$ echo "name=hello 123" | sed 's/.*name=\(.*\)\s.*/\1/'
This is the command used to extract the word hello (string between = and whitespace) from the given text. Now I want to replace hello with ***** using sed.
The output should be
name=***** 123
You need to perform substitution on matched portion only - sed doesn't provide a syntax for that, you could workaround using loop..
$ # tested on GNU sed, syntax might vary for other implementations
$ echo 'name=foo 123' | sed -E ':a s/(name=\**)[a-z]/\1*/; ta'
name=*** 123
$ echo 'name=hello 123' | sed -E ':a s/(name=\**)[a-z]/\1*/; ta'
name=***** 123
Or use perl which allows to use code in replacement section
$ echo 'name=foo 123' | perl -pe 's/name=\K[^ ]+/$&=~s|.|*|gr/e'
name=*** 123
$ echo 'name=hello 123' | perl -pe 's/name=\K[^ ]+/$&=~s|.|*|gr/e'
name=***** 123
$ # or construct a string based on length of matched portion
$ echo 'name=hello 123' | perl -pe 's/name=\K[^ ]+/"*" x length($&)/e'
name=***** 123
Using sed:
echo "name=hello" | sed 's/\(name\)=\(.\+\)/\1=*****/
If you want to replace each char with a *, see Sundeep's response.
Related
How to use sed and regex to replace the text between a variable number of one token?
Example of input:
/abc/bcd/cde/
Expected output:
/../../../
Tried:
Command: echo "/abc/bcd/cde/" | sed 's/\/.*\//\/..\//g' output: /../
Using perl and look around assertions :
$ perl -pe 's|(?<=/)\w{3}(?=/)|..|g' file
/../../../
Using sed :
$ echo "/abc/bcd/cde/" | sed -E 's|[a-z]{3}|..|g'
/../../../
Replace every substring of non-slashes ([^/]\+) with two dots:
$> echo "/abc/bcd/cde/" | sed 's$[^/]\+$..$g'
# => /../../../
Base on #Gilles Quenot implementation but, capturing any alpha numeric chars between //
$ echo "/abddc/bcqsdd/cdde/" | sed -E 's|(/)?[^/]+/|\1../|g'
I want to find the set of words that contain two consecutive e’s, and also contains two y’s.
So far i got to /eeyy/
Alteration with ERE:
$ echo evyyree | grep -E '.*ee.*yy|.*yy.*ee'
evyyree
$ echo eveeryy | grep -E '.*ee.*yy|.*yy.*ee'
eveeryy
If the match needs to be in the same word, you can do:
$ echo "eee yyyy" | grep -E 'ee[^[:space:]]*yy|yy[^[:space:]]*ee' # no match
$ echo "eeeyyyy" | grep -E 'ee[^[:space:]]*yy|yy[^[:space:]]*ee'
eeeyyyy
Then only that word:
$ echo 'eeeyy heelo' | grep -Eo 'ee[^[:space:]]*yy|yy[^[:space:]]*ee'
eeeyy
Pipe it:
$ echo eennmmyy | grep ee | grep yy
eennmmyy
awk approach to match all words that contain both ee and yy:
s="eennmmyy heello thees-whyy someyy"
echo $s | awk '{for(i=1;i<=NF;i++) if($i~/ee/ && $i~/yy/) print $i}'
The output:
eennmmyy
thees-whyy
The only sensible and extensible way to do this is with awk:
awk '/ee/&&/yy/' file
Imagine trying to do it the grep way if you also had to find zz. Here's awk:
awk '/ee/&&/yy/&&/zz/' file
and here's grep:
grep -E 'ee.*yy.*zz|ee.*zz.*yy|yy.*ee.*zz|yy.*zz.*ee|zz.*yy.*ee|zz.*ee.*yy' file
Now add a 4th additional string to search for and see what that looks like!
i have a range of strings which conform to one of the two following patters:
("string with spaces",4)
or
(string_without_spaces,4)
I need to extract the "string" via a bash command, and so far have found a pattern that works for each, but not for both.
echo "(\"string with spaces\",4)" | sed -n 's/("\(.*\)",.*)/\1/ip'
output:string with spaces
echo "(string_without_spaces,4)" | sed -n 's/(\(.*\),.*)/\1/ip'
output:string_without_spaces
I have tried using "\? however it does not match the " if it is there:
echo "(SIM,0)" | sed -n 's/("\?\(.*\)"\?,.*)/\1/ip'
output: SIM
echo "(\"SIM\",0)" | sed -n 's/("\?\(.*\)"\?,.*)/\1/ip'
output: SIM"
can anyone suggest a pattern that would extract the string in both scenarios? I am not tied to sed but would prefer to not have to install perl in this environment.
How about using [^"] instead of . to exclude " to be matched.
$ echo '("string with spaces",4)' | sed -n 's/("\?\([^"]*\)"\?,.*)/\1/p'
string with spaces
$ echo "(string_without_spaces,4)" | sed -n 's/("\?\([^"]*\)"\?,.*)/\1/p'
string_without_spaces
$ echo "(SIM,0)" | sed -n 's/("\?\([^"]*\)"\?,.*)/\1/p'
SIM
$ echo '("SIM",0)' | sed -n 's/("\?\([^"]*\)"\?,.*)/\1/p'
SIM
I'm am trying to replace a series of asterix symbols in a text file with a -999.9 using sed. However I can't figure out how to properly escape the wildcard symbol.
e.g.
$ echo "2006.0,1.0,************,-5.0" | sed 's/************/-999.9/g'
sed: 1: "s/************/-999.9/g": RE error: repetition-operator operand invalid
Doesn't work. And
$ echo "2006.0,1.0,************,-5.0" | sed 's/[************]/-999.9/g'
2006.0,1.0,-999.9-999.9-999.9-999.9-999.9-999.9-999.9-999.9-999.9-999.9-999.9-999.9,-5.0
puts a -999.9 for every * which isn't what I intended either.
Thanks!
Use this:
echo "2006.0,1.0,************,-5.0" | sed 's/[*]\+/-999.9/g'
Test:
$ echo "2006.0,1.0,************,-5.0" | sed 's/[*]\+/-999.9/g'
2006.0,1.0,-999.9,-5.0
Any of these (and more) is a regexp that will modify that line as you want:
$ echo "2006.0,1.0,************,-5.0" | sed 's/\*\**/999.9/g'
2006.0,1.0,999.9,-5.0
$ echo "2006.0,1.0,************,-5.0" | sed 's/\*\+/999.9/g'
2006.0,1.0,999.9,-5.0
$ echo "2006.0,1.0,************,-5.0" | sed -r 's/\*+/999.9/g'
2006.0,1.0,999.9,-5.0
$ echo "2006.0,1.0,************,-5.0" | sed 's/\*\{12\}/999.9/g'
2006.0,1.0,999.9,-5.0
$ echo "2006.0,1.0,************,-5.0" | sed -r 's/\*{12}/999.9/g'
2006.0,1.0,999.9,-5.0
$ echo "2006.0,1.0,************,-5.0" | sed 's/\*\{1,\}/999.9/g'
2006.0,1.0,999.9,-5.0
$ echo "2006.0,1.0,************,-5.0" | sed -r 's/\*{1,}/999.9/g'
2006.0,1.0,999.9,-5.0
sed operates on regular expressions, not strings, so you need to learn regular expression syntax if you're going to use sed and in particular the difference between BREs (which sed uses by default) and EREs (which some seds can be told to use instead) and PCREs (which sed never uses but some other tools and "regexp checkers" do). Only the first solution above is a BRE that will work on all seds on all platforms. Google is your friend.
* is a regex symbol that needs to be escaped.
You can even use BASH string replacement:
s="2006.0,1.0,************,-5.0"
echo "${s/\**,/-999.9,}"
2006.0,1.0,-999.9,-5.0
Using sed:
sed 's/\*\+/999.9/g' <<< "$s"
2006.0,1.0,999.9,-5.0
Ya, * are special meta character which repeats the previous token zero or more times. Escape * in-order to match literal * characters.
sed 's/\*\*\*\*\*\*\*\*\*\*\*\*/-999.9/g'
When this possibility was introduced into gawk I have no idea!
gawk -F, '{sub(/************/,"-999.9",$3)}1' OFS=, file
2006.0,1.0,-999.9,-5.0
Is there a way to tell sed to output only captured groups?
For example, given the input:
This is a sample 123 text and some 987 numbers
And pattern:
/([\d]+)/
Could I get only 123 and 987 output in the way formatted by back references?
The key to getting this to work is to tell sed to exclude what you don't want to be output as well as specifying what you do want. This technique depends on knowing how many matches you're looking for. The grep command below works for an unspecified number of matches.
string='This is a sample 123 text and some 987 numbers'
echo "$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.