Because picture (or example) is worth more than thousand words, I'll use an example:
RewriteRule ^/products/([0-9]+)$ /content.php?id=$1
In this RewriteRule example we've got simple regular expression. $1 is a reference to something that is captured by ([0-9]+), so it is reference to some number if the matching exists. Is it possible to do something like that in grep?
Let's say, some xml document contains the following :
<someTag>someValue</someTag>
I would like to extract only someValue, but input for second_bash_script for the following:
first_bash_script | grep "<someTag>\([[:digit::]]\)\+</someTag>" | second_bash_script
is someValue. Is it possible to extract only someValue using grep?
Thanks for any clue!
Those are two separate questions, right?
The answer to the first one would be: use sed, grep doesn't do substitutions.
sed 's_^/products/\([0-9]\+\)_/content.php?id=\1_g'
The second thing can be done with grep using Perl regexp:
$ echo '<someTag>42</someTag>' | grep -oP '(?<=<someTag>)\d+(?=</someTag>)'
42
Related
I have found on this answer the regex to find a string between two characters. In my case I want to find every pattern between ‘ and ’. Here's the regex :
(?<=‘)(.*?)(?=’)
Indeed, it works when I try it on https://regex101.com/.
The thing is I want to use it with grep but it doesn't work :
grep -E '(?<=‘)(.*?)(?=’)' file
Is there anything missing ?
Those are positive look-ahead and look behind assertions. You need to enable it using PCRE(Perl Compatible Regex) and perhaps its better to get only matching part using -o option in GNU grep:
grep -oP '(?<=‘)(.*?)(?=’)' file
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
Is there any way to do the opposite of showing only the matching part of strings in grep (the -o flag), that is, show everything except the part that matches the regex?
That is, the -v flag is not the answer, since that would not show files containing the match at all, but I want to show these lines, but not the part of the line that matches.
EDIT: I wanted to use grep over sed, since it can do "only-matching" matches on multi-line, with:
cat file.xml|grep -Pzo "<starttag>.*?(\n.*?)+.*?</starttag>"
This is a rather unusual requirement, I don't think grep would alternate the strings like that. You can achieve this with sed, though:
sed -n 's/$PATTERN//gp' file
EDIT in response to OP's edit:
You can do multiline matching with sed, too, if the file is small enough to load it all into memory:
sed -rn ':r;$!{N;br};s/<starttag>.*?(\n.*?)+.*?<\/starttag>//gp' file.xml
You can do that with a little help from sed:
grep "pattern" input_file | sed 's/pattern//g'
I don't think there is a way in grep.
If you use ack, you could output Perl's special variables $` and $' variables to show everything before and after the match, respectively:
ack string --output="\$`\$'"
Similarly if you wanted to output what did match along with other text, you could use $& which contains the matched string;
ack string --output="Matched: $&"
I'm using the operating systems dictionary file to scan. I'm creating a java program to allow a user to enter any concoction of letters to find words that contain those letters. How would I do this using grep commands?
To find words that contain only the given letters:
grep -v '[^aeiou]' wordlist
The above filters out the lines in wordlist that don't contain any characters except for those listed. It's sort of using a double negative to get what you want. Another way to do this would be:
grep '^[aeiou]+$' 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.