Made a regex that matches the field I want to assign to my variable in bash:
The regex is:
(\,?[ ]?(\.?\d{1,3}){4})+\ (.*)
and the substring I am interested about is $3 (group 3)
Could anyone please give me command line to assign the substring to my variable?
Example:
MYVARIABLE=$(echo $FULLSTRING | grep -oP '(\,?[ ]?(\.?\d{1,3}){4})+\ (.*)'
But this example obviously did not work
Thanks a lot
You may extract the Group 3 value using Bash regex matching:
text="1.23.23.45 This is what I want"
rx='(,? ?(\.?[0-9]{1,3}){4})+ (.*)'
if [[ $text =~ $rx ]]; then
echo "${BASH_REMATCH[3]}"
else
echo "No match!"
fi
See the online Bash demo printing This is what I want.
If there is a regex match (if [[ $text =~ $rx ]]), the contents of Group 3 are in "${BASH_REMATCH[3]}".
If you have Perl installed, then you can match against your regex and print the field you want:
MYVARIABLE=$(echo $FULLSTRING | perl -nE '/(\,?[ ]?(\.?\d{1,3}){4})+\ (.*)/;say $3')
Example:
FULLSTRING=', .123.4.5.6 matchthis'
MYVARIABLE=$(echo $FULLSTRING | perl -nE '/(\,?[ ]?(\.?\d{1,3}){4})+\ (.*)/;say $3')
echo $MYVARIABLE
Outputs: matchthis
Related
using =~ operator to match output of a command and grab group from it. Code is as follows:
Comamndout=$(cmd) Match=‘^hello world’ If $Comamndout =~ $Match; then
echo something fi
Commandout is in pattern
Something
Hello world
But if statement is failing.
Is bash regex support multiline search with everyline start with ^ and end with $.
No, the =~ operator doesn't perform a multiline search. A newline must be matched literally:
string=$(cmd)
regexp='(^|'$'\n'')hello world'
if [[ $string =~ $regexp ]]; then
echo matches
fi
=~ would treat multiple lines as one line.
if [[ $(echo -e "abc\nd") =~ ^a.*d$ ]]; then
echo "find a string '$(echo -e "abc\nd")' that starts with a and ends with d"
fi
Output:
find a string 'abc
d' that starts with a and ends with d
P.S.
When processing multiple lines, it is common to use grep or read with either re-direct or pipeline.
For a grep and pipeline example:
# to find a line start with either a or e
echo -e "abc\nd\ne" | grep -E "^[ae]"
Output:
abc
e
For a read and redirect example:
while read line; do
if [[ $line =~ ^a} ]] ; then
echo "find a line '${line}' start with a"
fi
done <<< $(echo -e "abc\nd\ne")
Output:
find a line 'abc' start with a
P.S.
-e of echo means translate following \n into new line. -E of grep means using the extended regular expression to match.
I want the name from fielnames like this:
abc-dirk-alt.avi
and I only want the part between the -- (dirk)
The normal regex is -(.*?)- but i dont know how to write this in a bash script.
how can I do this?
You may use a -([^-]*)- regex ([^-]* matches zero or more chars other than -) to avoid using lazy quantifiers and extract Group 1 value via ${BASH_REMATCH[1]} after a match is found:
s="abc-dirk-alt.avi"
rx="-([^-]*)-"
if [[ $s =~ $rx ]]; then
echo ${BASH_REMATCH[1]};
fi
See the online Bash demo.
Given any string in bash, e.g flaccid, I want to match all characters in the string but the first 3 (in this case I want to exclude "fla" and match only "ccid"). The regex also needs to work in sed.
I have tried positive look behind and the following regex expressions (as well as various other unsuccessful ones):
^.{3}+([a-z,A-Z]+)
sed -r 's/(?<=^....)(.[A-Z]*)/,/g'
Google hasn't been very helpful as it only produce results like "get first 3 characters .."
Thanks in advance!
If you want to get all characters but the first 3 from a string, you can use cut:
str="flaccid"
cut -c 4- <<< "$str"
or bash variable subsitution:
str="flaccid"
echo "${str:3}"
That will strip the first 3 characters out of your string.
You may just use a capturing group within an expression like ^.{3}(.*) / ^.{3}([a-zA-Z]+) and grab the ${BASH_REMATCH[1]} contents:
#!/bin/bash
text="flaccid"
rx="^.{3}(.*)"
if [[ $text =~ $rx ]]; then
echo ${BASH_REMATCH[1]};
fi
See online Bash demo
In sed, you should also be using capturing groups / backreferences to get what you need. To just keep the first 3 chars, you may use a simple:
echo "flaccid" | sed 's/.\{3\}//'
See this regex demo. The .\{3\} matches exactly any 3 chars and will remove them from the beginning only, since g modifier is not used.
Now, both the solutions above will output ccid, returning the first 3 chars only.
Using sed, just remove them
echo string | sed 's/^...//g'
How is it that no-one has named the most simple and portable solution:
shell "Parameter expansions":
str="flacid"
echo "${str#???}
For a regex (bash):
$ str="flaccid"
$ regex='^.{3}(.*)$'
$ [[ $str =~ $regex ]] && echo "${BASH_REMATCH[1]}"
ccid
Same regex in sed:
$ echo "flaccid" | sed -E "s/$regex/\1/"
ccid
Or sed (Basic Regex):
$ echo "flaccid" | sed 's/^.\{3\}\(.*\)$/\1/'
ccid
I am new to regex and I am trying to write a regex in a bash script .
I am trying to match line with a regex which has to return the second word in the line .
regex = "commit\s+(.*)"
line = "commit 5456eee"
if [$line =~ $regex]
then
echo $2
else
echo "No match"
fi
When I run this I get the following error:-
man.sh: line 1: regex: command not found
man.sh: line 2: line: command not found
I am new to bash scripting .
Can anyone please help me fix this .
I just want to write a regex to capture the word that follows commit
You don't want a regex, you want parameter expansion/substring extraction:
line="commit 5456eee"
first="${line% *}"
regex="${line#* }"
if [[ $line =~ $regex ]]
then
echo $2
else
echo "No match"
fi
$first == 'commit', $regex == '5456eee'. Bash provides all the tools you need.
If you really only need the second word you could also do it with awk
line = "commit 5456eee"
echo $line | awk '{ print $2 }'
or if you have a file:
cat filename | awk '{ print $2 }'
Even if it's no bash only solution, awk should be present on most linux os's.
You should remove the spaces around the equals sign, otherwise bash thinks you want to execute the regex command using = and "commit\s+(.*)" as arguments.
Then you should remove the spaces also in the if condition and quote the strings:
$ regex="commit\s+(.*)"
$ line="commit 5456eee"
$ if [ "$line"=~"$regex" ]
> then
> echo "Match"
> else
> echo "No match"
> fi
Match
maybe you didn't start your script with the
#!/bin/sh
or
#!/bin/bash
to define the language you're using... ?
It must be your first line.
then be careful, spaces are consistant in bash. In your "if" statement, it should be :
if [ $line =~ $regex ]
check this out and tell us more about the errors you get
if you make this script to a file like test.sh
and execute like that :
test.sh commit aaa bbb ccc
$0 $1 $2 $3 $4
you can get the arguments eassily by $0 $1...
A simple way to get the resulting capture group that was matched (if there is one) is to use BASH_REMATCH, which puts the match results into it's own array:
regex=$"commit (.*)"
line=$"commit 5456eee"
if [[ $line =~ $regex ]]
then
match=${BASH_REMATCH[1]}
echo $match
else
echo "No match"
fi
Since you have only one capture group it will be defined within the array as BASH_REMATCH[1]. In the above example I've assigned the variable $match to the result of BASH_REMATCH[1] which returns:
5456eee
How do I replace regex with $var in this command?
echo "$many_lines" | sed -n '/regex/{g;1!p;};h'
$var could look like fs2#auto-17.
The sed command will output the line immediately before a regex, but not the line containing the regex.
If all this can be done easier with a Perl one-liner, then it is fine with me.
It is not beautiful, but this gives me the previous line to $var which is want I wanted.
echo "$many_lines" | grep -B1 "$var" | grep -v "$var"
In Perl regexes, you can interpolate variable contents into regexes like /$foo/. However, the contents will be interpreted as a pattern. If you want to match the literal content of $foo, you have to escape the metacharacters: $safe = quotemeta $foo; /$safe/. This can be shortended to /\Q$foo\E/, which is what you usually want. A trailing \E is optional.
I don't know if the sed regex engine has a similar feature.
A Perl one-liner: perl -ne'$var = "..."; print $p if /\Q$var/; $p=$_'
Use double quotes instead of single quotes to allow variable expansion:
echo $many_lines | sed -n "/$var/"'{g;1!p;};h'
Since you are looking for a line before the regex, with a single one liner it will not be that trivial and beautiful, but here is how I will do it (Using Perl only):
echo "$many_lines" | perl -nle 'print $. if /\Q$var/' | while read line_no; do
export line_no
echo $many_lines | perl -nle 'print if $. - 1 == $ENV{line_no}'
done
or if you want to do in one line
echo "$many_lines" | perl -nle 'BEGIN {my $content = ""; } $content .= $_; END { while ($content =~ m#([^\n]+)\n[^\n]+\Q$var#gosm) { print $1 }}'
Or this one, should definitely match:
echo "$many_lines" | perl -nle 'BEGIN {my #contents; } push #contents, $_; if ($contents[-1] =~ m#\Q$var#o)') { print $contents[-2] if defined $contents[-2]; }
Or you can use here-documents too, if you don't want to escape the double quotes!
In Perl it looks like this:
$heredoc = <<HEREDOC;
here is your text and $var is your parameter
HEREDOC
Its important to end the heredoc with the same string you began, in my example its "HEREDOC" in a new line!