I have a txt file from which I want to get only the expressions of the type
'USA_word*' where * is whatever ( I don't want the whole line, only the expressions )
I try the command
grep -oP ''USA_word*''
But I get a list :
USA_word
USA_word
USA_word
.....
without the part signified by the *.
You may use
grep -o 'USA_word[^[:blank:]]*'
The [^[:blank:]]* part matches 0+ non-whitespace chars.
Besides, this does not use -P PCRE option, and uses a pure BRE POSIX regex making it compatible with the majority of grep implementations.
Related
I want to grep the shortest match and the pattern should be something like:
<car ... model=BMW ...>
...
...
...
</car>
... means any character and the input is multiple lines.
You're looking for a non-greedy (or lazy) match. To get a non-greedy match in regular expressions you need to use the modifier ? after the quantifier. For example you can change .* to .*?.
By default grep doesn't support non-greedy modifiers, but you can use grep -P to use the Perl syntax.
Actualy the .*? only works in perl. I am not sure what the equivalent grep extended regexp syntax would be. Fortunately you can use perl syntax with grep so grep -P would work but grep -E which is same as egrep would not work (it would be greedy).
See also: http://blog.vinceliu.com/2008/02/non-greedy-regular-expression-matching.html
grep
For non-greedy match in grep you could use a negated character class. In other words, try to avoid wildcards.
For example, to fetch all links to jpeg files from the page content, you'd use:
grep -o '"[^" ]\+.jpg"'
To deal with multiple line, pipe the input through xargs first. For performance, use ripgrep.
My grep that works after trying out stuff in this thread:
echo "hi how are you " | grep -shoP ".*? "
Just make sure you append a space to each one of your lines
(Mine was a line by line search to spit out words)
Sorry I am 9 years late, but this might work for the viewers in 2020.
So suppose you have a line like "Hello my name is Jello".
Now you want to find the words that start with 'H' and end with 'o', with any number of characters in between. And we don't want lines we just want words. So for that we can use the expression:
grep "H[^ ]*o" file
This will return all the words. The way this works is that: It will allow all the characters instead of space character in between, this way we can avoid multiple words in the same line.
Now you can replace the space character with any other character you want.
Suppose the initial line was "Hello-my-name-is-Jello", then you can get words using the expression:
grep "H[^-]*o" file
The short answer is using the next regular expression:
(?s)<car .*? model=BMW .*?>.*?</car>
(?s) - this makes a match across multiline
.*? - matches any character, a number of times in a lazy way (minimal
match)
A (little) more complicated answer is:
(?s)<([a-z\-_0-9]+?) .*? model=BMW .*?>.*?</\1>
This will makes possible to match car1 and car2 in the following text
<car1 ... model=BMW ...>
...
...
...
</car1>
<car2 ... model=BMW ...>
...
...
...
</car2>
(..) represents a capturing group
\1 in this context matches the sametext as most recently matched by
capturing group number 1
I know that its a bit of a dead post but I just noticed that this works. It removed both clean-up and cleanup from my output.
> grep -v -e 'clean\-\?up'
> grep --version grep (GNU grep) 2.20
It's not much but I didn't find a solution by searching here or experimenting with a regex-generator:
I want to search for lines in a textfile containing
due:Mo, due:Di, ... or due:So
Maybe I'm near the solution with
grep "due:+(Mo\|Di\|Mi\|Do\|Fr)" todo.txt --color=auto
But this one just highlights and searches for the 5 days (German days of the week). Thank you for your help!
You are really close, in a POSIX BRE pattern, the ( and ) capturing parentheses should be escaped and you do not need the + quantifier (that is actually parsed as a literal + symbol in your POSIX BRE pattern) after ::
grep "due:\(Mo\|Di\|Mi\|Do\|Fr\)\>"
I also suggest a word boundary check with \>.
You may get rid of some escapes using a POSIX ERE with -E option:
grep -E "due:(Mo|Di|Mi|Do|Fr)\>"
See the online demo.
Trying to use regex with grep in the command line to give me lines that start with either a whitespace or lowercase int followed by a space. From there, they must end with either a semi colon or a o.
I tried
grep ^[\s\|int]\s+[\;\o]$ fileName
but I don't get what I'm looking for. I also tried
grep ^\s*int\s+([a-z][a-zA-Z]*,\s*)*[a-z]A-Z]*\s*;
but nothing.
Let's consider this test file:
$ cat file
keep marco
polo
int keep;
int x
If I understand your rules correctly, two of the lines in the above should be kept and the other two discarded.
Let's try grep:
$ grep -E '^(\s|int\s).*[;o]$' file
keep marco
int keep;
The above uses \s to mean space. \s is supported by GNU grep. For other greps, we can use a POSIX character class instead. After reorganizing the code slightly to reduce typing:
grep -E '^(|int)[[:blank:]].*[;o]$' file
How it works
In a Unix shell, the single quotes in the command are critical: they stop the shell from interpreting or expanding any character inside the single quotes.
-E tells grep to use extended regular expressions. Thus reduces the need for backslashes.
Let's examine the regular expression, one piece at a time:
^ matches at the beginning of a line.
(\s|int\s) This matches either a space or int followed by a space.
.* matches zero or more of any character.
[;o] matches any character in the square brackets which means that it matches either ; or o.
$ matches at the end of a line.
I have the following string libVersion = '1.23.45.6' and I need to replace 1.23.45.6 with 1.23.45.7.
Obviously the version could be any number with similar format (it does not have to be 4 numbers).
I tried to use the following but doesn't work
echo "libVersion = '1.23.45.6'" |sed "s/([0-9\.]+)/1.23.45.7/g"
Basic sed, ie sed without any arguments uses BRE (Basic Regular Expression). In BRE, you have to escape +, to bring the power of regex + which repeats the previous token one or more times, likewise for the capturing groups \(regex\)
echo "libVersion = '1.23.45.6'" | sed "s/[0-9.]\+/1.23.45.7/"
You may also use a negated char class to replace all the chars exists within single quotes.
echo "libVersion = '1.23.45.6'" | sed "s/'[^']*'/'1.23.45.7'/"
Since the replacement should occur only one time, you don't need a g global modifier.
I want to grep the shortest match and the pattern should be something like:
<car ... model=BMW ...>
...
...
...
</car>
... means any character and the input is multiple lines.
You're looking for a non-greedy (or lazy) match. To get a non-greedy match in regular expressions you need to use the modifier ? after the quantifier. For example you can change .* to .*?.
By default grep doesn't support non-greedy modifiers, but you can use grep -P to use the Perl syntax.
Actualy the .*? only works in perl. I am not sure what the equivalent grep extended regexp syntax would be. Fortunately you can use perl syntax with grep so grep -P would work but grep -E which is same as egrep would not work (it would be greedy).
See also: http://blog.vinceliu.com/2008/02/non-greedy-regular-expression-matching.html
grep
For non-greedy match in grep you could use a negated character class. In other words, try to avoid wildcards.
For example, to fetch all links to jpeg files from the page content, you'd use:
grep -o '"[^" ]\+.jpg"'
To deal with multiple line, pipe the input through xargs first. For performance, use ripgrep.
My grep that works after trying out stuff in this thread:
echo "hi how are you " | grep -shoP ".*? "
Just make sure you append a space to each one of your lines
(Mine was a line by line search to spit out words)
Sorry I am 9 years late, but this might work for the viewers in 2020.
So suppose you have a line like "Hello my name is Jello".
Now you want to find the words that start with 'H' and end with 'o', with any number of characters in between. And we don't want lines we just want words. So for that we can use the expression:
grep "H[^ ]*o" file
This will return all the words. The way this works is that: It will allow all the characters instead of space character in between, this way we can avoid multiple words in the same line.
Now you can replace the space character with any other character you want.
Suppose the initial line was "Hello-my-name-is-Jello", then you can get words using the expression:
grep "H[^-]*o" file
The short answer is using the next regular expression:
(?s)<car .*? model=BMW .*?>.*?</car>
(?s) - this makes a match across multiline
.*? - matches any character, a number of times in a lazy way (minimal
match)
A (little) more complicated answer is:
(?s)<([a-z\-_0-9]+?) .*? model=BMW .*?>.*?</\1>
This will makes possible to match car1 and car2 in the following text
<car1 ... model=BMW ...>
...
...
...
</car1>
<car2 ... model=BMW ...>
...
...
...
</car2>
(..) represents a capturing group
\1 in this context matches the sametext as most recently matched by
capturing group number 1
I know that its a bit of a dead post but I just noticed that this works. It removed both clean-up and cleanup from my output.
> grep -v -e 'clean\-\?up'
> grep --version grep (GNU grep) 2.20