How do I lowercase all MD5 hashes inside of a file? - regex

I'm reposting this question with more context and examples because admittedly my last post was rushed. I'm trying to find ALL MD5 hashes (based off the regex) and simply lowercase them (regardless of format, and regardless of what else is on the line).
I previously posted a similar question related to lowercasing emails and this was the solved answer (maybe this can assist in finding the answer).
sed -e 's/^\([^#]*\)#/\L\1#/' file
The MD5 Regex I use is: [a-f0-9]{32}
I'll provide some examples now (each line is unique - there's NOT a set pattern)
Input data:
fwefwe:few32rfwe:3r2frewg:-::d3ewStack:D077F244DEF8A70E5EA758BD8352FCD8
fwefwe:few33rfwe:3r2frewg:-::dsasaewStack:06D80EB0C50B49A509B49F2424E8C805
fwefwe:few34rfwe:3r2f3213ef::2d3ewStack:F1BDF5ED1D7AD7EDE4E3809BD35644B0
fwefwe:few35rf32re4frewgre3frewg:-::d3ewStack:DDE2C7AD63AD86D6A18DE781205D194F
Output data:
fwefwe:few32rfwe:3r2frewg:-::d3ewStack:d077f244def8a70e5ea758bd8352fcd8
fwefwe:few33rfwe:3r2frewg:-::dsasaewStack:06d80eb0c50b49a509b49f2424e8c805
fwefwe:few34rfwe:3r2f3213ef::2d3ewStack:f1bdf5ed1d7ad7ede4e3809bd35644b0
fwefwe:few35rf32re4frewgre3frewg:-::d3ewStack:dde2c7ad63ad86d6a18de781205d194f

You may use
sed -E 's/[A-F0-9]{32}/\L&/g' file
If you also want to modify the existing file add -i option:
sed -i -E 's/[A-F0-9]{32}/\L&/g' file
The point is that you need to use an -E option to enable POSIX ERE regex syntax and write {...} quantifier without escaping. POSIX BRE equivalent will look like sed -i 's/[A-F0-9]\{32\}/\L&/g' file.
Also, I added g flag to modify all match occurrences on a line.

You do it the same way as you did when lowercasing the email. Use the regexp to match part of the line, then use a back-reference in the replacement to pick that up.
sed -e 's/[a-fA-F0-9]\{32\}/\L&/' file
In the replacement, & is replaced with whatever matched the regexp, and \L lowercases it.

Related

sed regular expression does not work as expected. Differs on pipe and file

I have a string in text file where i want to replace the version number. Quotation marks can vary from ' to ". Also spaces around = can be there and can be not as well:
$data['MODULEXXX_VERSION'] = "1.0.0";
For testing i use
echo "_VERSION'] = \"1.1.1\"" | sed "s/\(_VERSION.*\)[1-9]\.[1-9]\.[1-9]/\11.1.2/"
which works perfectly.
When i change it to search in the file (the file has the same string):
sed "s/\(_VERSION.*\)[1-9]\.[1-9]\.[1-9]/\11.1.2/" -i test.php
, it does not find anything.
After after playing with the search part of regex, i found one more odd thing:
sed "s/\(_VERSION.*\)[1-9]\./\1***/" -i test.php
works and changes the string to $data['MODULEXXX_VERSION'] = "***0.0";, but
sed "s/\(_VERSION.*\)[1-9]\.[1-9]/\1***/" -i test.php
does not find anything anymore. Why?
I am using Ubuntu 17.04 desktop.
Anyone can explain what am I doing wrong? What would be the best command for replacing version numbers in the file for the string $data['MODULEXXX_VERSION'] = "***0.0";?
The main problem is that [1-9] doesn't match the 0s in the version number. You need to use [0-9].
Besides that, you may use the following sed command:
sed -r 's/(.*_VERSION['\''"]]\s*=\s*).*/\1"1.0.1";/' conf.php
This doesn't look at the current value, it simply replaces everything after the =.
I've used -r which enables extended posix regular expressions which makes it a bit simpler to formulate the pattern.
Another, probably cleaner attempt is to store the conf.php as a template like conf.php.tpl and then use a template engine to render the file. Or if you really want to use sed, the file may look like:
$data['FOO_VERSION'] = "FOO_VERSION_TPL";
Then just use:
sed 's/FOO_VERSION_TPL/1.0.1/' conf.php.tpl > conf.php
If there are multiple values to replace:
sed \
-e 's/FOO/BAR/' \
-e 's/HELLO/WORLD/' \
conf.php.tpl > conf.php
But I recommend a template engine instead of sed. That becomes more important when the content of the variables to replace may contain characters special to regular expressions.

Regex command line change format of each line

I have a file that contains lines in a format similar to this...
/data/file.geojson?10,20,30,40
/data/file.geojson?bbox=-5.20751953125,49.05227025601607,3.0322265625,56.46249048388979
/data/file.geojson?bbox=-21.46728515625,45.99696161820381,19.2919921875,58.88194208135912
/data/file.geojson?bbox=-2.8482055664062496,54.38935426009769,-0.300750732421875,55.158473983815306
/data/file.geojson?bbox=-21.46728515625,45.99696161820381,19.2919921875,58.88194208135912
/data/file.geojson?bbox=-21.46728515625,45.99696161820381,19.2919921875,58.88194208135912
I've tried a combination of grep, sed, gawk, and |(pipes) to try and pattern match and then change the format to be more like this...
[10,40],[30,40],[30,20][10,20],
[-5.20751953125,56.46249048388979],[3.0322265625,56.46249048388979].....
Hopefully you get the idea from the first line so I don't have to type out all the examples manually!
I've got the hang of regex to match the co-ordinates. In fact the input file is the result of extracting from apache access logs. It might be easier to read/understand answers if they just match positive integer numbers, I will then be able to slot in a more complicated pattern to match the right range.
To be able to arrange the results like you which it is important to be able to access the last for values per line.
No pattern matching is required if you use awk. You can split the input strings by a set of delimiters and reassemble the resulting fields. 40 can be accessed as $(NF), 30 as $(NF-1) and so on.
awk -F'[?,=]' '
{printf "[%s,%s],[%s,%s],[%s,%s],[%s,%s]\n",
$(NF-3),$(NF),$(NF-1),$(NF),
$(NF-1),$(NF-2),$(NF-3),$(NF-2)
}' file
I'm using ?, , or = as the field delimiters. This makes it simple to access the columns of interest.
Output:
[10,40],[30,40],[30,20],[10,20]
[-5.20751953125,56.46249048388979],[3.0322265625,56.46249048388979],[3.0322265625,49.05227025601607],[-5.20751953125,49.05227025601607]
[-21.46728515625,58.88194208135912],[19.2919921875,58.88194208135912],[19.2919921875,45.99696161820381],[-21.46728515625,45.99696161820381]
[-2.8482055664062496,55.158473983815306],[-0.300750732421875,55.158473983815306],[-0.300750732421875,54.38935426009769],[-2.8482055664062496,54.38935426009769]
[-21.46728515625,58.88194208135912],[19.2919921875,58.88194208135912],[19.2919921875,45.99696161820381],[-21.46728515625,45.99696161820381]
[-21.46728515625,58.88194208135912],[19.2919921875,58.88194208135912],[19.2919921875,45.99696161820381],[-21.46728515625,45.99696161820381]
Btw, also sed can be used here:
sed -r 's/.*[?=]([^,]+),([^,]+),([^,]+),(.*)/[\1,\4],[\3,\4],[\3,\2],[\1,\2]/' file
The command is capturing the numbers at the end each in a separate capturing group and re-assembles them in the replacement part.
Not all versions of sed support the + quantifier. The most compatible version would look like this :)
sed 's/.*[?=]\([^,]\{1,\}\),\([^,]\{1,\}+\),\([^,]\{1,\}\),\(.*\)/[\1,\4],[\3,\4],[\3,\2],[\1,\2]/' file
sed strips off items prior to numbers, then awk splits on comma and outputs in different order. Assuming data is in a file called "td.txt"
sed 's/^[^0-9-]*//' td.txt|awk -F, '{print "["$1","$4"],["$3","$4"],["$3","$2"],["$1","$2"],"}'
This might work for you (GNU sed):
sed -r 's/^.*\?[^-0-9]*([^,]*),([^,]*),([^,]*),([^,]*)/[\1,\4],[\3,\4],[\3,\2],[\1,\2]/' file
Or with more toothpicks:
sed 's/^.*\?[^-0-9]*\([^,]*\),\([^,]*\),\([^,]*\),\([^,]*\)/[\1,\4],[\3,\4],[\3,\2],[\1,\2]/' file
You can use the following to match:
(\/data\/file\.geojson\?(?:bbox=)?)([0-9.-]+),([0-9.-]+),([0-9.-]+),([0-9.-]+)
And replace with the following:
$1[$2,$3],[$4,$5]
See DEMO

sed : match all instances of regex in infile1.txt, and output only these to outfile2.txt

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. :)

Grep/Sed between two tags with multiline

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).

Is there a truly universal wildcard in Grep? [duplicate]

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