I have yet another list of subdomain. I want to remove any Wildcard subdomain which include these special characters:
()!&$#*+?
Mostly, the data are prefixly random. Also, could be middle. Here's some sample of output data
(www.imgur.com
***************diet.blogspot.com
*-1.gbc.criteo.com
------------------------------------------------------------i.imgur.com
This has been quite an inconvenience while scanning through the list. As always, I'm trying sed to fix it:
sed -i "/[!()#$&?+]/d" foo.txt ###Didn't work
sed -i "/[\!\(\)\#\$\&\?\+]/d" ###Escaping char didn't work
Performing commands above still result in an unchanged list and the file still on original state. I'm thinking that; to fix this is to pipe series of sed command in order to remove it one by one:
cat foo.txt | sed -e "/!/d" -e "/#/d" -e "/\*/d" -e "/\$/d" -e "/(/d" -e "/)/d" -e "/+/d" -e "/\'/d" -e "/&/d" >> foo2.txt
cat foo.txt | sed -e "/\!/d" | sed -e "/\#/d" | sed -e "/\*/d" | sed -e "/\$/d" | sed -e "/\+/d" | sed -e "/\'/d" | sed -e "/\&/d" >> foo2.txt
If escaping all special char doesn't work, it must've been my false logic. Also tried with /g still doesn't increase my luck.
As a side note: I don't want - to be deleted as some valid subdomain can have - character:
line-apps.com
line-apps-beta.com
line-apps-rc.com
line-apps-dev.com
Any help would be cherished.
Using sed
$ sed '/[[:punct:]]/d' input_file
This should delete all lines with special characters, however, it would help if you provided sample data.
To do what you're trying to do in your answer (which adds [ and ] and more to the set of characters in your question) would be:
sed '/[][!?+,#$&*() ]/d'
or just:
grep -v '[][!?+,#$&*() ]'
Per POSIX to include ] in a bracket expression it must be the first character otherwise it indicates the end of the bracket expression.
Consider printing lines you want instead of deleting lines you do not want, though, e.g.:
grep '^[[:alnum:]_.-]$' file
to print lines that only contain letters, numbers, underscores, dashes, and/or periods.
I am running a linux bash script. From stout lines like: /gpx/trk/name=MyTrack1, I want to keep only the end of line after =.
I am struggling to understand why the following sed command is not working as I expect:
echo "/gpx/trk/name=MyTrack1" | sed -e "s/^*=//"
(I also tried)
echo "/gpx/trk/name=MyTrack1" | sed -e "s/^*\=//"
The return is always /gpx/trk/name=MyTrack1 and not MyTrack1
An even simpler way if this is the only structure you are concerned about:
echo "/gpx/trk/name=MyTrack1" | cut -d = -f 2
Simply try:
echo "/gpx/trk/name=MyTrack1" | sed 's/.*=//'
Solution 2nd: With another sed.
echo "/gpx/trk/name=MyTrack1" | sed 's/\(.*=\)\(.*\)/\2/'
Explanation: As per OP's request adding explanation for this code here:
s: Means telling sed to do substitution operation.
\(.*=\): Creating first place in memory to keep this regex's value which tells sed to keep everything in 1st place of memory from starting to till = so text /gpx/trk/name= will be in 1 place.
\(.*\): Creating 2nd place in memory for sed telling it to keep everything now(after the match of 1st one, so this will start after =) and have value in it as MyTrack1
/\2/: Now telling sed to substitute complete line with only 2nd memory place holder which is MyTrack1
Solution 3rd: Or with awk considering that your Input_file is same as shown samples.
echo "/gpx/trk/name=MyTrack1" | awk -F'=' '{print $2}'
Solution 4th: With awk's match.
echo "/gpx/trk/name=MyTrack1" | awk 'match($0,/=.*$/){print substr($0,RSTART+1,RLENGTH-1)}'
$ echo "/gpx/trk/name=MyTrack1" | sed -e "s/^.*=//"
MyTrack1
The regular expression ^.*= matches anything up to and including the last = in the string.
Your regular expression ^*= would match the literal string *= at the start of a string, e.g.
$ echo "*=/gpx/trk/name=MyTrack1" | sed -e "s/^*=//"
/gpx/trk/name=MyTrack1
The * character in a regular expression usually modifies the immediately previous expression so that zero or more of it may be matched. When * occurs at the start of an expression on the other hand, it matches the character *.
Not to take you off the sed track, but this is easy with Bash alone:
$ echo "$s"
/gpx/trk/name=MyTrack1
$ echo "${s##*=}"
MyTrack1
The ##*= pattern removes the maximal pattern from the beginning of the string to the last =:
$ s="1=2=3=the rest"
$ echo "${s##*=}"
the rest
The equivalent in sed would be:
$ echo "$s" | sed -E 's/^.*=(.*)/\1/'
the rest
Where #*= would remove the minimal pattern:
$ echo "${s#*=}"
2=3=the rest
And in sed:
$ echo "$s" | sed -E 's/^[^=]*=(.*)/\1/'
2=3=the rest
Note the difference in * in Bash string functions vs a sed regex:
The * in Bash (in this context) is glob like - itself means 'any character'
The * in a regex refers to the previous pattern and for 'any character' you need .*
Bash has extensive string manipulation functions. You can read about Bash string patterns in BashFAQ.
I have a file that has a line stating
version = "12.0.08-SNAPSHOT"
The word version and quoted strings can occur on multiple lines in that file.
I am looking for a single line bash statement that can output the following string:
12.0.08-SNAPSHOT
The version can have RELEASE tag too instead of SNAPSHOT.
So to summarize, given
version = "12.0.08-SNAPSHOT"
expected output: 12.0.08-SNAPSHOT
And given
version = "12.0.08-RELEASE"
expected output: 12.0.08-RELEASE
The following command prints strings enquoted in version = "...":
grep -Po '\bversion\s*=\s*"\K.*?(?=")' yourFile
-P enables perl regexes, which allow us to use features like \K and so on.
-o only prints matched parts instead of the whole lines.
\b ensures that version starts at a word boundary and we do not match things like abcversion.
\s stands for any kind of whitespace.
\K lets grep forget, that it matched the part before \K. The forgotten part will not be printed.
.*? matches as few chararacters as possible (the matching part will be printed) ...
(?=") ... until we see a ", which won't be included in the match either (this is called a lookahead).
Not all grep implementations support the -P option. Alternatively, you can use perl, as described in this answer:
perl -nle 'print $& if m{\bversion\s*=\s*"\K.*?(?=")}' yourFile
Seems like a job for cut:
$ echo 'version = "12.0.08-SNAPSHOT"' | cut -d'"' -f2
12.0.08-SNAPSHOT
$ echo 'version = "12.0.08-RELEASE"' | cut -d'"' -f2
12.0.08-RELEASE
Portable solution:
$ echo 'version = "12.0.08-RELEASE"' |sed -E 's/.*"(.*)"/\1/g'
12.0.08-RELEASE
or even:
$ perl -pe 's/.*"(.*)"/\1/g'.
$ awk -F"\"" '{print $2}'
I got my research result after using sed :
zcat file* | sed -e 's/.*text=\(.*\)status=[^/]*/\1/' | cut -f 1 - | grep "pattern"
But it only shows the part that I cut. How can I print all lines after a match ?
I'm using zcat so I cannot use awk.
Thanks.
Edited :
This is my log file :
[01/09/2015 00:00:47] INFO=54646486432154646 from=steve idfrom=55516654455457 to=jone idto=5552045646464 guid=100021623456461451463 n
um=6 text=hi my number is 0 811 22 1/12 status=new survstatus=new
My aim is to find all users that spam my site with their telephone numbers (using grep "pattern") then print all the lines to get all the information about each spam. The problem is there may be matches in INFO or id, so I use sed to get the text first.
Printing all lines after a match in sed:
$ sed -ne '/pattern/,$ p'
# alternatively, if you don't want to print the match:
$ sed -e '1,/pattern/ d'
Filtering lines when pattern matches between "text=" and "status=" can be done with a simple grep, no need for sed and cut:
$ grep 'text=.*pattern.* status='
You can use awk
awk '/pattern/,EOF'
n.b. don't be fooled: EOF is just an uninitialized variable, and by default 0 (false). So that condition cannot be satisfied until the end of file.
Perhaps this could be combined with all the previous answers using awk as well.
Maybe this is what you actually want? Find lines matching "pattern" and extract the field after text= up through just before status=?
zcat file* | sed -e '/pattern/s/.*text=\(.*\)status=[^/]*/\1/'
You are not revealing what pattern actually is -- if it's a variable, you cannot use single quotes around it.
Notice that \(.*\)status=[^/]* would match up through survstatus=new in your example. That is probably not what you want? There doesn't seem to be a status= followed by a slash anywhere -- you really should explain in more detail what you are actually trying to accomplish.
Your question title says "all line after a match" so perhaps you want everything after text=? Then that's simply
sed 's/.*text=//'
i.e. replace up through text= with nothing, and keep the rest. (I trust you can figure out how to change the surrounding script into zcat file* | sed '/pattern/s/.*text=//' ... oops, maybe my trust failed.)
The seldom used branch command will do this for you. Until you match, use n for next then branch to beginning. After match, use n to skip the matching line, then a loop copying the remaining lines.
cat file | sed -n -e ':start; /pattern/b match;n; b start; :match n; :copy; p; n ; b copy'
zcat file* | sed -e 's/.*text=\(.*\)status=[^/]*/\1/' | ***cut -f 1 - | grep "pattern"***
instead change the last 2 segments of your pipeline so that:
zcat file* | sed -e 's/.*text=\(.*\)status=[^/]*/\1/' | **awk '$1 ~ "pattern" {print $0}'**
I want to find files that have "abc" AND "efg" in that order, and those two strings are on different lines in that file. Eg: a file with content:
blah blah..
blah blah..
blah abc blah
blah blah..
blah blah..
blah blah..
blah efg blah blah
blah blah..
blah blah..
Should be matched.
Grep is an awkward tool for this operation.
pcregrep which is found in most of the modern Linux systems can be used as
pcregrep -M 'abc.*(\n|.)*efg' test.txt
where -M, --multiline allow patterns to match more than one line
There is a newer pcre2grep also. Both are provided by the PCRE project.
pcre2grep is available for Mac OS X via Mac Ports as part of port pcre2:
% sudo port install pcre2
and via Homebrew as:
% brew install pcre
or for pcre2
% brew install pcre2
pcre2grep is also available on Linux (Ubuntu 18.04+)
$ sudo apt install pcre2-utils # PCRE2
$ sudo apt install pcregrep # Older PCRE
Here is a solution inspired by this answer:
if 'abc' and 'efg' can be on the same line:
grep -zl 'abc.*efg' <your list of files>
if 'abc' and 'efg' must be on different lines:
grep -Pzl '(?s)abc.*\n.*efg' <your list of files>
Params:
-P Use perl compatible regular expressions (PCRE).
-z Treat the input as a set of lines, each terminated by a zero byte instead of a newline. i.e. grep treats the input as a one big line. Note that if you don't use -l it will display matches followed by a NUL char, see comments.
-l list matching filenames only.
(?s) activate PCRE_DOTALL, which means that '.' finds any character or newline.
I'm not sure if it is possible with grep, but sed makes it very easy:
sed -e '/abc/,/efg/!d' [file-with-content]
sed should suffice as poster LJ stated above,
instead of !d you can simply use p to print:
sed -n '/abc/,/efg/p' file
I relied heavily on pcregrep, but with newer grep you do not need to install pcregrep for many of its features. Just use grep -P.
In the example of the OP's question, I think the following options work nicely, with the second best matching how I understand the question:
grep -Pzo "abc(.|\n)*efg" /tmp/tes*
grep -Pzl "abc(.|\n)*efg" /tmp/tes*
I copied the text as /tmp/test1 and deleted the 'g' and saved as /tmp/test2. Here is the output showing that the first shows the matched string and the second shows only the filename (typical -o is to show match and typical -l is to show only filename). Note that the 'z' is necessary for multiline and the '(.|\n)' means to match either 'anything other than newline' or 'newline' - i.e. anything:
user#host:~$ grep -Pzo "abc(.|\n)*efg" /tmp/tes*
/tmp/test1:abc blah
blah blah..
blah blah..
blah blah..
blah efg
user#host:~$ grep -Pzl "abc(.|\n)*efg" /tmp/tes*
/tmp/test1
To determine if your version is new enough, run man grep and see if something similar to this appears near the top:
-P, --perl-regexp
Interpret PATTERN as a Perl regular expression (PCRE, see
below). This is highly experimental and grep -P may warn of
unimplemented features.
That is from GNU grep 2.10.
This can be done easily by first using tr to replace the newlines with some other character:
tr '\n' '\a' | grep -o 'abc.*def' | tr '\a' '\n'
Here, I am using the alarm character, \a (ASCII 7) in place of a newline.
This is almost never found in your text, and grep can match it with a ., or match it specifically with \a.
awk one-liner:
awk '/abc/,/efg/' [file-with-content]
If you are willing to use contexts, this could be achieved by typing
grep -A 500 abc test.txt | grep -B 500 efg
This will display everything between "abc" and "efg", as long as they are within 500 lines of each other.
You can do that very easily if you can use Perl.
perl -ne 'if (/abc/) { $abc = 1; next }; print "Found in $ARGV\n" if ($abc && /efg/); }' yourfilename.txt
You can do that with a single regular expression too, but that involves taking the entire contents of the file into a single string, which might end up taking up too much memory with large files.
For completeness, here is that method:
perl -e '#lines = <>; $content = join("", #lines); print "Found in $ARGV\n" if ($content =~ /abc.*efg/s);' yourfilename.txt
I don't know how I would do that with grep, but I would do something like this with awk:
awk '/abc/{ln1=NR} /efg/{ln2=NR} END{if(ln1 && ln2 && ln1 < ln2){print "found"}else{print "not found"}}' foo
You need to be careful how you do this, though. Do you want the regex to match the substring or the entire word? add \w tags as appropriate. Also, while this strictly conforms to how you stated the example, it doesn't quite work when abc appears a second time after efg. If you want to handle that, add an if as appropriate in the /abc/ case etc.
If you need both words are close each other, for example no more than 3 lines, you can do this:
find . -exec grep -Hn -C 3 "abc" {} \; | grep -C 3 "efg"
Same example but filtering only *.txt files:
find . -name *.txt -exec grep -Hn -C 3 "abc" {} \; | grep -C 3 "efg"
And also you can replace grep command with egrep command if you want also find with regular expressions.
I released a grep alternative a few days ago that does support this directly, either via multiline matching or using conditions - hopefully it is useful for some people searching here. This is what the commands for the example would look like:
Multiline:
sift -lm 'abc.*efg' testfile
Conditions:
sift -l 'abc' testfile --followed-by 'efg'
You could also specify that 'efg' has to follow 'abc' within a certain number of lines:
sift -l 'abc' testfile --followed-within 5:'efg'
You can find more information on sift-tool.org.
Possible with ripgrep:
$ rg --multiline 'abc(\n|.)+?efg' test.txt
3:blah abc blah
4:blah abc blah
5:blah blah..
6:blah blah..
7:blah blah..
8:blah efg blah blah
Or some other incantations.
If you want . to count as a newline:
$ rg --multiline '(?s)abc.+?efg' test.txt
3:blah abc blah
4:blah abc blah
5:blah blah..
6:blah blah..
7:blah blah..
8:blah efg blah blah
Or equivalent to having the (?s) would be rg --multiline --multiline-dotall
And to answer the original question, where they have to be on separate lines:
$ rg --multiline 'abc.*[\n](\n|.)*efg' test.txt
And if you want it "non greedy" so you don't just get the first abc with the last efg (separate them into pairs):
$ rg --multiline 'abc.*[\n](\n|.)*?efg' test.txt
https://til.hashrocket.com/posts/9zneks2cbv-multiline-matches-with-ripgrep-rg
Sadly, you can't. From the grep docs:
grep searches the named input FILEs (or standard input if no files are named, or if a single hyphen-minus (-) is given as file name) for lines containing a match to the given PATTERN.
While the sed option is the simplest and easiest, LJ's one-liner is sadly not the most portable. Those stuck with a version of the C Shell (instead of bash) will need to escape their bangs:
sed -e '/abc/,/efg/\!d' [file]
Which line unfortunately does not work in bash et al.
With silver searcher:
ag 'abc.*(\n|.)*efg' your_filename
similar to ring bearer's answer, but with ag instead. Speed advantages of silver searcher could possibly shine here.
#!/bin/bash
shopt -s nullglob
for file in *
do
r=$(awk '/abc/{f=1}/efg/{g=1;exit}END{print g&&f ?1:0}' file)
if [ "$r" -eq 1 ];then
echo "Found pattern in $file"
else
echo "not found"
fi
done
you can use grep incase you are not keen in the sequence of the pattern.
grep -l "pattern1" filepattern*.* | xargs grep "pattern2"
example
grep -l "vector" *.cpp | xargs grep "map"
grep -l will find all the files which matches the first pattern, and xargs will grep for the second pattern. Hope this helps.
If you have some estimation about the distance between the 2 strings 'abc' and 'efg' you are looking for, you might use:
grep -r . -e 'abc' -A num1 -B num2 | grep 'efg'
That way, the first grep will return the line with the 'abc' plus #num1 lines after it, and #num2 lines after it, and the second grep will sift through all of those to get the 'efg'.
Then you'll know at which files they appear together.
With ugrep released a few months ago:
ugrep 'abc(\n|.)+?efg'
This tool is highly optimized for speed. It's also GNU/BSD/PCRE-grep compatible.
Note that we should use a lazy repetition +?, unless you want to match all lines with efg together until the last efg in the file.
You have at least a couple options --
DOTALL method
use (?s) to DOTALL the . character to include \n
you can also use a lookahead (?=\n) -- won't be captured in match
example-text:
true
match me
false
match me one
false
match me two
true
match me three
third line!!
{BLANK_LINE}
command:
grep -Pozi '(?s)true.+?\n(?=\n)' example-text
-p for perl regular expressions
-o to only match pattern, not whole line
-z to allow line breaks
-i makes case-insensitive
output:
true
match me
true
match me three
third line!!
notes:
- +? makes modifier non-greedy so matches shortest string instead of largest (prevents from returning one match containing entire text)
you can use the oldschool O.G. manual method using \n
command:
grep -Pozi 'true(.|\n)+?\n(?=\n)'
output:
true
match me
true
match me three
third line!!
I used this to extract a fasta sequence from a multi fasta file using the -P option for grep:
grep -Pzo ">tig00000034[^>]+" file.fasta > desired_sequence.fasta
P for perl based searches
z for making a line end in 0 bytes rather than newline char
o to just capture what matched since grep returns the whole line (which in this case since you did -z is the whole file).
The core of the regexp is the [^>] which translates to "not the greater than symbol"
As an alternative to Balu Mohan's answer, it is possible to enforce the order of the patterns using only grep, head and tail:
for f in FILEGLOB; do tail $f -n +$(grep -n "pattern1" $f | head -n1 | cut -d : -f 1) 2>/dev/null | grep "pattern2" &>/dev/null && echo $f; done
This one isn't very pretty, though. Formatted more readably:
for f in FILEGLOB; do
tail $f -n +$(grep -n "pattern1" $f | head -n1 | cut -d : -f 1) 2>/dev/null \
| grep -q "pattern2" \
&& echo $f
done
This will print the names of all files where "pattern2" appears after "pattern1", or where both appear on the same line:
$ echo "abc
def" > a.txt
$ echo "def
abc" > b.txt
$ echo "abcdef" > c.txt; echo "defabc" > d.txt
$ for f in *.txt; do tail $f -n +$(grep -n "abc" $f | head -n1 | cut -d : -f 1) 2>/dev/null | grep -q "def" && echo $f; done
a.txt
c.txt
d.txt
Explanation
tail -n +i - print all lines after the ith, inclusive
grep -n - prepend matching lines with their line numbers
head -n1 - print only the first row
cut -d : -f 1 - print the first cut column using : as the delimiter
2>/dev/null - silence tail error output that occurs if the $() expression returns empty
grep -q - silence grep and return immediately if a match is found, since we are only interested in the exit code
This should work too?!
perl -lpne 'print $ARGV if /abc.*?efg/s' file_list
$ARGV contains the name of the current file when reading from file_list
/s modifier searches across newline.
The filepattern *.sh is important to prevent directories to be inspected. Of course some test could prevent that too.
for f in *.sh
do
a=$( grep -n -m1 abc $f )
test -n "${a}" && z=$( grep -n efg $f | tail -n 1) || continue
(( ((${z/:*/}-${a/:*/})) > 0 )) && echo $f
done
The
grep -n -m1 abc $f
searches maximum 1 matching and returns (-n) the linenumber.
If a match was found (test -n ...) find the last match of efg (find all and take the last with tail -n 1).
z=$( grep -n efg $f | tail -n 1)
else continue.
Since the result is something like 18:foofile.sh String alf="abc"; we need to cut away from ":" till end of line.
((${z/:*/}-${a/:*/}))
Should return a positive result if the last match of the 2nd expression is past the first match of the first.
Then we report the filename echo $f.
To search recursively across all files (across multiple lines within each file) with BOTH strings present (i.e. string1 and string2 on different lines and both present in same file):
grep -r -l 'string1' * > tmp; while read p; do grep -l 'string2' $p; done < tmp; rm tmp
To search recursively across all files (across multiple lines within each file) with EITHER string present (i.e. string1 and string2 on different lines and either present in same file):
grep -r -l 'string1\|string2' *
Here's a way by using two greps in a row:
egrep -o 'abc|efg' $file | grep -A1 abc | grep efg | wc -l
returns 0 or a positive integer.
egrep -o (Only shows matches, trick: multiple matches on the same line produce multi-line output as if they are on different lines)
grep -A1 abc (print abc and the line after it)
grep efg | wc -l (0-n count of efg lines found after abc on the same or following lines, result can be used in an 'if")
grep can be changed to egrep etc. if pattern matching is needed
This should work:
cat FILE | egrep 'abc|efg'
If there is more than one match you can filter out using grep -v