I got a string that looks like this SOMETHING00000076XYZ
How can I extract the number 76 out of the string using a shell script? Note that 76 is preceded by zeroes and followed by letters.
1st solution: If you are ok with awk could you please try following.
echo "SOMETHING00000076XYZ" | awk 'match($0,/0+[0-9]+/){val=substr($0,RSTART,RLENGTH);sub(/0+/,"",val);print val;val=""}'
In case you want to save this into a variable use following.
variable="$(echo "SOMETHING00000076XYZ" | awk '{sub(/.*[^1-9]0+/,"");sub(/[a-zA-Z]+/,"")} 1')"
2nd solution: Adding 1 more awk solution here(keeping your sample in mind).
echo "SOMETHING00000076XYZ" | awk '{sub(/.*[^1-9]0+/,"");sub(/[a-zA-Z]+/,"")} 1'
Here is a sed option:
echo "SOMETHING00000076XYZ" | sed -r 's/[^0-9]*0*([0-9]+).*/\1/g';
76
Here is an explanation of the regex pattern used:
[^0-9]* match zero or more non digits
0* match zero or more 0's
([0-9]+) match AND capture any quantity of non zero digits
.* match the remainder of the string
Then, we just replace with \1, which is the first (and only) capture group.
echo 'SOMETHING00000076XYZ' | grep -o '[1-9][0-9]*'
Using gnu grep:
grep -oP '0+\K\d+' <<< 'SOMETHING00000076XYZ'
76
\K resets any matched information.
Here is another variant of awk:
awk -F '0+' 'match($2, /^[0-9]+/){print substr($2, 1, RLENGTH)}' <<< 'SOMETHING00000076XYZ'
76
You can try Perl as well
$ echo "SOMETHING00000076XYZ" | perl -ne ' /\D+0+(\d+)/ and print $1 '
76
$ a=$(echo "SOMETHING00000076XYZ" | perl -ne ' /\D+0+(\d+)/ and print $1 ')
$ echo $a
76
$
$ echo 'SOMETHING00000076XYZ' | awk '{sub(/^[^0-9]+/,""); print $0+0}'
76
You can use sed as
echo "SOMETHING00000076XYZ" | sed "s/[a-zA-Z]//g" | sed "s/^0*//"
The first step is for removing all letters
The second step is for removing leading zeroes
Related
I want to Filter all content after match with the content and bring the first value after the "."
I have an output something like this:
Output:
product: 13.6.0.35_0
More specifically, I need only the first two digits and the first digit after the dot, remembering that we should not cling to the values in the issue, but rather on the method of filtering the content.
Expected:
13.6
I tried something like:
echo "product: 13.6.0.35_0" | grep -ow '\w*13\w*'
If you need to use grep with the current logic, you can use
echo "product: 13.6.0.35_0" | grep -ow '13\.[0-9]*' | head -1
where 13\.[0-9]* matches 13, . and zero or more digits (as whole word due to w option) and head -1 gets the first match.
You may also use sed or awk:
sed -En 's/.* ([0-9]+\.[0-9]+).*/\1/p' <<< "product: 13.6.0.35_0"
awk -F'[[:space:].]' '{print $2"."$3}' <<< "product: 13.6.0.35_0"
See the online demo.
The sed command matches any text up to space, then matches the space and captures the two subsequent dot-separated numbers into Group 1 (\1) and then the rest of the line is matched and replaced with Group 1 value that is printed (as the default line output is suppressed with -n).
In the awk command, the field separator is set to whitespace and . with -F'[[:space:].]' and the {print $2"."$3} part prints the second and third field values joined with a ..
A pure shell solution using the builtin read , Parameter Expansion and curly braces for command groupings.
echo "product: 13.6.0.35_0" | { read -r _ value; echo "${value%.*.*}" ; }
You can also use cut:
echo 'product: 13.6.0.35_0' | cut -d ' ' -f2 | cut -d '.' -f1-2
13.6
I reached the expected output, it's simple but it works:
var=$(echo "product: 13.6.0.35_0" | grep -Eo '[[:digit:]]+' | sed -n 1,2p)
echo ${var} | sed 's/ /./g'
$ acpi
Battery 0: Charging, 18%, 01:37:09 until charged
How to grep the battery level value without percentage character (18)?
This should do it but I'm getting an empty result:
acpi | grep -e '(?<=, )(.*)(?=%)'
Your regex is correct but will work with experimental -P or perl mode regex option in gnu grep. You will also need -o to show only matching text.
Correct command would be:
grep -oP '(?<=, )\d+(?=%)'
However, if you don't have gnu grep then you can also use sed like this:
sed -nE 's/.*, ([0-9]+)%.*/\1/p' file
18
Could you please try following, written and tested in link https://ideone.com/nzSGKs
your_command | awk 'match($0,/Charging, [0-9]+%/){print substr($0,RSTART+10,RLENGTH-11)}'
Explanation: Adding detailed explanation for above only for explanation purposes.
your_command | ##Running OP command and passing its output to awk as standrd input here.
awk ' ##Starting awk program from here.
match($0,/Charging, [0-9]+%/){ ##Using match function to match regex Charging, [0-9]+% in line here.
print substr($0,RSTART+10,RLENGTH-11) ##Printing sub string and printing from 11th character from starting and leaving last 11 chars here in matched regex of current line.
}'
Using awk:
awk -F"," '{print $2+0}'
Using GNU sed:
sed -rn 's/.*\, *([0-9]+)\%\,.*/\1/p'
You can use sed:
$ acpi | sed -nE 's/.*Charging, ([[:digit:]]*)%.*/\1/p'
18
Or, if Charging is not always in the string, you can look for the ,:
$ acpi | sed -nE 's/[^,]*, ([[:digit:]]*)%.*/\1/p'
Using bash:
s='Battery 0: Charging, 18%, 01:37:09 until charged'
res="${s#*, }"
res="${res%%%*}"
echo "$res"
Result: 18.
res="${s#*, }" removes text from the beginning to the first comma+space and "${res%%%*}" removes all text from end till (and including) the last occurrence of %.
Problem
I want to get any text that consists of 1 to three digits followed by a % but without the % using sed.
What I tried
So i guess the following regex should match the right pattern : [0-9]{1,3}%.
Then i can use this sed command to catch the three digits and only print them :
sed -nE 's/.*([0-9]{1,3})%.*/\1/p'
Example
However when i run it, it shows :
$ echo "100%" | sed -nE 's/.*([0-9]{1,3})%.*/\1/p'
0
instead of
100
Obviously, there's something wrong with my sed command and i think the problem comes from here :
[0-9]{1,3}
which apparently doesn't do what i want it to do.
edit:
Solution
The .* at the start of sed -nE 's/.*([0-9]{1,3})%.*/\1/p' "ate" the two first digits.
The right way to write it, according to Wicktor's answer, is :
sed -nE 's/(.*[^0-9])?([0-9]{1,3})%.*/\2/p'
The .* grabs all digits leaving just the last of the three digits in 100%.
Use
sed -nE 's/(.*[^0-9])?([0-9]{1,3})%.*/\2/p'
Details
(.*[^0-9])? - (Group 1) an optional sequence of any 0 or more chars up to the non-digit char including it
([0-9]{1,3}) - (Group 2) one to three digits
% - a % char
.* - the rest of the string.
The match is replaced with Group 2 contents, and that is the only value printed since n suppresses the default line output.
It will be easier to use a cut + grep option:
echo "abc 100%" | cut -d% -f1 | grep -oE '[0-9]{1,3}'
100
echo "100%" | cut -d% -f1 | grep -oE '[0-9]{1,3}'
100
Or else you may use this awk:
echo "100%" | awk 'match($0, /[0-9]{1,3}%/){print substr($0, RSTART, RLENGTH-1)}'
100
Or else if you have gnu grep then use -P (PCRE) option:
echo "abc 100%" | ggrep -oP '[0-9]{1,3}(?=%)'
100
This might work for you (GNU sed):
sed -En 's/.*\<([0-9]{1,3})%.*/\1/p' file
This is a filtering exercise, so use the -n option.
Use a back reference to capture 1 to 3 digits, followed by % and print the result if successful.
N.B. The \< ensures the digits start on a word boundary, \b could also be used. The -E option is employed to reduce the number of back slashes which would normally be necessary to quote (,),{ and } metacharacters.
I have an output like Johny-Smith, Juarez-Hugo, etc. and I need instead S, H, etc. Basically, I need the last uppercase letter in a string and that's it. If this is possible in any built in Linux tools (ex awk, sed, grep, etc.) it would be greatly appreciated.
Do you need like this ?
echo "Johny-Smith" | sed 's/^.*\([A-Z]\)[^A-Z]*$/\1/g'
Test:
$ echo "Johny-Smith-Hello Johny-Smith" | sed 's/.*\([A-Z]\)[^A-Z]*/\1/g'
S
With GNU grep and if PCRE option is available
$ echo 'Johny-Smith' | grep -oP '.*\K[A-Z]'
S
$ echo 'Juarez-Hugo' | grep -oP '.*\K[A-Z]'
H
-o prints only matched portion
-P Perl regular expression
.*\K positive lookbehind, not part of output
[A-Z] any uppercase character
with perl, see perldoc for command line options explanation
$ # prints the string within captured group
$ echo 'Johny-Smith' | perl -lne 'print /.*([A-Z])/'
S
$ echo 'Juarez-Hugo' | perl -lne 'print /.*([A-Z])/'
H
In Bash:
$ var="Johny-Smith-Hello Johny-Smith"; var="${var//[^[:upper:]]/}";echo "${var: -1}"
S
${var//[^[:upper:]]/} remove all non-upper case letter chars
echo ${var: -1} output the last one
Is there a way to tell sed to output only captured groups?
For example, given the input:
This is a sample 123 text and some 987 numbers
And pattern:
/([\d]+)/
Could I get only 123 and 987 output in the way formatted by back references?
The key to getting this to work is to tell sed to exclude what you don't want to be output as well as specifying what you do want. This technique depends on knowing how many matches you're looking for. The grep command below works for an unspecified number of matches.
string='This is a sample 123 text and some 987 numbers'
echo "$string" | sed -rn 's/[^[:digit:]]*([[:digit:]]+)[^[:digit:]]+([[:digit:]]+)[^[:digit:]]*/\1 \2/p'
This says:
don't default to printing each line (-n)
exclude zero or more non-digits
include one or more digits
exclude one or more non-digits
include one or more digits
exclude zero or more non-digits
print the substitution (p) (on one line)
In general, in sed you capture groups using parentheses and output what you capture using a back reference:
echo "foobarbaz" | sed 's/^foo\(.*\)baz$/\1/'
will output "bar". If you use -r (-E for OS X) for extended regex, you don't need to escape the parentheses:
echo "foobarbaz" | sed -r 's/^foo(.*)baz$/\1/'
There can be up to 9 capture groups and their back references. The back references are numbered in the order the groups appear, but they can be used in any order and can be repeated:
echo "foobarbaz" | sed -r 's/^foo(.*)b(.)z$/\2 \1 \2/'
outputs "a bar a".
If you have GNU grep:
echo "$string" | grep -Po '\d+'
It may also work in BSD, including OS X:
echo "$string" | grep -Eo '\d+'
These commands will match any number of digit sequences. The output will be on multiple lines.
or variations such as:
echo "$string" | grep -Po '(?<=\D )(\d+)'
The -P option enables Perl Compatible Regular Expressions. See man 3 pcrepattern or man 3 pcresyntax.
Sed has up to nine remembered patterns but you need to use escaped parentheses to remember portions of the regular expression.
See here for examples and more detail
you can use grep
grep -Eow "[0-9]+" file
run(s) of digits
This answer works with any count of digit groups. Example:
$ echo 'Num123that456are7899900contained0018166intext' \
| sed -En 's/[^0-9]*([0-9]{1,})[^0-9]*/\1 /gp'
123 456 7899900 0018166
Expanded answer.
Is there any way to tell sed to output only captured groups?
Yes. replace all text by the capture group:
$ echo 'Number 123 inside text' \
| sed 's/[^0-9]*\([0-9]\{1,\}\)[^0-9]*/\1/'
123
s/[^0-9]* # several non-digits
\([0-9]\{1,\}\) # followed by one or more digits
[^0-9]* # and followed by more non-digits.
/\1/ # gets replaced only by the digits.
Or with extended syntax (less backquotes and allow the use of +):
$ echo 'Number 123 in text' \
| sed -E 's/[^0-9]*([0-9]+)[^0-9]*/\1/'
123
To avoid printing the original text when there is no number, use:
$ echo 'Number xxx in text' \
| sed -En 's/[^0-9]*([0-9]+)[^0-9]*/\1/p'
(-n) Do not print the input by default.
(/p) print only if a replacement was done.
And to match several numbers (and also print them):
$ echo 'N 123 in 456 text' \
| sed -En 's/[^0-9]*([0-9]+)[^0-9]*/\1 /gp'
123 456
That works for any count of digit runs:
$ str='Test Num(s) 123 456 7899900 contained as0018166df in text'
$ echo "$str" \
| sed -En 's/[^0-9]*([0-9]{1,})[^0-9]*/\1 /gp'
123 456 7899900 0018166
Which is very similar to the grep command:
$ str='Test Num(s) 123 456 7899900 contained as0018166df in text'
$ echo "$str" | grep -Po '\d+'
123
456
7899900
0018166
About \d
and pattern: /([\d]+)/
Sed does not recognize the '\d' (shortcut) syntax. The ascii equivalent used above [0-9] is not exactly equivalent. The only alternative solution is to use a character class: '[[:digit:]]`.
The selected answer use such "character classes" to build a solution:
$ str='This is a sample 123 text and some 987 numbers'
$ echo "$str" | sed -rn 's/[^[:digit:]]*([[:digit:]]+)[^[:digit:]]+([[:digit:]]+)[^[:digit:]]*/\1 \2/p'
That solution only works for (exactly) two runs of digits.
Of course, as the answer is being executed inside the shell, we can define a couple of variables to make such answer shorter:
$ str='This is a sample 123 text and some 987 numbers'
$ d=[[:digit:]] D=[^[:digit:]]
$ echo "$str" | sed -rn "s/$D*($d+)$D+($d+)$D*/\1 \2/p"
But, as has been already explained, using a s/…/…/gp command is better:
$ str='This is 75577 a sam33ple 123 text and some 987 numbers'
$ d=[[:digit:]] D=[^[:digit:]]
$ echo "$str" | sed -rn "s/$D*($d+)$D*/\1 /gp"
75577 33 123 987
That will cover both repeated runs of digits and writing a short(er) command.
Give up and use Perl
Since sed does not cut it, let's just throw the towel and use Perl, at least it is LSB while grep GNU extensions are not :-)
Print the entire matching part, no matching groups or lookbehind needed:
cat <<EOS | perl -lane 'print m/\d+/g'
a1 b2
a34 b56
EOS
Output:
12
3456
Single match per line, often structured data fields:
cat <<EOS | perl -lape 's/.*?a(\d+).*/$1/g'
a1 b2
a34 b56
EOS
Output:
1
34
With lookbehind:
cat <<EOS | perl -lane 'print m/(?<=a)(\d+)/'
a1 b2
a34 b56
EOS
Multiple fields:
cat <<EOS | perl -lape 's/.*?a(\d+).*?b(\d+).*/$1 $2/g'
a1 c0 b2 c0
a34 c0 b56 c0
EOS
Output:
1 2
34 56
Multiple matches per line, often unstructured data:
cat <<EOS | perl -lape 's/.*?a(\d+)|.*/$1 /g'
a1 b2
a34 b56 a78 b90
EOS
Output:
1
34 78
With lookbehind:
cat EOS<< | perl -lane 'print m/(?<=a)(\d+)/g'
a1 b2
a34 b56 a78 b90
EOS
Output:
1
3478
I believe the pattern given in the question was by way of example only, and the goal was to match any pattern.
If you have a sed with the GNU extension allowing insertion of a newline in the pattern space, one suggestion is:
> set string = "This is a sample 123 text and some 987 numbers"
>
> set pattern = "[0-9][0-9]*"
> echo $string | sed "s/$pattern/\n&\n/g" | sed -n "/$pattern/p"
123
987
> set pattern = "[a-z][a-z]*"
> echo $string | sed "s/$pattern/\n&\n/g" | sed -n "/$pattern/p"
his
is
a
sample
text
and
some
numbers
These examples are with tcsh (yes, I know its the wrong shell) with CYGWIN. (Edit: For bash, remove set, and the spaces around =.)
Try
sed -n -e "/[0-9]/s/^[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\).*$/\1 \2 \3 \4 \5 \6 \7 \8 \9/p"
I got this under cygwin:
$ (echo "asdf"; \
echo "1234"; \
echo "asdf1234adsf1234asdf"; \
echo "1m2m3m4m5m6m7m8m9m0m1m2m3m4m5m6m7m8m9") | \
sed -n -e "/[0-9]/s/^[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\)[^0-9]*\([0-9]*\).*$/\1 \2 \3 \4 \5 \6 \7 \8 \9/p"
1234
1234 1234
1 2 3 4 5 6 7 8 9
$
You need include whole line to print group, which you're doing at the second command but you don't need to group the first wildcard. This will work as well:
echo "/home/me/myfile-99" | sed -r 's/.*myfile-(.*)$/\1/'
It's not what the OP asked for (capturing groups) but you can extract the numbers using:
S='This is a sample 123 text and some 987 numbers'
echo "$S" | sed 's/ /\n/g' | sed -r '/([0-9]+)/ !d'
Gives the following:
123
987
I want to give a simpler example on "output only captured groups with sed"
I have /home/me/myfile-99 and wish to output the serial number of the file: 99
My first try, which didn't work was:
echo "/home/me/myfile-99" | sed -r 's/myfile-(.*)$/\1/'
# output: /home/me/99
To make this work, we need to capture the unwanted portion in capture group as well:
echo "/home/me/myfile-99" | sed -r 's/^(.*)myfile-(.*)$/\2/'
# output: 99
*) Note that sed doesn't have \d
You can use ripgrep, which also seems to be a sed replacement for simple substitutions, like this
rg '(\d+)' -or '$1'
where ripgrep uses -o or --only matching and -r or --replace to output only the first capture group with $1 (quoted to be avoid intepretation as a variable by the shell) two times due to two matches.