I am trying to select the line that has the name "paul" in it.
!grep -w '^paul' some_file
This also returns the lines starting with paul-henri. How do I select the single line that starts with the word 'paul' only?
(In other words, dash - or slash / and dot . are getting selected if followed by the word paul)
Update:
Thanks to Tim, this worked:
grep -w '^paul' some_file | grep -vE 'paul[-./?]'
You could match on the pattern ^paul[^-]:
!grep -w '^paul[^-]' some_file
This would match any line starting with paul, which is then followed by one or more characters other than dash. If you need to also match possible lines starting with and containing only paul, then you might need to use a negative lookahead:
^paul(?!-)
But, this would require an extended version of grep, and your version of grep might not support it.
Related
I have a need to perform multiple grep matches as part of the same grep command. When I run them individually, they work fine. But not when together. I hope someone could either show me a solution or perhaps can help me find a work-around. Here is sample stream:
(string start..) RollUp:"V" Enzyme:"ENZA ENZB ENZD ENZE" (..string end)
In the first command I am needing to isolate all RollUp substrings.Value is always A or V:
grep -o "RollUp:\"[AV]\""
In the second command I am needing to isolate all combinations of Enzyme values (1-20 total, spaces in between, don't know values names). This command works:
grep -oE 'Enzyme:[[:space:]]*"[^"]+"'
However, I need to match both patterns as part of same stream. When I try:
grep -oE "RollUp:\"[AV]\""\|Enzyme:[[:space:]]*"[^"]+""
, nothing is returned. I would be grateful for any ideas for getting this double grep pattern match to work. Thank you!
regex someting[^"]+ : this means string something followed by anything till next " is seen. Here + sign means , at least one or more match.
grep -oE 'RollUp:"[^"]+|Enzyme:[[:space:]]*"[^"]+"' file
I am trying to grep a file for the exact occurrence of a match, but I get also longer spurious matches:
grep CAT1717O99 myfile.txt -F -w
Output:
CAT1717O99
CAT1717O99.5
I would like to output only the first exactly matching line. Is there any way to get rid of the second line?
Thanks in advance.
Arturo
This is the file 'myfile.txt':
CAT1717O99
CAT1717O99.5
This will do the work for you.
grep -Fx "CAT1717O99" textfile
-F means Fixed
-x mean exact
Use the power of Perl-compatible regular expression (PCRE) and search the matches to the given pattern:
grep -Po "\bCAT1717O99(\s|$)" myfile.txt
(\s|$) - alternative group, ensures matching substring CAT1717O99 if it's followed by whitespace or placed at the end of the line
-P option, allows regular expressions
-o option, prints only matched parts of matching lines
You'll need explicitly request spaces in order to ignore special chars.
grep -E '(^| )CAT1717O99( |$)' myFile.txt
from grep manual :
-w, --word-regexp
Select only those lines containing matches that form whole words. The test is that the matching substring must either be at the beginning of the line, or preceded by a non-word constituent character. Similarly, it must be either at the end of the line or followed by a non-word constituent character. Word-constituent characters are letters, digits, and the underscore.
I am trying to grep the output of a command that outputs unknown text and a directory per line. Below is an example of what I mean:
.MHuj.5.. /var/log/messages
The text and directory may be different from time to time or system to system. All I want to do though is be able to grep the directory out and send it to a variable.
I have looked around but cannot figure out how to grep to the end of a word. I know I can start the search phrase looking for a "/", but I don't know how to tell grep to stop at the end of the word, or if it will consider the next "/" a new word or not. The directories listed could change, so I can't assume the same amount of directories will be listed each time. In some cases, there will be multiple lines listed and each will have a directory list in it's output. Thanks for any help you can provide!
If your directory paths does not have spaces then you can do:
$ echo '.MHuj.5.. /var/log/messages' | awk '{print $NF}'
/var/log/messages
It's not clear from a single example whether we can generalize that e.g. the first occurrence of a slash marks the beginning of the data you want to extract. If that holds, try
grep -o '/.*' file
To fetch everything after the last space, try
grep -o '[^ ]*$' file
For more advanced pattern matching and extraction, maybe look at sed, or Awk or Perl or Python.
Your line can be described as:
^\S+\s+(\S+)$
That's assuming whitespace is your delimiter between the random text and the directory. It simply separates the whitespace from the non-whitespace and captures the second part.
Or you might want to look into the word boundary character class: \b.
I know you said to use grep, but I can't help to mention that this is trivially done using awk:
awk '{ print $NF }' input.txt
This is assuming that a whitespace is the delimiter and that the path does not contain any whitespaces.
grep (GNU grep) 2.14
Hello,
I have a log file that I want to filter on a selected word. However, it tends to filter on many for example.
tail -f gateway-* | grep "P_SIP:N_iptB1T1"
This will also find words like this:
"P_SIP:N_iptB1T10"
"P_SIP:N_iptB1T11"
"P_SIP:N_iptB1T12"
etc
However, I don't want to display anything after the 1. grep is picking up 11, 12, 13, etc.
Many thanks for any suggestions,
You can restrict the word to end at 1:
tail -f gateway-* | grep "P_SIP:N_iptB1T1\>"
This will work assuming that you have a matching case which is only "P_SIP:N_iptB1T1".
But if you want to extract from P_SIP:N_iptB1T1x, and display only once, then you need to restrict to show only first match.
grep -o "P_SIP:N_iptB1T1"
-o, --only-matching show only the part of a line matching PATTERN
More info
At least two approaches can be tried:
grep -w pattern matches for full words. Seems to work for this case too, even though the pattern has punctuation.
grep pattern -m 1 to restrict the output to first match. (Also doable with grep xxx | head -1)
If the lines contains the quotes as in your example, just use the -E option in grep and match the closing quote with \". For example:
grep -E "P_SIP:N_iptB1T1\"" file
If these quotes aren't in the text file, and there's blank spaces or endlines after the word, you can match these too:
# The word is followed by one or more blanks
grep -E "P_SIP:N_iptB1T1\s+" file
# Match lines ending with the interesting word
grep -E "P_SIP:N_iptB1T1$" file
I'm using the operating systems dictionary file to scan. I'm creating a java program to allow a user to enter any concoction of letters to find words that contain those letters. How would I do this using grep commands?
To find words that contain only the given letters:
grep -v '[^aeiou]' wordlist
The above filters out the lines in wordlist that don't contain any characters except for those listed. It's sort of using a double negative to get what you want. Another way to do this would be:
grep '^[aeiou]+$' wordlist
which searches the whole line for a sequence of one or more of the selected letters.
To find words that contain all of the given letters is a bit more lengthy, because there may be other letters in between the ones we want:
cat wordlist | grep a | grep e | grep i | grep o | grep u
(Yes, there is a useless use of cat above, but the symmetry is better this way.)
You can use a single grep to solve the last problem in Greg's answer, provided your grep supports PCRE. (Based on this excellent answer, boiled down a bit)
grep -P "(?=.*a)(?=.*e)(?=.*i)(?=.*o)(?=.*u)" wordlist
The positive lookahead means it will match anything with an "a" anywhere, and an "e" anywhere, and.... etc etc.