I've scraped a large amount (10GB) of PDFs and converted them to text files, but due to the format of the original PDFs, there is an issue:
Many of the words which break across lines have a dash in them that artificially breaks up the word, like this:
You can see that this happened because the original PDFs files have breaks:
What would be the cleanest and fastest way to "join" every word instance that matches this pattern inside of a .txt file?
Perhaps some sort of Regex search, like for a [a-z]\-\s \w of some kind (word character followed by dash followed by space) would work?
Or would some sort of sed replacement work better?
Currently, I'm trying to get a sed regex to work, but I'm not sure how to translate this to use capture groups to replace the selected text:
sed -n '\%\w\- [a-z]%p' Filename.txt
My input text would look like this:
The dog rolled down the st- eep hill and pl- ayed outside.
And the output would be:
The dog rolled down the steep hill and played outside.
Ideally, the expression would also work for words split up by a newline, like this:
The rule which provided for the consid-
eration of the resolution, was agreed to earlier by a
To this:
The rule which provided for the consideration
of the resolution, was agreed to earlier by a
It's straightforward in sed:
sed -e ':a' -e '/-$/{N;s/-\n//;ba
}' -e 's/- //g' filename
This translates roughly as "if the line ends with a dash, read in the next line as well (so that you have a line with a carriage return in the middle) then excise the dash and carriage return, and loop back the beginning just in case this new line also ends with a dash. Then remove any instances of - ".
You may use this gnu-awk code:
cat file
The dog rolled down the st- eep hill and pl- ayed outside.
The rule which provided for the consid-
eration of the resolution, was agreed to earlier by a
Then use awk like this:
awk 'p != "" {
w = $1
$1 = ""
sub(/^[[:blank:]]+/, ORS)
$0 = p w $0
p = ""
}
{
$0 = gensub(/([_[:alnum:]])-[[:blank:]]+([_[:alnum:]])/, "\\1\\2", "g")
}
/-$/ {
p = $0
sub(/-$/, "", p)
}
p == ""' file
The dog rolled down the steep hill and played outside.
The rule which provided for the consideration
of the resolution, was agreed to earlier by a
If you can consider perl then this may also work for you:
Then use:
perl -0777 -pe 's/(\w)-\h+(\w)/$1$2/g; s/(\w)-\R(\w+)\s+/$1$2\n/g' file
You simply add backslash-parentheses (or use the -r or -E option if available to do away with the requirement to put backslashes before capturing parentheses) and recall the matched text with \1 for the first capturing parenthesis, \2 for the second, etc.
sed 's/\(\w\)\- \([a-z]\)/\1\2/g' Filename.txt
The \w escape is not standard sed but if it works for you, feel free to use it. Otherwise, it is easy to replace with [A-Za-z0-9_#] or whatever else you want to call "word characters".
I'm guessing not all of the matches will be hyphenated words so perhaps run the result through a spelling checker or something to verify whether the result is an English word. (I would probably switch to a more capable scripting language like Python for that, though.)
I'm trying to use sed to clean up lines of URLs to extract just the domain.
So from:
http://www.suepearson.co.uk/product/174/71/3816/
I want:
http://www.suepearson.co.uk/
(either with or without the trailing slash, it doesn't matter)
I have tried:
sed 's|\(http:\/\/.*?\/\).*|\1|'
and (escaping the non-greedy quantifier)
sed 's|\(http:\/\/.*\?\/\).*|\1|'
but I can not seem to get the non-greedy quantifier (?) to work, so it always ends up matching the whole string.
Neither basic nor extended Posix/GNU regex recognizes the non-greedy quantifier; you need a later regex. Fortunately, Perl regex for this context is pretty easy to get:
perl -pe 's|(http://.*?/).*|\1|'
In this specific case, you can get the job done without using a non-greedy regex.
Try this non-greedy regex [^/]* instead of .*?:
sed 's|\(http://[^/]*/\).*|\1|g'
With sed, I usually implement non-greedy search by searching for anything except the separator until the separator :
echo "http://www.suon.co.uk/product/1/7/3/" | sed -n 's;\(http://[^/]*\)/.*;\1;p'
Output:
http://www.suon.co.uk
this is:
don't output -n
search, match pattern, replace and print s/<pattern>/<replace>/p
use ; search command separator instead of / to make it easier to type so s;<pattern>;<replace>;p
remember match between brackets \( ... \), later accessible with \1,\2...
match http://
followed by anything in brackets [], [ab/] would mean either a or b or /
first ^ in [] means not, so followed by anything but the thing in the []
so [^/] means anything except / character
* is to repeat previous group so [^/]* means characters except /.
so far sed -n 's;\(http://[^/]*\) means search and remember http://followed by any characters except / and remember what you've found
we want to search untill the end of domain so stop on the next / so add another / at the end: sed -n 's;\(http://[^/]*\)/' but we want to match the rest of the line after the domain so add .*
now the match remembered in group 1 (\1) is the domain so replace matched line with stuff saved in group \1 and print: sed -n 's;\(http://[^/]*\)/.*;\1;p'
If you want to include backslash after the domain as well, then add one more backslash in the group to remember:
echo "http://www.suon.co.uk/product/1/7/3/" | sed -n 's;\(http://[^/]*/\).*;\1;p'
output:
http://www.suon.co.uk/
Simulating lazy (un-greedy) quantifier in sed
And all other regex flavors!
Finding first occurrence of an expression:
POSIX ERE (using -r option)
Regex:
(EXPRESSION).*|.
Sed:
sed -r 's/(EXPRESSION).*|./\1/g' # Global `g` modifier should be on
Example (finding first sequence of digits) Live demo:
$ sed -r 's/([0-9]+).*|./\1/g' <<< 'foo 12 bar 34'
12
How does it work?
This regex benefits from an alternation |. At each position engine tries to pick the longest match (this is a POSIX standard which is followed by couple of other engines as well) which means it goes with . until a match is found for ([0-9]+).*. But order is important too.
Since global flag is set, engine tries to continue matching character by character up to the end of input string or our target. As soon as the first and only capturing group of left side of alternation is matched (EXPRESSION) rest of line is consumed immediately as well .*. We now hold our value in the first capturing group.
POSIX BRE
Regex:
\(\(\(EXPRESSION\).*\)*.\)*
Sed:
sed 's/\(\(\(EXPRESSION\).*\)*.\)*/\3/'
Example (finding first sequence of digits):
$ sed 's/\(\(\([0-9]\{1,\}\).*\)*.\)*/\3/' <<< 'foo 12 bar 34'
12
This one is like ERE version but with no alternation involved. That's all. At each single position engine tries to match a digit.
If it is found, other following digits are consumed and captured and the rest of line is matched immediately otherwise since * means
more or zero it skips over second capturing group \(\([0-9]\{1,\}\).*\)* and arrives at a dot . to match a single character and this process continues.
Finding first occurrence of a delimited expression:
This approach will match the very first occurrence of a string that is delimited. We can call it a block of string.
sed 's/\(END-DELIMITER-EXPRESSION\).*/\1/; \
s/\(\(START-DELIMITER-EXPRESSION.*\)*.\)*/\1/g'
Input string:
foobar start block #1 end barfoo start block #2 end
-EDE: end
-SDE: start
$ sed 's/\(end\).*/\1/; s/\(\(start.*\)*.\)*/\1/g'
Output:
start block #1 end
First regex \(end\).* matches and captures first end delimiter end and substitues all match with recent captured characters which
is the end delimiter. At this stage our output is: foobar start block #1 end.
Then the result is passed to second regex \(\(start.*\)*.\)* that is same as POSIX BRE version above. It matches a single character
if start delimiter start is not matched otherwise it matches and captures the start delimiter and matches the rest of characters.
Directly answering your question
Using approach #2 (delimited expression) you should select two appropriate expressions:
EDE: [^:/]\/
SDE: http:
Usage:
$ sed 's/\([^:/]\/\).*/\1/g; s/\(\(http:.*\)*.\)*/\1/' <<< 'http://www.suepearson.co.uk/product/174/71/3816/'
Output:
http://www.suepearson.co.uk/
Note: this will not work with identical delimiters.
sed does not support "non greedy" operator.
You have to use "[]" operator to exclude "/" from match.
sed 's,\(http://[^/]*\)/.*,\1,'
P.S. there is no need to backslash "/".
sed - non greedy matching by Christoph Sieghart
The trick to get non greedy matching in sed is to match all characters excluding the one that terminates the match. I know, a no-brainer, but I wasted precious minutes on it and shell scripts should be, after all, quick and easy. So in case somebody else might need it:
Greedy matching
% echo "<b>foo</b>bar" | sed 's/<.*>//g'
bar
Non greedy matching
% echo "<b>foo</b>bar" | sed 's/<[^>]*>//g'
foobar
Non-greedy solution for more than a single character
This thread is really old but I assume people still needs it.
Lets say you want to kill everything till the very first occurrence of HELLO. You cannot say [^HELLO]...
So a nice solution involves two steps, assuming that you can spare a unique word that you are not expecting in the input, say top_sekrit.
In this case we can:
s/HELLO/top_sekrit/ #will only replace the very first occurrence
s/.*top_sekrit// #kill everything till end of the first HELLO
Of course, with a simpler input you could use a smaller word, or maybe even a single character.
HTH!
This can be done using cut:
echo "http://www.suepearson.co.uk/product/174/71/3816/" | cut -d'/' -f1-3
another way, not using regex, is to use fields/delimiter method eg
string="http://www.suepearson.co.uk/product/174/71/3816/"
echo $string | awk -F"/" '{print $1,$2,$3}' OFS="/"
sed certainly has its place but this not not one of them !
As Dee has pointed out: Just use cut. It is far simpler and much more safe in this case. Here's an example where we extract various components from the URL using Bash syntax:
url="http://www.suepearson.co.uk/product/174/71/3816/"
protocol=$(echo "$url" | cut -d':' -f1)
host=$(echo "$url" | cut -d'/' -f3)
urlhost=$(echo "$url" | cut -d'/' -f1-3)
urlpath=$(echo "$url" | cut -d'/' -f4-)
gives you:
protocol = "http"
host = "www.suepearson.co.uk"
urlhost = "http://www.suepearson.co.uk"
urlpath = "product/174/71/3816/"
As you can see this is a lot more flexible approach.
(all credit to Dee)
sed 's|(http:\/\/[^\/]+\/).*|\1|'
There is still hope to solve this using pure (GNU) sed. Despite this is not a generic solution in some cases you can use "loops" to eliminate all the unnecessary parts of the string like this:
sed -r -e ":loop" -e 's|(http://.+)/.*|\1|' -e "t loop"
-r: Use extended regex (for + and unescaped parenthesis)
":loop": Define a new label named "loop"
-e: add commands to sed
"t loop": Jump back to label "loop" if there was a successful substitution
The only problem here is it will also cut the last separator character ('/'), but if you really need it you can still simply put it back after the "loop" finished, just append this additional command at the end of the previous command line:
-e "s,$,/,"
sed -E interprets regular expressions as extended (modern) regular expressions
Update: -E on MacOS X, -r in GNU sed.
Because you specifically stated you're trying to use sed (instead of perl, cut, etc.), try grouping. This circumvents the non-greedy identifier potentially not being recognized. The first group is the protocol (i.e. 'http://', 'https://', 'tcp://', etc). The second group is the domain:
echo "http://www.suon.co.uk/product/1/7/3/" | sed "s|^\(.*//\)\([^/]*\).*$|\1\2|"
If you're not familiar with grouping, start here.
I realize this is an old entry, but someone may find it useful.
As the full domain name may not exceed a total length of 253 characters replace .* with .\{1, 255\}
This is how to robustly do non-greedy matching of multi-character strings using sed. Lets say you want to change every foo...bar to <foo...bar> so for example this input:
$ cat file
ABC foo DEF bar GHI foo KLM bar NOP foo QRS bar TUV
should become this output:
ABC <foo DEF bar> GHI <foo KLM bar> NOP <foo QRS bar> TUV
To do that you convert foo and bar to individual characters and then use the negation of those characters between them:
$ sed 's/#/#A/g; s/{/#B/g; s/}/#C/g; s/foo/{/g; s/bar/}/g; s/{[^{}]*}/<&>/g; s/}/bar/g; s/{/foo/g; s/#C/}/g; s/#B/{/g; s/#A/#/g' file
ABC <foo DEF bar> GHI <foo KLM bar> NOP <foo QRS bar> TUV
In the above:
s/#/#A/g; s/{/#B/g; s/}/#C/g is converting { and } to placeholder strings that cannot exist in the input so those chars then are available to convert foo and bar to.
s/foo/{/g; s/bar/}/g is converting foo and bar to { and } respectively
s/{[^{}]*}/<&>/g is performing the op we want - converting foo...bar to <foo...bar>
s/}/bar/g; s/{/foo/g is converting { and } back to foo and bar.
s/#C/}/g; s/#B/{/g; s/#A/#/g is converting the placeholder strings back to their original characters.
Note that the above does not rely on any particular string not being present in the input as it manufactures such strings in the first step, nor does it care which occurrence of any particular regexp you want to match since you can use {[^{}]*} as many times as necessary in the expression to isolate the actual match you want and/or with seds numeric match operator, e.g. to only replace the 2nd occurrence:
$ sed 's/#/#A/g; s/{/#B/g; s/}/#C/g; s/foo/{/g; s/bar/}/g; s/{[^{}]*}/<&>/2; s/}/bar/g; s/{/foo/g; s/#C/}/g; s/#B/{/g; s/#A/#/g' file
ABC foo DEF bar GHI <foo KLM bar> NOP foo QRS bar TUV
Have not yet seen this answer, so here's how you can do this with vi or vim:
vi -c '%s/\(http:\/\/.\{-}\/\).*/\1/ge | wq' file &>/dev/null
This runs the vi :%s substitution globally (the trailing g), refrains from raising an error if the pattern is not found (e), then saves the resulting changes to disk and quits. The &>/dev/null prevents the GUI from briefly flashing on screen, which can be annoying.
I like using vi sometimes for super complicated regexes, because (1) perl is dead dying, (2) vim has a very advanced regex engine, and (3) I'm already intimately familiar with vi regexes in my day-to-day usage editing documents.
Since PCRE is also tagged here, we could use GNU grep by using non-lazy match in regex .*? which will match first nearest match opposite of .*(which is really greedy and goes till last occurrence of match).
grep -oP '^http[s]?:\/\/.*?/' Input_file
Explanation: using grep's oP options here where -P is responsible for enabling PCRE regex here. In main program of grep mentioning regex which is matching starting http/https followed by :// till next occurrence of / since we have used .*? it will look for first / after (http/https://). It will print matched part only in line.
echo "/home/one/two/three/myfile.txt" | sed 's|\(.*\)/.*|\1|'
don bother, i got it on another forum :)
sed 's|\(http:\/\/www\.[a-z.0-9]*\/\).*|\1| works too
Here is something you can do with a two step approach and awk:
A=http://www.suepearson.co.uk/product/174/71/3816/
echo $A|awk '
{
var=gensub(///,"||",3,$0) ;
sub(/\|\|.*/,"",var);
print var
}'
Output:
http://www.suepearson.co.uk
Hope that helps!
Another sed version:
sed 's|/[:alnum:].*||' file.txt
It matches / followed by an alphanumeric character (so not another forward slash) as well as the rest of characters till the end of the line. Afterwards it replaces it with nothing (ie. deletes it.)
#Daniel H (concerning your comment on andcoz' answer, although long time ago): deleting trailing zeros works with
s,([[:digit:]]\.[[:digit:]]*[1-9])[0]*$,\1,g
it's about clearly defining the matching conditions ...
You should also think about the case where there is no matching delims. Do you want to output the line or not. My examples here do not output anything if there is no match.
You need prefix up to 3rd /, so select two times string of any length not containing / and following / and then string of any length not containing / and then match / following any string and then print selection. This idea works with any single char delims.
echo http://www.suepearson.co.uk/product/174/71/3816/ | \
sed -nr 's,(([^/]*/){2}[^/]*)/.*,\1,p'
Using sed commands you can do fast prefix dropping or delim selection, like:
echo 'aaa #cee: { "foo":" #cee: " }' | \
sed -r 't x;s/ #cee: /\n/;D;:x'
This is lot faster than eating char at a time.
Jump to label if successful match previously. Add \n at / before 1st delim. Remove up to first \n. If \n was added, jump to end and print.
If there is start and end delims, it is just easy to remove end delims until you reach the nth-2 element you want and then do D trick, remove after end delim, jump to delete if no match, remove before start delim and and print. This only works if start/end delims occur in pairs.
echo 'foobar start block #1 end barfoo start block #2 end bazfoo start block #3 end goo start block #4 end faa' | \
sed -r 't x;s/end//;s/end/\n/;D;:x;s/(end).*/\1/;T y;s/.*(start)/\1/;p;:y;d'
If you have access to gnu grep, then can utilize perl regex:
grep -Po '^https?://([^/]+)(?=)' <<< 'http://www.suepearson.co.uk/product/174/71/3816/'
http://www.suepearson.co.uk
Alternatively, to get everything after the domain use
grep -Po '^https?://([^/]+)\K.*' <<< 'http://www.suepearson.co.uk/product/174/71/3816/'
/product/174/71/3816/
The following solution works for matching / working with multiply present (chained; tandem; compound) HTML or other tags. For example, I wanted to edit HTML code to remove <span> tags, that appeared in tandem.
Issue: regular sed regex expressions greedily matched over all the tags from the first to the last.
Solution: non-greedy pattern matching (per discussions elsewhere in this thread; e.g. https://stackoverflow.com/a/46719361/1904943).
Example:
echo '<span>Will</span>This <span>remove</span>will <span>this.</span>remain.' | \
sed 's/<span>[^>]*>//g' ; echo
This will remain.
Explanation:
s/<span> : find <span>
[^>] : followed by anything that is not >
*> : until you find >
//g : replace any such strings present with nothing.
Addendum
I was trying to clean up URLs, but I was running into difficulty matching / excluding a word - href - using the approach above. I briefly looked at negative lookarounds (Regular expression to match a line that doesn't contain a word) but that approach seemed overly complex and did not provide a satisfactory solution.
I decided to replace href with ` (backtick), do the regex substitutions, then replace ` with href.
Example (formatted here for readability):
printf '\n
<a aaa h href="apple">apple</a>
<a bbb "c=ccc" href="banana">banana</a>
<a class="gtm-content-click"
data-vars-link-text="nope"
data-vars-click-url="https://blablabla"
data-vars-event-category="story"
data-vars-sub-category="story"
data-vars-item="in_content_link"
data-vars-link-text
href="https:example.com">Example.com</a>\n\n' |
sed 's/href/`/g ;
s/<a[^`]*`/\n<a href/g'
apple
banana
Example.com
Explanation: basically as above. Here,
s/href/` : replace href with ` (backtick)
s/<a : find start of URL
[^`] : followed by anything that is not ` (backtick)
*` : until you find a `
/<a href/g : replace each of those found with <a href
Unfortunately, as mentioned, this it is not supported in sed.
To overcome this, I suggest to use the next best thing(actually better even), to use vim sed-like capabilities.
define in .bash-profile
vimdo() { vim $2 --not-a-term -c "$1" -es +"w >> /dev/stdout" -cq! ; }
That will create headless vim to execute a command.
Now you can do for example:
echo $PATH | vimdo "%s_\c:[a-zA-Z0-9\\/]\{-}python[a-zA-Z0-9\\/]\{-}:__g" -
to filter out python in $PATH.
Use - to have input from pipe in vimdo.
While most of the syntax is the same. Vim features more advanced features, and using \{-} is standard for non-greedy match. see help regexp.
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
I have a problem that because of PHP version, I need to change my code from $array[stringindex] to $array['stringindex'];
So I want to find all the text using regex, and replace them all. How to find all strings that look like this? $array[stringindex].
Here's a solution in PHP:
$re = "/(\\$[[:alpha:]][[:alnum:]]+\\[)([[:alpha:]][[:alnum:]]+)(\\])/";
$str = "here is \$array[stringindex] but not \$array['stringindex'] nor \$3array[stringindex] nor \$array[4stringindex]";
$subst = "$1'$2'$3";
$result = preg_replace($re, $subst, $str);
You can try it out interactively here. I search for variables beginning with a letter, otherwise things like $foo[42] would be converted to $foo['42'], which might not be desirable.
Note that all the solutions here will not handle every case correctly.
Looking at the Sublime Text regex help, it would seem you could just paste (\\$[[:alpha:]][[:alnum:]]+\\[)([[:alpha:]][[:alnum:]]+)(\\]) into the Search box and $1'$2'$3 into the Replace field.
It depends of the tool you want to use to do the replacement.
with sed for exemple, it would be something like that:
sed "s/\(\$array\)\[\([^]]*\)\]/\1['\2']/g"
If sed is allowed you could simply do:
sed -i "s/(\$[^[]*[)([^]]*)]/\1'\2']/g" file
Explanation:
sed "s/pattern/replace/g" is a sed command which searches for pattern and replaces it with replace. The g options means replace multiple times per line.
(\$[^[]*[)([^]]*)] this pattern consists of two groups (in between brackets). The first is a dollar followed by a series of non [ chars. Then an opening square bracket follows, followed by a series of non closing brackets which is then followed by a closing square bracket.
\1'\2'] the replacement string: \1 means insert the first captured group (analogous for \2. Basically we wrap \2 in quotes (which is what you wanted).
the -i options means that the changes should be applied to the original file, which is supplied at the end.
For more information, see man sed.
This can be combined with the find command, as follows:
find . -name '*.php' -exec sed -i "s/(\$[^[]*[)([^]]*)]/\1'\2']/g" '{}' \;
This will apply the sed command to all php files found.
I'm trying to grep a code base to find alpha numeric codes between quotes. So, for example my code base might contain the line
some stuff "A234DG3" maybe more stuff
And I'd like to output: A234DG3
I'm lucky in that I know my string is 7 long and only integers and the letters A-Z, a-z.
After a bit of playing I've come up with the following, but it's just not coming out with what I'd like
grep -ro '".*"' . | grep [A-Za-z0-9]{7} | less
Where am I going wrong here? It feels like grep should give me what I want, but am I better off using something else? Cheers!
The problem is that an RE is pretty much required to match the longest sequence it can. So, given something like:
a "bcd" efg "hij" klm "nop" q
A pattern of ".*" should match: "bcd" efg "hij" klm "nop" (everything from the first quote to the last quote), not just "bcd".
You probably want a pattern more like "[^"]*" to match the open-quote, an arbitrary number of other things, then a close quote.
Using basic or extended POSIX regular expressions there is no way to extract the value between the quotes with grep. Since that I would use sed for a portable solution:
sed -n 's/.*\"\([^"]\+\)".*/\1/p' <<< 'some stuff "A234DG3" maybe more stuff'
However, having GNU goodies, GNU grep will support PCRE expressions with the -P command line option. You can use this:
grep -oP '.*?"\K[^"]+(?=")' <<< 'some stuff "A234DG3" maybe more stuff'
.*" matches everything until the first quote - including it. The \K option clears the matching buffer and therefore works like a handy, dynamic lookbehind assertion. (I could have used a real lookbehind but I like \K). [^"]+ matches the text between the quotes. (?=") is a lookahead assertion the ensure after the match will follow a " - without including it into the match.
So after more playing about I've come up with this which gives me what I'm after:
grep -r -E -o '"[A-Za-z0-9]{7}"' . | less
With the -E allowing the use of the {7} length matcher