I have many files from which I need to get information.
Example of my files:
first file content:
"test This info i need grep</singleline>"
and
second file content (with two lines):
"test This info=
i need grep too</singleline>"
in results I need grep this text: from first file - "This info i need grep" and from second file - "This info= i need grep too"
In first file I use:
grep -o 'test .*</singleline>' * | sed -e 's/test \(.*\)<\/singleline>/\1/'
and successfully get "This info i need grep" but I can not get the information from the second file by using the same command.
Please help rewrite the command or write what the other.
Or, if you insist to use grep, you can:
grep -Pzo 'test(\n|.)*(?=</singleline>)' test.txt
To understand the meaning of each flag, use grep --help:
-P, --perl-regexp
PATTERN is a Perl regular expression
-o, --only-matching
show only the part of a line matching PATTERN
-z, --null-data
a data line ends in 0 byte, not newline
I'd use pcregrep, which can match multiline regexes:
pcregrep -Mo 'test \K((?s).)*?(?=</singleline>)' filename
The tricks are:
-M allows pcregrep to match on more than one line,
-o makes it print only the match,
\K throws away the part of the match that comes before it,
(?=</singleline>) is a lookahead term that matches an empty string if (and only if) it is followed by </singleline>, and
((?s).)*? to match any characters non-greedily, which is to say that if you have several occurrences of </singleline> in the file, it will match until the closest rather than the furthest. If this is not desired, remove the ?. (?s) enables the s option locally for the term to make . match newlines in it; it wouldn't do that by default.
Thanks to #CasimiretHippolyte for pointing out the ((?s).) alternative to (.|\n).
It looks like you're parsing quoted-printable encoded text, where a "soft" line break (one that is an artifact from fixed-line-width formatting) is indicated with a line-terminating = (directly before the \n).
Since in a later comment you also expressed the desire to print each match as a single line, I suggest the following 2-pass appraoch:
use awk to remove the soft line breaks
then use grep on the result
awk '/=$/ { printf "%s", substr($0, 1, length($0)-2); next } 1' file |
grep -Po 'test .*?(?=</singleline>)'
Tip of the hat to Wintermute's helpful answer for the non-greedy quantifier, *?, and both Wintermute's and Maroun Maroun's helpful answer for the positive look-ahead assertion, (?=...).
Not that the awk command removes the line-ending = (along with the newline); replace the substr call with just $0 to retain it.
Since strings of interest are first converted back their original single-line representations:
The matches are printed in their original form.
You can use regular (GNU) grep with line-by-line matching; contrast this with
needing to read the entire file at once, as in Maroun Maroun's helpful answer.
Note that, as of this writing, * must be replaced with *? in his answer to work correctly work in files with multiple matches.
needing to install another utility, pcregrep, as in Wintermute's helpful answer.
additionally, the matches would have to be cleaned up to be single-line (something you didn't originally state as a requirement).
Related
I have a large binary file. I want to extract certain strings from it and copy them to a new text file.
For example, in:
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZM-^G
I want to take the number '7' (after the #^#^#E) and every character after it stopping at the Z ('ignoring the M-^G).
I want to copy this 7cacscKLrrok9bwC3Z64NTnZ to a new file.
There will be multiple such strings in one file. The end will always be denoted by the M- (which I don't want copied). The start will always be denoted by a 7 (which I do want copied).
Unfortunately, my knowledge of grep, sed, etc, does not extend to this level. Can someone please suggest a viable way to achieve this?
cat -v filename | grep [7][A-Z,a-z] will show all strings with a '7' followed by a letter but that's not much.
Thank you.
I've noticed that my requirements are rather more complicated.
(I've performed the correct - I hope - formatting this time). Thanks to 'tshiono' for his (?) answer to the earlier submission.
I want to check the ending of a string and, if it ends in M-, grep another string that follows it (with junk in between). If the string does not end in M-, then I don't want it copied (let alone any other strings).
So what I would like is:
grep -a -Po "7[[:alnum:]]+(?=M-)" file_name and if the ending is M- then grep -a -Po "5x[[:alnum:]]+(?=\^)" file_name to copy the string that starts with 5x and ends with a ^.
In this example:
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZM-^GwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
The outcome would be:
7cacscKLrrok9bwC3Z64NTnZ
5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk
However, if the ending is not M- (more precisely, if the ending is ^S), then do not try the second grep and do not record anything at all.
In this example:
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZ^SGwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
The outcome would be null (nothing copied) as the 7cacs... string ends in ^S.
Is grep the correct tool? Grep a file and if the condition in the grep command is 'yes' then issue a different grep command but if the condition is 'no' then do nothing.
Thanks again.
I have noticed one addition modification.
Can one add an OR command to the second part? Grep if the second string starts with 5x OR 6x?
In the example below, grep -aPo "7[[:alnum:]]+M-.*?5x[[:alnum:]]+\^" filename | grep -aPo "7[[:alnum:]]+(?=M-)|5x[[:alnum:]]+(?=\^)" will extract the strings starting with 7 and the strings starting with 5x.
How can one change the 5x to 5x or 6x?
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZM-^GwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7AAAAAscKLrrok9bwC3Z64NTnZM-^GwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM6x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
In this example, the desired outcome would be:
7cacscKLrrok9bwC3Z64NTnZ
5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk
7AAAAAscKLrrok9bwC3Z64NTnZ
6x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk
UPDATE MARCH 09:
I need to create a series of complex grep (or perl) commands to extract strings from a series of binary files.
I need two strings from the binary file.
The first string will always start with a 1.
The first string will end with a letter or number. The next letter will always be a lower case k. I do not want this k character.
The difficulty is that the ending k will not always be the first k in the string. It might be the first k but it might not.
After the k, there is a second string. The second string will always start with an A or a B.
The ending of the second string will be in one of two forms:
a) it will end with a space then display the first three characters from the first string in lower case followed by a )
b) it will end with a ^K then display the first three characters from the first string in lower case.
For example:
1pppsx9YPar8Rvs75tJYWZq3eo8PgwbckB4m4zT7Yg042KIDYUE82e893hY ppp)
Should be:
1pppsx9YPar8Rvs75tJYWZq3eo8Pgwbc and B4m4zT7Yg042KIDYUE82e893hY - delete the k and the space then ppp.
For example:
1zzzsx9YPkr8Rvs75tJYWZq3eo8PgwbckA2m4zT7Yg042KIDYUE82e893hY^Kzzz
Should be:
1zzzsx9YPkar8Rvs75tJYWZq3eo8Pgwbc and A4m4zT7Yg042KIDYUE82e893hY - delete the second k and the ^Kzzz.
In the second example, we see that the first k is part of the first string. It is the k before the A that breaks up the first and second strings.
I hope there is a super grep expert who can help! Many thanks!
If your grep supports -P option, would you please try:
grep -a -Po "7[[:alnum:]]+(?=M-)" file
The -a option forces grep to read the input as a text file.
The -P option enables the perl-compatible regex.
The -o option tells grep to print only the matched substring(s).
The pattern (?=M-) is a zero-width lookahead assertion (introduced in
Perl) without including it in the result.
Alternatively you can also say with sed:
sed 's/M-/\n/g' file | sed -n 's/.*\(7[[:alnum:]]\+\).*/\1/p'
The first sed command splits the input file into miltiple lines by
replacing the substring M- with a newline.
It has two benefits: it breaks the lines to allow multiple matches with
sed and excludes the unnecessary portion M- from the input.
The next sed command extracts the desired pattern from the input.
It assumes your sed accepts \n in the replacement, which is
a GNU extension (not POSIX compliant). Otherwise please try (in case you are working on bash):
sed 's/M-/\'$'\n''/g' file | sed -n 's/.*\(7[[:alnum:]]\+\).*/\1/p'
[UPDATE]
(The requirement has been updated by the OP and the followings are solutions according to it.)
Let me assume the string which starts with 7 and ends with M- is always followed
by another (no more and no less than one) string which starts with 5x and ends
with ^ (ascii caret character) with junks in between.
Then would you please try the following:
grep -aPo "7[[:alnum:]]+M-.*?5x[[:alnum:]]+\^" file | grep -aPo "7[[:alnum:]]+(?=M-)|5x[[:alnum:]]+(?=\^)"
It executes the task in two steps (two cascaded greps).
The 1st grep narrows down the input data into the candidate substring
which will include the desired two sequences and junks in between.
The regex .*? in between matches any (ascii or binary) characters
except for a newline character.
The trailing ? enables the shortest match
which avoids the overrun due to the greedy nature of regex. The regex is intended to match junks in between.
The 2nd grep includes two regex's merged with a pipe | meaning logical OR.
Then it extracts two desired sequences.
A potential problem of grep solution is that grep is a line oriented command
and cannot include the newline character in the matched string.
If a newline character is included in the junks in between (I'm not sure about the possibility), the above solution will fail.
As a workaround, perl will provide flexible manipulations with binary data.
perl -0777 -ne '
while (/(7[[:alnum:]]+)M-.*?(5x[[:alnum:]]+)\^/sg) {
printf("%s\n%s\n", $1, $2);
}
' file
The regex is mostly same as that of grep because the -P option of grep means
perl-compatible.
It can capture multiple patterns at once in variables $1 and $2 hence just one regex is enough.
The -0777 option to the perl command tells perl to slurp all data
at once.
The s option at the end the regex makes a dot match a newline character.
The g option enables the global (multiple) match.
[UPDATE2]
In order to make the regex match either 5x or 6x, replace 5x with (5|6)x.
Namely:
grep -aPo "7[[:alnum:]]+M-.*?(5|6)x[[:alnum:]]+\^" file | grep -aPo "7[[:alnum:]]+(?=M-)|(5|6)x[[:alnum:]]+(?=\^)"
As mentioned before, the pipe | means OR. The OR operator has the lowest priority in the evaluation, hence you need to enclose them with parens in this case.
If there is a possibility any other number than 5 or 6 may appear, it will be safer to put [[:digit:]] instead, which matches any one digit betweeen 0 and 9:
grep -aPo "7[[:alnum:]]+M-.*?[[:digit:]]x[[:alnum:]]+\^" file | grep -aPo "7[[:alnum:]]+(?=M-)|[[:digit:]]x[[:alnum:]]+(?=\^)"
[UPDATE3]
(Answering the OP's requirement on March 9th)
Let me start with a perl code which regex will be relatively easier
to explain.
perl -0777 -ne 'while (/(1(.{3}).+)k([AB].*)[\013 ]\2/g){print "$1 $3\n"}' file
Output:
1pppsx9YPar8Rvs75tJYWZq3eo8Pgwbc B4m4zT7Yg042KIDYUE82e893hY
1zzzsx9YPkr8Rvs75tJYWZq3eo8Pgwbc A2m4zT7Yg042KIDYUE82e893hY
[Explanation of regex]
(1(.{3}).+)k([AB].*)[\013 ]\2
( start of the 1st capture group referred by $1 later
1 literal "1"
( start of the 2nd capture group referred by \2 later
.{3} a sequence of the identical three characters such as ppp or zzz
) end of the 2nd capture group
.+ followed by any characters with "greedy" match which may include the 1st "k"
) end of the 1st capture group
k literal "k"
( start of the 3rd capture group referred by $3 later
[AB].* the character "A" or "B" followed by any characters
) end of the 3rd capture group
[\013 ] followed by ^K or a whitespace
\2 followed by the capture group 2 previously assigned
When implementing it with grep, we will encounter a limitation of grep.
Although we want to extract multiple patterns from the input file,
the -e option (which can specify multiple search patterns) does not
work with -P option. Then we need to split the regex into two patterns
such as:
grep -Po "(1(.{3}).+)(?=k([AB].*)[\013 ]\2)" file
grep -Po "(1(.{3}).+)k\K([AB].*)(?=[\013 ]\2)" file
And the result will be:
1pppsx9YPar8Rvs75tJYWZq3eo8Pgwbc
1zzzsx9YPkr8Rvs75tJYWZq3eo8Pgwbc
B4m4zT7Yg042KIDYUE82e893hY
A2m4zT7Yg042KIDYUE82e893hY
Please be noted the order of output is not same as the order of appearance in the original file.
Another option will be to introduce ripgrep or rg which is a fast
and versatile version of grep. You may need to install ripgrep with
sudo apt install ripgrep or using other package handling tool.
An advantage of ripgrep is it supports -r (replace) option in which
you can make use of the backreferences:
rg -N -Po "(1(.{3}).+)k([AB].*)[\013 ]\2" -r '$1 $3' file
The -r '$1 $3' option prints the 1st and the 3rd capture groups and the result will be the same as perl.
In the general case, you can use the strings utility to pluck out ASCII from binary files; then of course you can try to grep that output for patterns that you find interesting.
Many traditional Unix utilities like grep have internal special markers which might get messed up by binary input. For example, the character \xFF was used for internal purposes by some versions of GNU grep so you can't grep for that character even if you can figure out a way to represent it in the shell (Bash supports $'\xff' for example).
A traditional approach would be to run hexdump or a similar utility, and then grep that for patterns. However, more modern scripting languages like Perl and Python make it easy to manipulate arbitrary binary data.
perl -ne 'print if m/\xff\xff/' </dev/urandom
This might work for you (GNU sed):
sed -En '/\n/!{s/M-\^G/\n/;s/7[^\n]*\n/\n&/};/^7[^\n]*/P;D' file
Split each line into zero or more lines that begin with 7 and end just before M-^G and only print such lines.
I'm trying to use Perl to reorder the content of an md5 file. For each line, I want the filename without the path then the hash. The best command I've come up with is:
$ perl -pe 's|^([[:alnum:]]+).*?([^/]+)$|$2 $1|' DCIM.md5
The input file (DCIM.md5) is produced by md5sum on Linux. It looks like this:
e26ff03dc1bac80226e200c0c63d17a2 ./Path1/IMG_20150201_160548.jpg
01f92572e4c6f2ea42bd904497e4f939 ./Path 2/IMG_20150204_190528.jpg
afce027c977944188b4f97c5dd1bd101 ./Path3/Path 4/IMG_20151011_193008.jpg
The hash is matched by the first group ([[:alnum:]]+) in the
regular expression.
Then the spaces and the path to the file are
matched by .*?.
Then the filename is matched by ([^/]+).
The expression is enclosed with ^ (apparently non-necessary here)
and $. Without the $, the expression does not output what I expect.
I use | rather than / as a separator to avoid escaping it in file paths.
That command returns:
IMG_20150201_160548.jpg
e26ff03dc1bac80226e200c0c63d17a2IMG_20150204_190528.jpg
01f92572e4c6f2ea42bd904497e4f939IMG_20151011_193008.jpg
afce027c977944188b4f97c5dd1bd101IMG_20151011_195133.jpg
The matching is correct, the output sequence is correct (filename without path then hash) but the spacing is not: there's a newline after the filename. I expect it after the hash, like this:
IMG_20150201_160548.jpg e26ff03dc1bac80226e200c0c63d17a2
IMG_20150204_190528.jpg 01f92572e4c6f2ea42bd904497e4f939
IMG_20151011_193008.jpg afce027c977944188b4f97c5dd1bd101
It seems to me that my command outputs the newline character, but I don't know how to change this behavior.
Or possibly the problem comes from the shell, not the command?
Finally, some version information:
$ perl -version
This is perl 5, version 22, subversion 1 (v5.22.1) built for i686-linux-gnu-thread-multi-64int
(with 69 registered patches, see perl -V for more detail)
[^/]+ matches newlines, so the ones in your input are part of $2, which gets put first in your transformed $_ (And there's no newline in $1 so there's no newline at the end of $_...)
Solution: Read up on the -l option from perlrun. In particular:
-l[octnum]
enables automatic line-ending processing. It has two separate effects. First, it automatically chomps $/ (the input record separator) when used with -n or -p. Second, it assigns $\ (the output record separator) to have the value of octnum so that any print statements will have that separator added back on. If octnum is omitted, sets $\ to the current value of $/ .
Alternate solution, which uses lots of concepts from other answers, and comments ...
$ perl -pe 's|(\p{hex}+).*?([^/]+?)$|$2 $1|' DCIM.md5
... and explanation.
After investigating all the answers and trying to figure them out, I've decided that the base of the problem is that the [^/]+ is greedy. Its greediness causes it to capture the newline; it ignores the $ anchor.
This was hard for me to figure out, since I did a lot of parsing using sed before using Perl, and even a greedy wildcard won't capture a newline in sed. Hopefully this post will help those who (being used to sed as I am) are also wondering (as I did) why the $ isn't acting "as I expect it to."
We can see the "greedy" issue by trying what I'll post as another, alternate answer.
Write the file:
$ cat > DCIM.md5<<EOF
> e26ff03dc1bac80226e200c0c63d17a2 ./Path1/IMG_20150201_160548.jpg
> 01f92572e4c6f2ea42bd904497e4f939 ./Path 2/IMG_20150204_190528.jpg
> afce027c977944188b4f97c5dd1bd101 ./Path3/Path 4/IMG_20151011_193008.jpg
> EOF
Get rid of the greedy [^/]+ by changing it to [^/]+?. Parse.
$ perl -pe 's|([[:alnum:]]+).*?([^/]+?)$|$2 $1|' DCIM.md5
IMG_20150201_160548.jpg e26ff03dc1bac80226e200c0c63d17a2
IMG_20150204_190528.jpg 01f92572e4c6f2ea42bd904497e4f939
IMG_20151011_193008.jpg afce027c977944188b4f97c5dd1bd101
Desired output accomplished.
The accepted answer, by #Shawn,
$ perl -lpe 's|^([[:alnum:]]+).*?([^/]+)$|$2 $1|' DCIM.md5
basically changes the $ anchor so as to behave the way a sed person would expect it to.
The answer by #CrafterKolyan takes care of the greedy [^/] capturing the newline by saying you can't have a forward-slash or a newline. This answer still needs the $ anchor to prevent the following situation
1) .* captures the empty string (0 or more of any character)
2) [^/\n]+ captures . .
The answer by #Borodin takes a quite different approach, but it's a great concept.
#Borodin, in addition, made a great comment that allows a more-precise/more-exact version of this answer, which is the version I put at the top of this post.
Finally, if one wants to follow the Perl programming model, here's another alternative.
$ perl -pe 's|([[:xdigit:]]+).*?([^/]+?)(\n\|\Z)|$2 $1$3|' DCIM.md5
P.S. Because sed isn't quite like perl (no non-greedy wildcards,) here's a sed example that shows the behavior I discuss.
$ sed 's|^\([[:alnum:]]\+\).*/\([^/]\+\)$|\2 \1|' DCIM.md5
This is basically a "direct translation" of the perl expression except for the extra '/' before the [^/] stuff. I hope it will help those comparing sed and perl.
use [^/\n] instead of [^/]:
perl -pe 's|^([[:alnum:]]+).*?([^/\n]+)$|$2 $1|' DCIM.md5
Doing a substitution leaves you having to write a regex pattern that matches everything you don't want as well as everything you do. It's usually much better to match just the parts you need and build another string from them
Like this
for ( <> ) {
die unless m< (\w++) .*? ([^/\s]+) \s* \z >x;
print "$2 $1\n";
}
or if you must have a one-liner
perl -ne 'die unless m< (\w++) .*? ([^/\s]+) \s*\z >x; print "$2 $1\n";' myfile.md5
output
IMG_20150201_160548.jpg e26ff03dc1bac80226e200c0c63d17a2
IMG_20150204_190528.jpg 01f92572e4c6f2ea42bd904497e4f939
IMG_20151011_193008.jpg afce027c977944188b4f97c5dd1bd101
I have a text file infile1 with 1,000's of lines.
I wish to use sed to extract the occuring instances of a regex pattern match to outfile2.
NB
Each instance of the regex pattern match may occur more than once on each line of infile1.
Each instance of the extracted regex pattern should be printed to a new line in outfile2.
Does anyone know the syntax within sed to place the regex into?
ps the regex pattern is
\(Google[ ]{1,3}“[a-zA-Z0-9 ]{1,100}[., ]{0,3}”\)
Thank you :)
I think you want
grep -oE 'Google[ ]{1,3}"[a-zA-Z0-9 ]{1,100}[., ]{0,3}"' filename
-o tells grep to print only the matches, each on a line of its own, and -E instructs it to interpret the regex in extended POSIX syntax, which your regex appears to be.
Note that [ ] could be replaced with just a space, and you might want to use [[:alnum:] ] instead of [a-zA-Z0-9 ] to cover umlauts and suchlike if they exist in the current locale.
Addendum: It is also possible to do this with sed. I don't recommend it, but you could write (using GNU sed):
sed -rn 's/Google[ ]{1,3}"[A-Za-z0-9 ]{1,100}[., ]{0,3}"/\n&\n/g; s/[^\n]*\n([^\n]*\n)/\1/g; s/\n[^\n]*$//p' filename
To make this work with older versions of BSD sed, use -En instead of -rn. -r and -E enable extended regex syntax. -r was historically used by GNU sed, -E by BSD sed; newer versions of them support both for compatibility. -n disables auto-printing.
The code works as follows:
# mark all occurrences of the regex by circumscribing them with newlines
s/Google[ ]{1,3}"[A-Za-z0-9 ]{1,100}[., ]{0,3}"/\n&\n/g
# Isolate every other line from the pattern space (the matches). This will
# leave the part behind the last match...
s/[^\n]*\n([^\n]*\n)/\1/g
# ...so we remove it afterwards and print the result of the transformation if it
# happened (the s///p flag does that). The transformation will not happen if
# there were no matches in the line (because then no newlines will have been
# inserted), so in those cases nothing will be printed.
s/\n[^\n]*$//p
It can be done with sed too, but it isn't pretty:
sed -n ':start /foo/{ h; s/\(foo\).*/\1/; s/.*\(foo\)/\1/; p; g; s/foo\(.*\)/\1/; b start; }' infile1 >outfile2
-- provided that you replace the four occurences of foo above with your pattern Google {1,3}“[a-zA-Z0-9 ]{1,100}[., ]{0,3}”.
Yeah, I told you it isn't pretty. :)
I have a file that looks like this:
#"Afghanistan.png",
#"Albania.png",
#"Algeria.png",
#"American_Samoa.png",
I want it to look like this
#"Afghanistan.png",
#"Afghanistan",
#"Albania.png",
#"Albania",
#"Algeria.png",
#"Algeria",
#"American_Samoa.png",
#"American_Samoa",
I thought I could use sed to do this but I can't figure out how to store something in a buffer and then modify it.
Am I even using the right tool?
Thanks
You don't have to get tricky with regular expressions and replacement strings: use sed's p command to print the line intact, then modify the line and let it print implicitly
sed 'p; s/\.png//'
Glenn jackman's response is OK, but it also doubles the rows which do not match the expression.
This one, instead, doubles only the rows which matched the expression:
sed -n 'p; s/\.png//p'
Here, -n stands for "print nothing unless explicitely printed", and the p in s/\.png//p forces the print if substitution was done, but does not force it otherwise
That is pretty easy to do with sed and you not even need to use the hold space (the sed auxiliary buffer). Given the input file below:
$ cat input
#"Afghanistan.png",
#"Albania.png",
#"Algeria.png",
#"American_Samoa.png",
you should use this command:
sed 's/#"\([^.]*\)\.png",/&\
#"\1",/' input
The result:
$ sed 's/#"\([^.]*\)\.png",/&\
#"\1",/' input
#"Afghanistan.png",
#"Afghanistan",
#"Albania.png",
#"Albania",
#"Algeria.png",
#"Algeria",
#"American_Samoa.png",
#"American_Samoa",
This commands is just a replacement command (s///). It matches anything starting with #" followed by non-period chars ([^.]*) and then by .png",. Also, it matches all non-period chars before .png", using the group brackets \( and \), so we can get what was matched by this group. So, this is the to-be-replaced regular expression:
#"\([^.]*\)\.png",
So follows the replacement part of the command. The & command just inserts everything that was matched by #"\([^.]*\)\.png", in the changed content. If it was the only element of the replacement part, nothing would be changed in the output. However, following the & there is a newline character - represented by the backslash \ followed by an actual newline - and in the new line we add the #" string followed by the content of the first group (\1) and then the string ",.
This is just a brief explanation of the command. Hope this helps. Also, note that you can use the \n string to represent newlines in some versions of sed (such as GNU sed). It would render a more concise and readable command:
sed 's/#"\([^.]*\)\.png",/&\n#"\1",/' input
I prefer this over Carles Sala and Glenn Jackman's:
sed '/.png/p;s/.png//'
Could just say it's personal preference.
or one can combine both versions and apply the duplication only on lines matching the required pattern
sed -e '/^#".*\.png",/{p;s/\.png//;}' input
This question already has answers here:
How do I match any character across multiple lines in a regular expression?
(26 answers)
Closed 3 years ago.
Really basic question here. So I'm told that a dot . matches any character EXCEPT a line break. I'm looking for something that matches any character, including line breaks.
All I want to do is to capture all the text in a website page between two specific strings, stripping the header and the footer. Something like HEADER TEXT(.+)FOOTER TEXT and then extract what's in the parentheses, but I can't find a way to include all text AND line breaks between header and footer, does this make sense? Thanks in advance!
When I need to match several characters, including line breaks, I do:
[\s\S]*?
Note I'm using a non-greedy pattern
You could do it with Perl:
$ perl -ne 'print if /HEADER TEXT/ .. /FOOTER TEXT/' file.html
To print only the text between the delimiters, use
$ perl -000 -lne 'print $1 while /HEADER TEXT(.+?)FOOTER TEXT/sg' file.html
The /s switch makes the regular expression matcher treat the entire string as a single line, which means dot matches newlines, and /g means match as many times as possible.
The examples above assume you're cranking on HTML files on the local disk. If you need to fetch them first, use get from LWP::Simple:
$ perl -MLWP::Simple -le '$_ = get "http://stackoverflow.com";
print $1 while m!<head>(.+?)</head>!sg'
Please note that parsing HTML with regular expressions as above does not work in the general case! If you're working on a quick-and-dirty scanner, fine, but for an application that needs to be more robust, use a real parser.
By definition, grep looks for lines which match; it reads a line, sees whether it matches, and prints the line.
One possible way to do what you want is with sed:
sed -n '/HEADER TEXT/,/FOOTER TEXT/p' "$#"
This prints from the first line that matches 'HEADER TEXT' to the first line that matches 'FOOTER TEXT', and then iterates; the '-n' stops the default 'print each line' operation. This won't work well if the header and footer text appear on the same line.
To do what you want, I'd probably use perl (but you could use Python if you prefer). I'd consider slurping the whole file, and then use a suitably qualified regex to find the matching portions of the file. However, the Perl one-liner given by '#gbacon' is an almost exact transliteration into Perl of the 'sed' script above and is neater than slurping.
The man page of grep says:
grep, egrep, fgrep, rgrep - print lines matching a pattern
grep is not made for matching more than a single line. You should try to solve this task with perl or awk.
As this is tagged with 'bbedit' and BBedit supports Perl-Style Pattern Modifiers you can allow the dot to match linebreaks with the switch (?s)
(?s).
will match ANY character. And yes,
(?s).+
will match the whole text.
As pointed elsewhere, grep will work for single line stuff.
For multiple-lines (in ruby with Regexp::MULTILINE, or in python, awk, sed, whatever), "\s" should also capture line breaks, so
HEADER TEXT(.*\s*)FOOTER TEXT
might work ...
here's one way to do it with gawk, if you have it
awk -vRS="FOOTER" '/HEADER/{gsub(/.*HEADER/,"");print}' file