This can be so easy to a people who know. I am almost finishing this command
echo VERSION=1.0 | sed 's/^VERSION=\([0-9]\).\([0-9]\)/VERSION=\1.\2+1/'
I only want to write VERSION=1.1 . How can I evaluate \2 to integer and sum +1..
of couse sed can do that. that's what e for. you can pass matched/replaced string to shell command using "e"
see the example based on your sed line:
kent$ echo VERSION=1.0 | sed 's/^VERSION=\([0-9]\).\([0-9]\)/echo "VERSION=\1.$((\2+1))"/e'
VERSION=1.1
You can use the bc command:
echo VERSION=`echo "1.0 + 0.1" | bc`
Results in:
VERSION=1.1
man bc
echo "VERSION="`echo "v=1.0; v+=0.1; v" | bc` > myFile.txt
cat myFile.txt
VERSION=1.1
Crpytic answer - how to use the whole toolkit:
x='VERSION=1.0'
echo -n $x | sed 's/\..*/./'; expr `echo $x | grep -o '\..*' | cut -c 2-` + 1
Related
#!/bin/bash
MA=$(bt-device -l | cut -d " " -f 3)
MAC=${MA:1: -1}
bluetoothctl connect $MAC
Expected Result
98:9E:63:18:00:88
Actual result
(98:9E:63:18:00:88
A few alternatives:
$ echo 'Denny’s Tunez (98:9E:63:18:00:88)' | sed -En 's/^[^(]*\(([^)]*)\).*/\1/p'
98:9E:63:18:00:88
$ echo 'Denny’s Tunez (98:9E:63:18:00:88)' | cut -d'(' -f2 | cut -d')' -f1
98:9E:63:18:00:88
$ echo 'Denny’s Tunez (98:9E:63:18:00:88)' | awk -F'[)(]' '{print $2}'
98:9E:63:18:00:88
$ echo 'Denny’s Tunez (98:9E:63:18:00:88)' | grep -Eow '(..)(:..){5}'
98:9E:63:18:00:88
$ x='Denny’s Tunez (98:9E:63:18:00:88)'
$ y="${x//*\(/}"
$ y="${y//\)*}"
$ echo $y
98:9E:63:18:00:88
With GNU bash and its Parameter Expansion:
s="(98:9E:63:18:00:88)"
s="${s/#?/}" # remove first character
s="${s/%?/}" # remove last character
echo "$s"
Output:
98:9E:63:18:00:88
Using sed it can be done in a single step:
s='Denny’s Tunez (98:9E:63:18:00:88)'
echo "$s" | sed -E 's/.* \(|)//g'
98:9E:63:18:00:88
So for your example you can use:
mac=$(bt-device -l | sed -E 's/.* \(|)//g')
You can use parameter expansion:
offset and length
echo ${MA:1: -1}
prefix and suffix removal
tmp=${MA#(}
echo ${tmp%)}
parameter matching
tmp=${MA/#\(}
echo ${tmp/%\)}
Another approach is to:
whitelist what you do want
echo "$MA" | tr -dC '[0-9A-F:]'
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 "${var#refs/heads/}"
testing/branch
I have a black list to save tag id list, e.g. 1-3,7-9, actually it represents 1,2,3,7,8,9. And could expand it by below shell
for i in {1..3,7..9}; do for j in {$i}; do echo -n "$j,"; done; done
1,2,3,7,8,9
but first I should convert - to ..
echo -n "1-3,7-9" | sed 's/-/../g'
1..3,7..9
then put it into for expression as a parameter
echo -n "1-3,7-9" | sed 's/-/../g' | xargs -I # for i in {#}; do for j in {$i}; do echo -n "$j,"; done; done
zsh: parse error near `do'
echo -n "1-3,7-9" | sed 's/-/../g' | xargs -I # echo #
1..3,7..9
but for expression cannot parse it correctly, why is so?
Because you didn't do anything to stop the outermost shell from picking up the special keywords and characters ( do, for, $, etc ) that you mean to be run by xargs.
xargs isn't a shell built-in; it gets the command line you want it to run for each element on stdin, from its arguments. just like any other program, if you want ; or any other sequence special to be bash in an argument, you need to somehow escape it.
It seems like what you really want here, in my mind, is to invoke in a subshell a command ( your nested for loops ) for each input element.
I've come up with this; it seems to to the job:
echo -n "1-3,7-9" \
| sed 's/-/../g' \
| xargs -I # \
bash -c "for i in {#}; do for j in {\$i}; do echo -n \"\$j,\"; done; done;"
which gives:
{1..3},{7..9},
Could use below shell to achieve this
# Mac newline need special treatment
echo "1-3,7-9" | sed -e 's/-/../g' -e $'s/,/\\\n/g' | xargs -I# echo 'for i in {#}; do echo -n "$i,"; done' | bash
1,2,3,7,8,9,%
#Linux
echo "1-3,7-9" | sed -e 's/-/../g' -e 's/,/\n/g' | xargs -I# echo 'for i in {#}; do echo -n "$i,"; done' | bash
1,2,3,7,8,9,
but use this way is a little complicated maybe awk is more intuitive
# awk
echo "1-3,7-9,11,13-17" | awk '{n=split($0,a,","); for(i=1;i<=n;i++){m=split(a[i],a2,"-");for(j=a2[1];j<=a2[m];j++){print j}}}' | tr '\n' ','
1,2,3,7,8,9,11,13,14,15,16,17,%
echo -n "1-3,7-9" | perl -ne 's/-/../g;$,=",";print eval $_'
I need to grep for values with + or - symbol
eg: -3 or +3
I tried the following , doesnt help me
$ echo $accessTime | grep "^((\+)|(\-))[0-9]+$"
$ echo $accessTime | grep "+[0-9]+"
$ echo $accessTime | grep "\+[0-9]+"
$ echo $accessTime | grep "'+'[0-9]+"
$ echo $accessTime | grep "^'\+'[0-9]+"
$ echo $accessTime | grep "^(\+)[0-9]+"
$ echo $accessTime | grep "^(\+)[0-9]+"
Can you guys pls help me ....Im learning bash for past few days only..Thanks
Its easier than what you think, you should try
echo $accessTime | grep [+-]
You can use:
accessTime='+3'
echo "$accessTime" | grep "+[0-9]\+"
+3
Quantifier + needs to be escaped in normal grep and literal + must not be escaped.
With grep -E it is exactly reverse:
echo "$accessTime" | grep -E "\+[0-9]+"
+3
Quantifier + must not be escaped in grep -E (extended regex) and literal + . needs to be escaped.
I'm trying to make a small function that removes all the chars that are not digits.
123a45a ---> will become ---> 12345
I've came up with :
temp=$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