression like [:,] in the patterns)
really splits only on the specified separator(s) and not - like some other solutions presented here on e.g. spaces too.
is POSIX compatible
doesn't suffer from any subtle issues that might arise when bash’s nocasematch is on and a separator that has lower/upper case versions is used in a match like with ${parameter/pattern/string} or case
beware that:
it does however work on the variable itself and pop each element from it - if that is not desired, a helper variable is needed
it assumes var to be set and would fail if it's not and set -u is in effect
while true; do
x="${var%%,*}"
echo $x
#x is not really needed here, one can of course directly use "${var%%:*}"
if [ -z "${var##*,*}" ] && [ -n "${var}" ]; then
var="${var#*,}"
else
break
fi
done
Beware that separators that would be special characters in patterns (e.g. a literal *) would need to be quoted accordingly.
Here's my pure bash solution that doesn't change IFS, and can take in a custom regex delimiter.
loop_custom_delimited() {
local list=$1
local delimiter=$2
local item
if [[ $delimiter != ' ' ]]; then
list=$(echo $list | sed 's/ /'`echo -e "\010"`'/g' | sed -E "s/$delimiter/ /g")
fi
for item in $list; do
item=$(echo $item | sed 's/'`echo -e "\010"`'/ /g')
echo "$item"
done
}
Try this one.
#/bin/bash
testpid="abc,def,ghij"
count=`echo $testpid | grep -o ',' | wc -l` # this is not a good way
count=`expr $count + 1`
while [ $count -gt 0 ] ; do
echo $testpid | cut -d ',' -f $i
count=`expr $count - 1 `
done