After running
test=$(security 2>&1 >/dev/null find-generic-password -ga test) echo
$test
The output is
password: "1234"
I would like to remove everything and only leave the actual password, in these case just
1234
Any help?
If the format is always password: "value", you can do this efficiently without any external utility using just substring expansion:
echo "${test:11:-1}" #output: 1234
If you insist on using other utilities:
awk -F'"' '{ print $2 }' <<< "$test" #output: 1234
cut -d'"' -f2 <<< "$test" #output: 1234
Both commands above will fail if you have a " character in your password. Another solution using sed working fine with " in your password:
sed 's/^password: "\(.*\)"$/\1/' <<< "$test" #output: 1234
You can solve it by:
command:
echo 'password: "1234"' | awk -F"\"" '{print $2}'
output:
1234
Use a Perl-Compatible Regular Expression with Grep
If your system has a grep compiled with the PCRE engine (e.g. pcregrep, or support for grep -P) then you can drop pcregrep -o ': "\K[^"]+' into your pipeline. For example, to send just the material between the quotes to standard output:
$ echo 'password: "1234"' | pcregrep -o ': "\K[^"]+'
1234
$ echo 'password: "ABCD"' | pcregrep -o ': "\K[^"]+'
ABCD
$ echo 'password: "1A2B3C4D"' | pcregrep -o ': "\K[^"]+'
1A2B3C4D
Related
I have a string,
var=refs/heads/testing/branch
I want to get rid of refs/heads/ in the string using shell script, such that I have only:
var=testing/branch
Commands I tried (one per line):
echo $(var) | awk -F\\ {'print $2'}
echo $var | sed -e s,refs/heads/,,
echo "refs/heads/testing/branch" | grep -oP '(?<=refs/heads/\)\w+'
echo "refs/heads/testing/branch" | LC_ALL=C sed -e 's/.*\\//'
echo "refs/heads/testing/branch" | cut -d'\' -f2
echo refs/heads/testing/branch | sed -e s,refs/heads/,,
there are lots of options out there ,try easy ones:
echo $var | cut -d "/" -f 3,4
echo $var | awk -F"/" '{print $3"/"$4}'
Shell parameter expansion: remove the prefix "refs/heads/" from the variable contents
$ var=refs/heads/testing/branch
$ echo "${var#refs/heads/}"
testing/branch
I have this shell script variable, var. It keeps 3 entries separated by new line. From this variable var, I want to extract 2, and 0.078688. Just these two numbers.
var="USER_ID=2
# 0.078688
Suhas"
These are the code I tried:
echo "$var" | grep -o -P '(?<=\=).*(?=\n)' # For extracting 2
echo "$var" | awk -v FS="(# |\n)" '{print $2}' # For extracting 0.078688
None of the above working. What is the problem here? How to fix this ?
Just use tr alone for retaining the numerical digits, the dot (.) and the white-space and remove everything else.
tr -cd '0-9. ' <<<"$var"
2 0.078688
From the man page, of tr for usage of -c, -d flags,
tr [OPTION]... SET1 [SET2]
-c, -C, --complement
use the complement of SET1
-d, --delete
delete characters in SET1, do not translate
To store it in variables,
IFS=' ' read -r var1 var2 < <(tr -cd '0-9. ' <<<"$var")
printf "%s\n" "$var1"
2
printf "%s\n" "$var2"
2
0.078688
Or in an array as
IFS=' ' read -ra numArray < <(tr -cd '0-9. ' <<<"$var")
printf "%s\n" "${numArray[#]}"
2
0.078688
Note:- The -cd flags in tr are POSIX compliant and will work on any systems that has tr installed.
echo "$var" |grep -oP 'USER_ID=\K.*'
2
echo "$var" |grep -oP '# \K.*'
0.078688
Your solution is near to perfect, you need to chance \n to $ which represent end of line.
echo "$var" |awk -F'# ' '/#/{print $2}'
0.078688
echo "$var" |awk -F'=' '/USER_ID/{print $2}'
2
You can do it with pure bash using a regex:
#!/bin/bash
var="USER_ID=2
# 0.078688
Suhas"
[[ ${var} =~ =([0-9]+).*#[[:space:]]([0-9\.]+) ]] && result1="${BASH_REMATCH[1]}" && result2="${BASH_REMATCH[2]}"
echo "${result1}"
echo "${result2}"
With awk:
First value:
echo "$var" | grep 'USER_ID' | awk -F "=" '{print $2}'
Second value:
echo "$var" | grep '#' | awk '{print $2}'
Assuming this is the format of data as your sample
# For extracting 2
echo "$var" | sed -e '/.*=/!d' -e 's///'
echo "$var" | awk -F '=' 'NR==1{ print $2}'
# For extracting 0.078688
echo "$var" | sed -e '/.*#[[:blank:]]*/!d' -e 's///'
echo "$var" | awk -F '#' 'NR==2{ print $2}'
I created a script that was using
cut -d',' -f- --output-delimiter=$'\n'
to add a newline for each command separated value in RHEL 5, for e.g.
[root]# var="hi,hello how,are you,doing"
[root]# echo $var
hi,hello how,are you,doing
[root]# echo $var|cut -d',' -f- --output-delimiter=$'\n'
hi
hello how
are you
doing
But unfortunately when I run the same command in Solaris 10, it doesn't work at all :( !
bash-3.00# var="hi,hello how,are you,doing"
bash-3.00# echo $var
hi,hello how,are you,doing
bash-3.00# echo $var|cut -d',' -f- --output-delimiter=$'\n'
cut: illegal option -- output-delimiter=
usage: cut -b list [-n] [filename ...]
cut -c list [filename ...]
cut -f list [-d delim] [-s] [filename]
I checked the man page for 'cut' and alas there is no ' --output-delimiter ' in there !
So how do I achieve this in Solaris 10 (bash)? I guess awk would be a solution, but I'm unable to frame up the options properly.
Note: The comma separated variables might have " " space in them.
What about using tr for this?
$ tr ',' '\n' <<< "$var"
hi
hello how
are you
doing
or
$ echo $var | tr ',' '\n'
hi
hello how
are you
doing
With sed:
$ sed 's/,/\n/g' <<< "$var"
hi
hello how
are you
doing
Or with awk:
$ awk '1' RS=, <<< "$var"
hi
hello how
are you
doing
Perhaps do it in bash itself?
var="hi,hello how,are you,doing"
printf "$var" | (IFS=, read -r -a arr; printf "%s\n" "${arr[#]}")
hi
hello how
are you
doing
I'm trying to make a small function that removes all the chars that are not digits.
123a45a ---> will become ---> 12345
I've came up with :
temp=$word | grep -o [[:digit:]]
echo $temp
But instead of 12345 I get 1 2 3 4 5. How to I get rid of the spaces?
Pure bash:
word=123a45a
number=${word//[^0-9]}
Here's a pure bash solution
var='123a45a'
echo ${var//[^0-9]/}
12345
is this what you are looking for?
kent$ echo "123a45a"|sed 's/[^0-9]//g'
12345
grep & tr
echo "123a45a"|grep -o '[0-9]'|tr -d '\n'
12345
I would recommend using sed or perl instead:
temp="$(sed -e 's/[^0-9]//g' <<< "$word")"
temp="$(perl -pe 's/\D//g' <<< "$word")"
Edited to add: If you really need to use grep, then this is the only way I can think of:
temp="$( grep -o '[0-9]' <<< "$word" \
| while IFS= read -r ; do echo -n "$REPLY" ; done
)"
. . . but there's probably a better way. (It uses grep -o, like your solution, then runs over the lines that it outputs and re-outputs them without line-breaks.)
Edited again to add: Now that you've mentioned that you use can use tr instead, this is much easier:
temp="$(tr -cd 0-9 <<< "$word")"
What about using sed?
$ echo "123a45a" | sed -r 's/[^0-9]//g'
12345
As I read you are just allowed to use grep and tr, this can make the trick:
$ echo "123a45a" | grep -o [[:digit:]] | tr -d '\n'
12345
In your case,
temp=$(echo $word | grep -o [[:digit:]] | tr -d '\n')
tr will also work:
echo "123a45a" | tr -cd '[:digit:]'
# output: 12345
Grep returns the result on different lines:
$ echo -e "$temp"
1
2
3
4
5
So you cannot remove those spaces during the filtering, but you can afterwards, since $temp can transform itself like this:
temp=`echo $temp | tr -d ' '`
$ echo "$temp"
12345
My bash script is:
output=$(curl -s http://www.espncricinfo.com/england-v-south-africa-2012/engine/current/match/534225.html | sed -nr 's/.*<title>(.*?)<\/title>.*/\1/p')
score=echo"$output" | awk '{print $1}'
echo $score
The above script prints just a newline in my console whereas my required output is
$ curl -s http://www.espncricinfo.com/england-v-south-africa-2012/engine/current/match/534225.html | sed -nr 's/.*<title>(.*
?)<\/title>.*/\1/p' | awk '{print $1}'
SA
So, why am I not getting the output from my bash script whereas it works fine in terminal am I using echo"$output" in the wrong way.
#!/bin/bash
output=$(curl -s http://www.espncricinfo.com/england-v-south-africa-2012/engine/current/match/534225.html | sed -nr 's/.*<title>(.*?)<\/title>.*/\1/p')
score=$( echo "$output" | awk '{ print $1 }' )
echo "$score"
Score variable was probably empty, since your syntax was wrong.