matching text in quotes (newbie) - regex

I'm getting totally lost in shell programming, mainly because every site I use offers different tool to do pattern matching. So my question is what tool to use to do simple pattern matching in piped stream.
context: I have named.conf file, and i need all zones names in a simple file for further processing. So I do ~$ cat named.local | grep zone and get totally lost here. My output is ~hundred or so newlines in form 'zone "domain.tld" {' and I need text in double quotes.
Thanks for showing a way to do this.
J

I think what you're looking for is sed... it's a stream editor which will let you do replacements on a line-by-line basis.
As you're explaining it, the command `cat named.local | grep zone' gives you an output a little like this:
zone "domain1.tld" {
zone "domain2.tld" {
zone "domain3.tld" {
zone "domain4.tld" {
I'm guessing you want the output to be something like this, since you said you need the text in double quotes:
"domain1.tld"
"domain2.tld"
"domain3.tld"
"domain4.tld"
So, in reality, from each line we just want the text between the double-quotes (including the double-quotes themselves.)
I'm not sure you're familiar with Regular Expressions, but they are an invaluable tool for any person writing shell scripts. For example, the regular expression /.o.e/ would match any line where there's a word with the 2nd letter was a lower-case o, and the 4th was e. This would match string containing words like "zone", "tone", or even "I am tone-deaf."
The trick there was to use the . (dot) character to mean "any letter". There's a couple of other special characters, such as * which means "repeat the previous character 0 or more times". Thus a regular expression like a* would match "a", "aaaaaaa", or an empty string: ""
So you can match the string inside the quotes using: /".*"/
There's another thing you would know about sed (and by the comments, you already do!) - it allows backtracking. Once you've told it how to recognize a word, you can have it use that word as part of the replacement. For example, let's say that you wanted to turn this list:
Billy "The Kid" Smith
Jimmy "The Fish" Stuart
Chuck "The Man" Norris
Into this list:
The Kid
The Fish
The Man
First, you'd look for the string inside the quotes. We already saw that, it was /".*"/.
Next, we want to use what's inside the quotes. We can group it using parens: /"(.*)"/
If we wanted to replace the text with the quotes with an underscore, we'd do a replace: s/"(.*)"/_/, and that would leave us with:
Billy _ Smith
Jimmy _ Stuart
Chuck _ Norris
But we have backtracking! That'll let us recall what was inside the parens, using the symbol \1. So if we do now: s/"(.*)"/\1/ we'll get:
Billy The Kid Smith
Jimmy The Fish Stuart
Chuck The Man Norris
Because the quotes weren't in the parens, they weren't part of the contents of \1!
To only leave the stuff inside the double-quotes, we need to match the entire line. To do that we have ^ (which means "beginning of line"), and $ (which means "end of line".)
So now if we use s/^.*"(.*)".*$/\1/, we'll get:
The Kid
The Fish
The Man
Why? Let's read the regular expression s/^.*"(.*)".*$/\1/ from left-to-right:
s/ - Start a substitution regular expression
^ - Look for the beginning of the line. Start from there.
.* - Keep going, reading every character, until...
" - ... until you reach a double-quote.
( - start a group a characters we might want to recall later when backtracking.
.* - Keep going, reading every character, until...
) - (pssst! close the group!)
" - ... until you reach a double-quote.
.* - Keep going, reading every character, until...
$ - The end of the line!
/ - use what's after this to replace what you matched
\1 - paste the contents of the first group (what was in the parens) matched.
/ - end of regular expression
In plain English: "Read the entire line, copying aside the text between the double-quotes. Then replace the entire line with the content between the double qoutes."
You can even add double-quote around the replacing text s/^.*"(.*)".*$/"\1"/, so we'll get:
"The Kid"
"The Fish"
"The Man"
And that can be used by sed to replace the line with the content from within the quotes:
sed -e "s/^.*\"\(.*\)\".*$/\"\1\"/"
(This is just shell-escaped to deal with the double-quotes and slashes and stuff.)
So the whole command would be something like:
cat named.local | grep zone | sed -e "s/^.*\"\(.*\)\".*$/\"\1\"/"

Well, nobody mentioned cut yet, so, to prove that there are many ways to do something with the shell:
% grep '^zone' /etc/bind/named.conf | cut -d' ' -f2
"gennic.net"
"generic-nic.net"
"dyn.generic-nic.net"
"langtag.net"

1.
zoul#naima:etc$ cat named.conf | grep zone
zone "." IN {
zone "localhost" IN {
file "localhost.zone";
zone "0.0.127.in-addr.arpa" IN {
2.
zoul#naima:etc$ cat named.conf | grep ^zone
zone "." IN {
zone "localhost" IN {
zone "0.0.127.in-addr.arpa" IN {
3.
zoul#naima:etc$ cat named.conf | grep ^zone | sed 's/.*"\([^"]*\)".*/\1/'
.
localhost
0.0.127.in-addr.arpa
The regexp is .*"\([^"]*\)".*, which matches:
any number of any characters: .*
a quote: "
starts to remember for later: \(
any characters except quote: [^"]*
ends group to remember: \)
closing quote: "
and any number of characters: .*
When calling sed, the syntax is 's/what_to_match/what_to_replace_it_with/'. The single quotes are there to keep your regexp from being expanded by bash. When you “remember” something in the regexp using parens, you can recall it as \1, \2 etc. Fiddle with it for a while.

You should have a look at awk.

As long as someone is pointing out sed/awk, I'm going to point out that grep is redundant.
sed -ne '/^zone/{s/.*"\([^"]*\)".*/\1/;p}' /etc/bind/named.conf
This gives you what you're looking for without the quotes (move the quotes inside the parenthesis to keep them). In awk, it's even simpler with the quotes:
awk '/^zone/{print $2}' /etc/bind/named.conf
I try to avoid pipelines as much as possible (but not more). Remember, Don't pipe cat. It's not needed. And, insomuch as awk and sed duplicating grep's work, don't pipe grep, either. At least, not into sed or awk.
Personally, I'd probably have used perl. But that's because I probably would have done the rest of whatever you're doing in perl, making it a minor detail (and being able to slurp the whole file in and regex against everything simultaneously, ignoring \n's would be a bonus for cases where I don't control /etc/bind, such as on a shared webhost). But, if I were to do it in shell, one of the above two would be the way I'd approach it.

Related

How to replace spaces after a certain pattern with commas?

I am new to coding and I'm trying to format some bioinformatics data. I am trying to remove all the spaces after GT:GL:GOF:GQ:NR:NV with commas, but not anything outside of the format xx:xx:xx:xx:xx (like the example). I know I need to use sed with regex option but I'm not very familiar with how to use it. I've never actually used sed before and got confused trying so any help would be appreciated. Sorry if I formatted this poorly (this is my first post).
EDIT 2: I got actual data from the file this time which may help solve the problem. Removed the bad example.
New Example: I pulled this data from my actual file (this is just two samples), and it is surrounded by other data. Essentially the line has a bunch of data followed by "GT:GL:GOF:GQ:NR:NV ", after this there is more data in the format shown below, and finally there is some more random data. Unfortunately I can't post a full line of the data because it is extremely long and will not fit.
Input
0/1:-1,-1,-1:146:28:14,14:4,0 0/1:-1,-1,-1:134:6:2,2:1,0
Output
0/1:-1,-1,-1:146:28:14,14:4,0,0/1:-1,-1,-1:134:6:2,2:1,0
With Basic Regular Expressions, you can use character classes and backreferences to accomplish your task, e.g.
$ sed 's/\([0-9][0-9]*:[0-9][0-9]*\)[ ]\([0-9][0-9]*:[0-9][0-9]*\)/\1,\2/g' file
1/0 ./. 0/1 GT:GL:GOF:GQ:NR:NV 1:12:314,213:132:13:31,14:31:31 AB GT BB
1/0 ./. 0/1 GT:GL:GOF:GQ:NR:NV 10:13:12,41:41:1:13,13:131:1:1 AB GT RT
1/0 ./. 0/1 GT:GL:GOF:GQ:NR:NV 1:12:314,213:132:13:31,14:31:31 AB GT
Which basically says:
find and capture any [0-9][0-9]* one or more digits,
separated by a :, and
followed by [0-9][0-9]* one or more digits -- as capture group 1,
match a space following capture group 1 followed by capture group 2 (which is the same as capture group 1),
then replace the space separating the capture groups with a comma reinserting the capture group text using backreference 1 and 2 (e.g. \1 and \2), finally
make the replacement global (e.g. g) to replace all matching occurrences.
Edit Based On New Input Posted
If you still need all of the original commas added, and you now want to add a comma between ,0 0/ (where there is a comma precedes a single-digit followed by the space to be replaced with a comma, followed by a single-digit and a forward-slash), then all you need to do is make your capture groups conditional (on either capturing the original data as above -or- capturing this new segment. You do that by including an OR (e.g. \| in basic regex terms) between the conditions.
For instance by adding \|,[0-9] at the end of the first capture group and \|[0-9][/] at the end of the second, e.g.
$ sed 's/\([0-9][0-9]*:[0-9][0-9]*\|,[0-9]\)[ ]\([0-9][0-9]*:[0-9][0-9]*\|[0-9][/]\)/\1,\2/g' file
0/1:-1,-1,-1:146:28:14,14:4,0,0/1:-1,-1,-1:134:6:2,2:1,0
If you have other caveats in your file, I suggest you post several complete lines of input, and if they are too long, then create a zip, gzip, bzip or xz file and post it to a site like pastebin and add the link to your question.
If all you really care about now is the space in ,0 0/, then you can shorten the sed command to:
$ sed 's/\(,[0-9]\)[[:space:]]\([0-9][/]\)/\1,\2/g' file
0/1:-1,-1,-1:146:28:14,14:4,0,0/1:-1,-1,-1:134:6:2,2:1,0
(note: I've included [[:space:]] to handle any whitespace (space, tab, ...) instead of just the literal [ ] (space) in the new example)
Let me know if this fixes the issue.
I'm assuming that the xx:xx:xx or xx:xx:xx:xx can have any number of parts, since some have 3, and some have 4.
This is quite difficult to do reliably with sed, as it does not support lookarounds, which seem like they might be needed for this example.
You can try something like:
perl -pe 's/(?<=\d) (?=\d+(:\d+){2,})/,/g' input.txt
If you've got your heart set on sed, you can try this, but it may miss some cases:
sed -r 's/(:[0-9]+) ([0-9]+:)/\1,\2/g' input.txt
Could you please try following. This will take care of printing those values also which are NOT coming in match of regex. Also we would have made regex mentioned in match a bit shorter by doing it as [0-9]+\.{4} etc since this is tested on old awk so couldn't test it.
awk '
BEGIN{
OFS=","
}
match($0,/GT:GL:GOF:GQ:NR:NV [0-9]+:[0-9]+:[0-9]+:[0-9]+:[0-9]+/){
value=substr($0,RSTART!=1?1:RSTART,RSTART+RLENGTH-1)
value1=substr($0,RSTART+RLENGTH+1)
gsub(/[[:space:]]+/,",",value1)
print value,value1
next
}
1
' Input_file
You may also achieve your desired result without regex, using awk:
awk '{printf "%s", $1FS$2FS$3FS$4FS$5","$6","$7; for (i=8;i<=NF;i++) printf "%s", FS$i; print ""}' input.txt
Basically, it outputs from field 1 to 5 with the default field separator ("space"), then from field 5 to 7 with the comma separator, then from field 8 onwards with default separator again.
perl myscript.pl '0/1:-1,-1,-1:146:28:14,14:4,0 0/1:-1,-1,-1:134:6:2,2:1,0'
myscript.pl,
#!/usr/local/ActivePerl-5.20/bin/env perl
my $input = $ARGV[0];
$input =~ s/ /\,/g;
print $input, "\n";
__DATA__
output
0/1:-1,-1,-1:146:28:14,14:4,0,0/1:-1,-1,-1:134:6:2,2:1,0
This will remove all spaces, not just the space in question

Wrap yaml data in single quotes

I would like to wrap all of my YAML data (in a large file) with single quotes. I tried sed, but it did not work:
sed "s/\(.*: \)\(.*\)/\1'\2'/" <data.yml >datanew.yml
This took lines like this:
location_id: 25
street:
text: This is text: it contains colons
And produced lines like:
' location_id: '25
' street: '
' text: This is text: 'it contains colons
... but I would like them to look like:
location_id: '25'
street: ''
text: 'This is text: it contains colons'
Is this possible in sed (or awk or perl or ...)? From my research, it seems like sed may have trouble picking up the first colon since it matches greedy. I am running Ubuntu 14.04.
Additional Information
Note YAML has optional leading whitespace, a token followed by a colon and everything else on the line (which could include one or more additional colons) all of which needs to be wrapped in quotes.
You can test with the above three lines.
More
Thank you all for your suggestions. I am assuming most of them actually work, but not for me. Here is a snapshot from my terminal using one of the suggested patterns. Unfortunately they all fail for me in roughly the same way.
Even more frustrating, when I open the file in vim and run search and replace with that same pattern, it works perfectly. I tried to use that technique for my whole file, but vim wasn't pleased with the 4M lines.
Is my sed somehow broken??
This regex:
^\s*([^:]+)(:\s)(.*?)\s*$
Does what you want. Working Demo
It is easiest to express in Perl.
Given:
$ echo "$tgt"
location_id: 25
street:
text: This is text: it contains colons
In Perl:
$ echo "$tgt" | perl -lne "print if s/^\s*([^:]+)(:\s)(.*?)\s*$/\1\2'\3'/"
location_id: '25'
street: ''
text: 'This is text: it contains colons'
Little change in your sed
sed "s/\([^:]*: \)\(.*\)/\1'\2'/" <data.yml >datanew.yml
Here is an awk you can use:
cat file
Some other data
location_id: 25
street:
awk -v f="'" -F": *" 'NF==2 {$NF=f $NF f}1' file
Some other data
location_id '25'
street ''
It test if line has :, and if then it wraps ' around last filed, empty or not.
The following seems to work for the test cases you provided, as well as some cases I came up with:
sed "s/\([^:]*:\s*\)\(.*\)/\1'\2'/g"
The way it works is by doing a non-greedy match for text up until a colon, then followed by a colon and optional whitespace with [^:]*:\s* . All of that is put into a capture group. The reason I need the "then followed by a colon" is because some lines such as "%YAML 1.1" which appear in the file would match the regex, even though they should not be included. By adding the additional constraint that there be a colon, such lines are excluded from replacement.
The next part is relatively easy, just match any text following the previous capture group. This an be achieved with .* (which also included colons, as you mentioned above in your question).
The sed s command is used to replace the regex matched with the first capture group,\1, which is all the text up until the first colon and optional whitespace, followed by the 2nd capture group \2 which is all the text after the colon and whitespace, in single quotes.
Here's a demo of it:
regex test

How to substitute a string even it contains regex meta characters using Shell or Perl?

I want to substitue a word which maybe contains regex meta characters to another word, for example, substitue the .Precilla123 as .Precill, I try to use below solution:
sed 's/.Precilla123/.Precill/g'
but it will change below line
"Precilla123";"aaaa aaa";"bbb bbb"
to
.Precill";"aaaa aaa";"bbb bbb"
This side effect is not I wanted. So I try to use:
perl -pe 's/\Q.Precilla123\E/.Precill/g'
The above solution can disable interpreted regex meta characters, it will not have the side effect.
Unfortunately, if the word contains $ or #, this solution still cannot work, because Perl keep $ and # as variable prefix.
Can anybody help this? Many thanks.
Please note that the word I want to substitute is NOT hard coded, it comes from a input file, you can consider it as variable.
Unfortunately, if the word contains $ or #, this solution still cannot work, because Perl keep $ and # as variable prefix.
This is not true.
If the value that you want to replace is in a Perl variable, then quotemeta will work on the variable's contents just fine, including the characters $ and #:
echo 'pre$foo to .$foobar' | perl -pe 'my $from = q{.$foo}; s/\Q$from\E/.to/g'
Outputs:
pre$foo to .tobar
If the words that you want to replace are in an external file, then simply load that data in a BEGIN block before composing your regular expressions for replacement.
sed 's/\.Precilla123/.Precill/g'
Escape the meta character with \.
Be carrefull, mleta charactere are not the same for search pattern that are mainly regex []{}()\.*+^$ where replacement is limited to &\^$ (+ the separator that depend of your first char after the s in both pattern)

Vim Regex Capture Groups [bau -> byau : ceu -> cyeu]

I have a list of words:
bau
ceu
diu
fou
gau
I want to turn that list into:
byau
cyeu
dyiu
fyou
gyau
I unsuccessfully tried the command:
:%s/(\w)(\w\w)/\1y\2/g
Given that this doesn't work, what do I have to change to make the regex capture groups work in Vim?
One way to fix this is by ensuring the pattern is enclosed by escaped parentheses:
:%s/\(\w\)\(\w\w\)/\1y\2/g
Slightly shorter (and more magic-al) is to use \v, meaning that in the pattern after it all ASCII characters except '0'-'9', 'a'-'z', 'A'-'Z' and '_' have a special meaning:
:%s/\v(\w)(\w\w)/\1y\2/g
See:
:help \(
:help \v
You can also use this pattern which is shorter:
:%s/^./&y
%s applies the pattern to the whole file.
^. matches the first character of the line.
&y adds the y after the pattern.
If you don't want to escape the capturing groups with backslashes (this is what you've missed), prepend \v to turn Vim's regular expression engine into very magic mode:
:%s/\v(\w)(\w\w)/\1y\2/g
You also have to escape the Grouping paranthesis:
:%s/\(\w\)\(\w\w\)/\1y\2/g
That does the trick.
In Vim, on a selection, the following
:'<,'>s/^\(\w\+ - \w\+\).*/\1/
or
:'<,'>s/\v^(\w+ - \w+).*/\1/
parses
Space - Commercial - Boeing
to
Space - Commercial
Similarly,
apple - banana - cake - donuts - eggs
is parsed to
apple - banana
Explanation
^ : match start of line
\-escape (, +, ) per the first regex (accepted answer) -- or prepend with \v (#ingo-karkat's answer)
\w\+ finds a word (\w will find the first character): in this example, I search for a word followed by - followed by another word)
.* after the capturing group is needed to find / match / exclude the remaining text
Addendum. This is a bit off topic, but I would suggest that Vim is not well-suited for the execution of more complex regex expressions / captures. [I am doing something similar to the following, which is how I found this thread.]
In those instances, it is likely better to dump the lines to a text file and edit it "in place"
sed -i ...
or in a redirect
sed ... > out.txt
In a terminal (or BASH script, ...):
echo 'Space Sciences - Private Industry - Boeing' | sed -r 's/^((\w+ ){1,2}- (\w+ ){1,2}).*/\1/'
Space Sciences - Private Industry
cat in.txt
Space Sciences - Private Industry - Boeing
sed -r 's/^((\w+ ){1,2}- (\w+ ){1,2}).*/\1/' ~/in.txt > ~/out.txt
cat ~/out.txt
Space Sciences - Private Industry
## Caution: if you forget the > redirect, you'll edit your source.
## Subsequent > redirects also overwrite the output; use >> to append
## subsequent iterations to the output (preserving the previous output).
## To edit "in place" (`-i` argument/flag):
sed -i -r 's/^((\w+ ){1,2}- (\w+ ){1,2}).*/\1/' ~/in.txt
cat in.txt
Space Sciences - Private Industry
sed -r 's/^((\w+ ){1,2}- (\w+ ){1,2}).*/\1/'
(note the {1,2}) allows the flexibility of finding {x,y} repetitions of a word(s) -- see https://www.gnu.org/software/sed/manual/html_node/Regular-Expressions.html .
Here, since my phrases are separated by -, I can simply tweak those parameters to get what I want.

Why doesn't this simple regex match what I think it should?

I have a data file that looks like the following example. I've added '%' in lieu of \t, the tab control character.
1234:56% Alice Worthington
alicew% Jan 1, 2010 10:20:30 AM% Closed% Development
Digg:
Reddit:
Update%% file-one.txt% 1.1% c:/foo/bar/quux
Add%% file-two.txt% 2.5.2% c:/foo/bar/quux
Remove%% file-three.txt% 3.4% c:/bar/quux
Update%% file-four.txt% 4.6.5.3% c:/zzz
... many more records of the above form
The records I'm interested in are the lines beginning with "Update", "Add", "Remove", and so on. I won't know what the lines begin with ahead of time, or how many lines precede them. I do know that they always begin with a string of letters followed by two tabs. So I wrote this regex:
generate-report-for 1234:56 | egrep "^[[:alpha:]]+\t\t.+"
But this matches zero lines. Where did I go wrong?
Edit: I get the same results whether I use '...' or "..." for the egrep expression, so I'm not sure it's a shell thing.
Apparently \t isn't a special character for egrep. You can either use grep -P to enable Perl-compatible regex engine, or insert literal tabs with CtrlvCtrli
Even better, you could use the excellent ack
It looks like the shell is parsing "\t\t" before it is sent to egrep. Try "\\t\\t" or '\t\t' instead. That is 2 slashes in double quotes and one in single quotes.
The file might not be exactly what you see. Maybe there are control characters hidden. It happens, sometimes. My suggestion is that you debug this. First, reduce to the minimum regex pattern that matches, and then keep adding stuff one by one, until you find the problem:
egrep "[[:alpha:]]"
egrep "[[:alpha:]]+"
egrep "[[:alpha:]]+\t"
egrep "[[:alpha:]]+\t\t"
egrep "[[:alpha:]]+\t\t.+"
egrep "^[[:alpha:]]+\t\t.+"
There are variations on that sequence, depending on what you find out at each step. Also, the first step can really be skipped, but this is just for the sake of showing the technique.
you can use awk
awk '/^[[:alpha:]]\t\t/' file