From ps -ef command output -Dorg.xxx.yyy=/home/user/aaa/server.log.
I'd like to extract the file path /home/user/aaa/server.log (can be any name.file).
Now, I'm using command:
ps -ef | grep -Po '(?<=-Dorg.xxx.yyy=)[^\s]*'
It will display two matched results:
/home/user/aaa/server.log
)[^\s]*
It looks like it counts the command as well for the 2nd matched result. How can I remove it? Or is there other suggestions? (I can not use -m1).
If you just need the file name, use \K operator:
org\.xxx\.yyy=\K[^\s]*
ps -ef | grep -Po 'org\.xxx\.yyy=\K[^\s]*'
It will match the whole string, but will only print the file name matched with [^\s]*.
From perlre:
There is a special form of this construct, called \K (available since
Perl 5.10.0), which causes the regex engine to "keep" everything it
had matched prior to the \K and not include it in $& . This
effectively provides variable-length look-behind.
Use that:
grep -Po '(?<=-[D]org.xxx.yyy=)[^\s]*'
Just put one of the characters in square brackets ([D]). The meaning of the regex hasn't changed and the pattern doesn't match itself anymore.
Related
I have a text file with the following pattern written to it:
TIME[32.468ms] -(3)-............."TEXT I WANT TO KEEP"
I would like to discard the first part of each line containing
TIME[32.468ms] -(3)-.............
To test the regular expression I've tried the following:
cat myfile.txt | egrep "^TIME\[.*\]\s\s\-\(3\)\-\.+"
This identifies correctly the lines I want. Now, to delete the pattern I've tried:
cat myfile.txt | sed s/"^TIME\[.*\]\s\s\-\(3\)\-\.+"//
but it just seems to be doing the cat, since it shows the content of the complete file and no substitution happens.
What am I doing wrong?
OS: CentOS 7
With your shown samples, please try following grep command. Written and tested with GNU grep.
grep -oP '^TIME\[\d+\.\d+ms\]\s+-\(\d+\)-\.+\K.*' Input_file
Explanation: Adding detailed explanation for above code.
^TIME\[ ##Matching string TIME from starting of value here.
\d+\.\d+ms\] ##Matching digits(1 or more occurrences) followed by dot digits(1 or more occurrences) followed by ms ] here.
\s+-\(\d+\)-\.+ ##Matching spaces91 or more occurrences) followed by - digits(1 or more occurrences) - and 1 or more dots.
\K ##Using \K option of GNU grep to make sure previous match is found in line but don't consider it in printing, print next matched regex part only.
.* ##to match till end of the value.
2nd solution: Adding awk program here.
awk 'match($0,/^TIME\[[0-9]+\.[0-9]+ms\][[:space:]]+-\([0-9]+\)-\.+/){print substr($0,RSTART+RLENGTH)}' Input_file
Explanation: using match function of awk, to match regex ^TIME\[[0-9]+\.[0-9]+ms\][[:space:]]+-\([0-9]+\)-\.+ which will catch text which we actually want to remove from lines. Then printing rest of the text apart from matched one which is actually required by OP.
This awk using its sub() function:
awk 'sub(/^TIME[[][^]]*].*\.+/,"")' file
"TEXT I WANT TO KEEP"
If there is replacement, sub() returns true.
$ cut -d'"' -f2 file
TEXT I WANT TO KEEP
You may use:
s='TIME[32.468ms] -(3)-............."TEXT I WANT TO KEEP"'
sed -E 's/^TIME\[[^]]*].*\.+//'
"TEXT I WANT TO KEEP"
The \s regex extension may not be supported by your sed.
In BRE syntax (which is what sed speaks out of the box) you do not backslash round parentheses - doing that turns them into regex metacharacters which do not match themselves, somewhat unintuitively. Also, + is just a regular character in BRE, not a repetition operator (though you can turn it into one by similarly backslashing it: \+).
You can try adding an -E option to switch from BRE syntax to the perhaps more familiar ERE syntax, but that still won't enable Perl regex extensions, which are not part of ERE syntax, either.
sed 's/^TIME\[[^][]*\][[:space:]][[:space:]]-(3)-\.*//' myfile.txt
should work on any reasonably POSIX sed. (Notice also how the minus character does not need to be backslash-escaped, though doing so is harmless per se. Furthermore, I tightened up the regex for the square brackets, to prevent the "match anything" regex you had .* from "escaping" past the closing square bracket. In some more detail, [^][] is a negated character class which matches any character which isn't (a newline or) ] or [; they have to be specified exactly in this order to avoid ambiguity in the character class definition. Finally, notice also how the entire sed script should normally be quoted in single quotes, unless you have specific reasons to use different quoting.)
If you have sed -E or sed -r you can use + instead of * but then this complicates the overall regex, so I won't suggest that here.
A simpler one for sed:
sed 's/^[^"]*//' myfile.txt
If the "text you want to keep" always surrounded by the quote like this and only them having the quote in the line starting with "TIME...", then:
sed -n '/^TIME/p' file | awk -F'"' '{print $2}'
should get the line starting with "TIME..." and print the text within the quotes.
Thanks all, for your help.
By the end, I've found a way to make it work:
echo 'TIME[32.468ms] -(3)-.............TEXT I WANT TO KEEP' | grep TIME | sed -r 's/^TIME\[[0-9]+\.[0-9]+ms\]\s\s-\(3\)-\.+//'
More generally,
grep TIME myfile.txt | sed -r ‘s/^TIME\[[0-9]+\.[0-9]+ms\]\s\s-\(3\)-\.+//’
Cheers,
Pedro
I want to extract Jira ticket number from the branch name with sed.
This is what I have
echo "PTW-123-branch-name" | sed 's/.*\([A-Z]+-[0-9]+[^-]\).*/\1/'
expected result: PTW-123
What is wrong with the regexp?
You may use this sed:
echo "PTW-123-branch-name" | sed 's/\([0-9]\)-.*$/\1/'
PTW-123
Details:
\([0-9]\)-: Matches a digit and captures it in group #1 followed by hyphen
.*$: Match remaining string until end
\1: Is replacement that puts captured digit back in output
Alternatively you can use cut also:
echo "PTW-123-branch-name" | cut -d- -f1,2
PTW-123
In case you are ok with GNU grep please try following then. Simple explanation would be passing echo command's output as a standard input to grep command. Then in grep command using -oP option to print only matched portion and enabling PCRE regex capabilities here. In match section of grep then using non-greedy match to match till digits which should be followed by -, then if a match is found it will print it.
echo "PTW-123-branch-name" | grep -oP '^.*?\d+(?=-)'
I want to find a list of words that contain six or more consonants in a row from a number of text files.
I'm pretty new to the Unix terminal, but this is what I have tried:
cat *.txt | grep -Eo "\w+" | grep -i "[^AEOUIaeoui]{6}"
I use the cat command here because it will otherwise include the file names in the next pipe. I use the second pipe to get a list of all the words in the text files.
The problem is the last pipe, I want to somehow get it to grep 6 consonants in a row, it doesn't need to be the same one. I would know one way of solving the problem, but that would create a command longer that this entire post.
For the last grep you also need the -E switch - or you need to escape the curly braces:
cat *.txt | grep -Eo "\w+" | grep -Ei "[^AEOUIaeoui]{6}"
cat *.txt | grep -Eo "\w+" | grep -i "[^AEOUIaeoui]\{6\}"
I use the cat command here because it will otherwise include the file names in the next pipe
You can disable this using the -h flag:
grep -hEo "\w+" *.txt | grep -Ei "[^AEOUIaeoui]{6}"
You can use
grep -hEio '[[:alpha:]]*[b-df-hj-np-tv-z]{6}[[:alpha:]]*' *.txt
Regex details
[[:alpha:]]* - any zero or more letter
[b-df-hj-np-tv-z]{6} - six English consonant letters on end
[[:alpha:]]* - any zero or more letter.
The grep options make the regex search case insensitive (i) and grep shows the matched texts only (with o) without displaying the filenames (h). The -E option allows the POSIX ERE syntax, else, if you do not specify it, you would need to escape {6} as \{6\},
Use this Perl one-liner:
perl -lne 'print for grep { /[^aeoui]{6}/i } /\b([a-z]+)\b/ig' in_file.txt
Example:
cat > in_file.txt <<EOF
the abcdfghi aBcdfghi.
ABCDFGHI234
abcdEfgh
EOF
perl -lne 'print for grep { /[^aeoui]{6}/i } /\b([a-z]+)\b/ig' in_file.txt
Output:
abcdfghi
aBcdfghi
The Perl one-liner uses these command line flags:
-e : Tells Perl to look for code in-line, instead of in a file.
-n : Loop over the input one line at a time, assigning it to $_ by default.
-l : Strip the input line separator ("\n" on *NIX by default) before executing the code in-line, and append it when printing.
The regex uses these modifiers:
/g : Multiple matches.
/i : Case-insensitive matches.
/\b([a-z]+)\b/ig : Match words that consist of 1 or more letters only ([a-z]+), with words boundary \b on both sides. This way, ABCDFGHI234 does not match, but all 3 words in line 1 (the, abcdfghi, aBcdfghi) match. This may be important for some applications. Note that not all answers in this thread use the word boundary around letters, and thus do not make the distinction shown in this example.
/[^aeoui]{6}/i : Match 6 or more consecutive non-vowels. Non-vowels here resolve exactly to consonants, because the previous regex selected for words made of letters only, that is, vowels and consonants.
SEE ALSO:
perldoc perlrun: how to execute the Perl interpreter: command line switches
perldoc perlre: Perl regular expressions (regexes)
perldoc perlre: Perl regular expressions (regexes): Quantifiers; Character Classes and other Special Escapes; Assertions; Capture groups
perldoc perlrequick: Perl regular expressions quick start
Get all words containing 6 or more consonants in a row in a given directory
cat *.txt | grep -Eo "\w+" | grep -E "[^AEOUIaeoui]{6,}"
We can use grep -Eo (-E Extended regex, -o output ONLY matching)
cat *.txt will output all of the data from all txt files in the current directory
grep -Eo "\w+" will output all of the words from an input in the form of one word per line
We can use Regex to search for strings that contain a pattern:
[^LISTOFCHARACTERS] Any character but LISTOFCHARACTERS
{6,} 6 or more
Here is the line in a file I want to extract information from:
backup-initiation-time="00:00" backup-directory-path="/store/backup" backup-retention-period-days="2"
My command is:
grep "backup-directory-path" test.txt | sed 's/.*backup-directory-path="\(.*?\)" /\1/'
I just want /store/backup that's it. I don't know what I'm doing wrong.
You can use \K (keep) and simply:
grep -oP '.*backup-directory-path=\K([^ ])+'
This will display only the captured part after the "keep".
In order to remove the quotes you have there, just modify it:
grep -oP '.*backup-directory-path="\K([^"])+'
I couldn't find any documentation on non-greedy matching in sed so I'm not sure if this was implemented.
Instead of your non-greedy match using .*? you could use [^"]* if you know the last or one past last character you want to match, in your case ".
This command produces the expected output:
grep "backup-directory-path" test.txt | sed 's|.* backup-directory-path="\([^"]*\)".*|\1|'
I'd like to grab the digits in a string like so :
"sample_2341-43-11.txt" to 2341-43-11
And so I tried the following command:
echo "sample_2341-43-11.txt" | sed -n -r 's|[0-9]{4}\-[0-9]{2}\-[0-9]{2}|\1|p'
I saw this answer, which is where I got the idea.
Use sed to grab a string, but it doesn't work on my machine:
it gives an error "illegal option -r".
it doesn't like the \1, either.
I'm using sed on MacOSX yosemite.
Is this the easiest way to extract that information from the file name?
You need to set your grouping and match the rest of the line to remove it with the group. Also the - does not need to be escaped. And the -n will inhibit the output (It just returns exit level for script conditionals).
echo "sample_2341-43-11.txt" | sed -r 's/^.*([0-9]{4}-[0-9]{2}-[0-9]{2}).*$/\1/'
Enhanced regular expressions are not supported in the Mac version of sed.
You can use grep instead:
echo "sample_2341-43-11.txt" | grep -Eo "((\d+|-)+)"
OUTPUT
2341-43-11
echo "one1sample_2341-43-11.txt" \
| sed 's/[^[:digit:]-]\{1,\}/ /g;s/ \{1,\}/ /g;s/^ //;s/ $//'
1 2341-43-11
Extract all numbers(digit) completed with - (thus allow here --12 but can be easily treated)
posix compliant
all number of the line are on same line (if several) separate by a space character (could be changed to new line if wanted)
You can try this ways also
sed 's/[^_]\+_\([^.]\+\).*/\1/' <<< sample_2341-43-11.txt
OutPut:
2341-43-11
Explanation:
[^_]\+ - Match the content untile _ ( sample_)
\([^.]\+\) - Match the content until . and capture the pattern (2341-43-11)
.* - Discard remaining character (.txt)
You can go with what the poster above said. Well, making use of this
pattern "\d+-\d+-\d+" would match what you are looking for. See demo here
https://regex101.com/r/kO2cZ1/3