I am new to Linux and trying to learn Regular Expressions with egrep currently. I am trying to pull matches for apartments from a list of street addresses from a .txt file, particularly matching on either ('apartment', 'apt', '#').
I have tried:
egrep '[apt,#]' file.txt
egrep '[a.p.t.],[#]' file.txt
I need to match only words that contain the letters a,p,t consecutively, no matter 'apartment'/'apt' and also any lines with '#'
Any insight is appreciated
Related
Hi i have a few archive of FW log and occasionally im required to compare them with a series of IP addresses (thousand of them) to get the date and time if the ip addresses matches. my current script is as follow:
#input the list of ip into array
mapfile -t -O 1 var < ip.txt while true
do
#check array is not null
if [[-n "${var[i]}"]] then
zcat /.../abc.log.gz | grep "${var[i]}"
((i++))
It does work but its way too slow and i would think that grep-ping a line with multiple strings would be faster than zcat on every ip line. So my question is is there a way to generate a 'long grep search string' from the ip.txt? or is there a better way to do this
Sure. One thing is that using cat is usually slightly inefficient. I'd recommend using zgrep here instead. You could generate a regex as follows
IP=`paste -s -d ' ' ip.txt`
zgrep -E "(${IP// /|})" /.../abc.log.gz
The first line loads the IP addresses into IP as a single line. The second line builds up a regex that looks something like (127.0.0.1|8.8.8.8) by replacing spaces with |'s. It then uses zgrep to search through abc.log.gz once, with that -Extended regex.
However, I recommend that you do not do this. Firstly, you should escape strings put into a regex. Even if you know that ip.txt really contains IP addresses (e.g. not controlled by a malicious user), you should still escape the periods. But rather than building up a search string and then escape it, just use the -Fixed strings and -file features of grep. Then you get the simple and fast one-liner:
zgrep -F -f ip.txt /.../abc.log.gz
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 have a file with a list of word and I want to identify only the word in the file which exactly matches another word?
So, for example, if I have in the file, the words "BEBE, BEBÉ, BEBÉS", and I look for "BEBE", I want it to return just the first one, which is the exact match.
I tried using grep -w "BEBE" filename.txt, but it doesn't work. It still gives me back all three of them.
Use -o to only display the part that matches with -w, also use -F for fixed string if you're not regex matching:
$ cat file
BEBE, BEBÉ, BEBÉS
$ grep -woF 'BEBÉ' file
BEBÉ
$ grep -woF 'BEBÉS' file
BEBÉS
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.