String:"gamma021AH00999NAK41"
last two digit may vary.It may be 3 digit 4 digit ,etc...
"NAK" in the given string can be any other string but it contains only characters.
So my intention is to extract last numbers(example 41 in the given string) until first character.
Thanks in advance
Using only shell builtins (no external commands like sed or awk, thus much faster if you're going to be repeating this over and over, f/e, once per line):
s=gamma021AH00999NAK41
result=${s##*[[:alpha:]]}
echo "$result"
${var##pattern} is a parameter expansion which removes the longest possible match for pattern from the front of the value of var before returning it. *[[:alpha:]], as a wildcard followed by an alpha character, will thus remove everything before the K in your string.
You can replace all the alphabetic characters by for example "#" and then take the last field based on the "#" separator:
echo "gamma021AH00999NAK41" | sed "s/[aA-zZ]/#/g" | awk -F'#' '{print $NF}'
NOTE: This won't work if you have other than alphanumeric symbols in your string.
EDIT: Only without awk (Thanks #CharlesDuffy):
echo "gamma021AH00999NAK41" | awk -F'[[:alpha:]]' '{print $NF}'
I see no mention of varying length, so this command will work:
echo "gamma021AH00999NAK41" | cut -b '19-'
Answer : 41
I have a lines like these:
ORIGINAL
sometext1 sometext2 word:A12 B34 C56 sometext3 sometext4
sometext5 sometext6 word:A123 B45 C67 sometext7 sometext8
sometext9 sometext10 anotherword:(someword1 someword2 someword3) sometext11 sometext12
EDITED
asdjfkklj lkdsjfic kdiw:A12 B34 C56 lksjdfioe sldkjflkjd
lknal niewoc kdiw:A123 B45 C678 oknes lkwid
cnqule nkdal anotherword:(kdlklks inlqok mncvmnx) unqieo lksdnf
Desired output:
asdjfkklj lkdsjfic kdiw:A12-B34-C56 lksjdfioe sldkjflkjd
lknal niewoc kdiw:A123-B45-C678 oknes lkwid
cnqule nkdal anotherword:(kdlklks-inlqok-mncvmnx) unqieo lksdnf
EDITED: Would this be more explicit? But frankly this is much more difficult to read and answer than writing sometext#. I do not know people's preference.
I only want to replace the whitespaces with dashes after A alphabet letter followed by some digits AND replace the whitespaces with dashes between the words between the two parentheses. And not any other whitespaces in the line. Would appreciate an explanation of the syntax too.
Thanks!
This might work for you (GNU sed):
sed -r ':a;s/(A[0-9]+(-[A-Z][0-9]+)*) ([A-Z][0-9]+)/\1-\3/;ta;s/(\(\S+(-\S+)*) (\S+( \S+)*\))/\1-\3/;ta' file
Iteratively replace the space(s) in the required strings using a regexp and back references.
This code work good
darby#Debian:~/Scrivania$ cat test.txt | sed -r 's#\s+([A-Z][0-9]+)#-\1#g' | sed ':l s/\(([^ )]*\)[ ]/\1-/;tl'
asdjfkklj lkdsjfic kdiw:A12-B34-C56 lksjdfioe sldkjflkjd
lknal niewoc kdiw:A123-B45-C678 oknes lkwid
cnqule nkdal anotherword:(kdlklks-inlqok-mncvmnx) unqieo lksdnf
Explain my regex
In the first regex
Options
-r Enable regex extended
Pattern
\s+ One or more space characters
([A-Z][0-9]+) Submatch a uppercase letter and one or more digits
Replace
- Dash character
\1 Previous submatch
Note
The g after delimiters ///g is for global substitution.
In the second regex
Pattern
:l label branched to by t or b
tl jump to label if any substitution has been made on the pattern space since the most recent reading of input line or execution of command 't'. If label is not specified, then jump to the end of the script. This is a conditional branch
\(([^ )]*\) match all in round brackets and stop to first space found
[ ] one space character
Replace
\1 Previous submatch
- Add a dash
You need capture the first Alphanumeric group using () and the second group. Then you can simply replace all using backreferences \1 and \2 :
using sed twice
sed -E 's/(\b[A-Za-z][0-9]+) ([A-Z])/\1-\2/g' | sed -E 's/(\b[A-Za-z][0-9]+) ([A-Z])/\1-\2/g'
or using perl (with lookahead (?=...)the regex don't capture the 2nd group)
perl -pe 's/(\b[A-Za-z][0-9]+) (?=[A-Z])/\1-/g'
\b work boundary
[A-Za-z] 1 letter
[0-9]+ 1 or more digits
sed doesn't support lookahead and lookbehind fonctionality
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.
My issue is changing a part of a date in mm/dd/yyyy to mm/dd/2016 or for learning purposes let's say mm/dd/yyyy to mm/02/yyyy.
In my file I'm going to cat in:
05/06/1989
05/06/2001
01/03/2015
Using sed to replace that file, I am running commands such as:
sed 's|[0-9][0-9]/[0-9][0-9]/[0-9][0-9]|[0-9][0-9]/[0-9][0-9]/2016|g'
This printed out the exact same thing.
So then I tried maybe changing the year by doing:
sed 's/[0-9][0-9][0-9][0-9]/2016/g'
but this didn't do anything either.
Using sed only:
echo "01/03/2015" | sed -e 's|\([0-9][0-9]\)/\([0-9][0-9]\)/\([0-9]\{4\}\)|year \3 month \1 day \2|'
When you need to skip the first 12 fields, you can use cut -F, -13- or use
echo "1,2,3,4,5,6,7,8,9,10,11,12,01/03/2015" | sed -e 's|\([^,]*,\)\{12\}\([0-9][0-9]\)/\([0-9][0-9]\)/\([0-9]\{4\}\)|year \4 month \2 day \3|'
Explanation:
You can mark matches with the construction using (something_to_match) to mark a match and a number to show what is marked. The () and numbers are all treated special, so they all need to be escaped with backslashes.
sed 's/\(match1\)......\(match2\)/and now \1 and \2/'
When you write it this way, don't forget that the characters between the matches should match too (the dots are actualy wildcards for one char each).
[0-9][0-9] you understand, but you can also say repeat [0-9] two (or four) times. Give the number in curly brackets, and the brackets are special so escape them.
When you want to use the curly brackets more often, the first line can be changed in
echo "01/03/2015" | sed -e 's|\([0-9]\{2\}\)/\([0-9]{2\}\)/\([0-9]\{4\}\)|year \3 month \1 day \2|'
Parsing the csv is easy with cut. Using the sed solution is just a challenge for learning sed better. What is that \([^,]*,\)?
Yes you are right, the \(\) is for matching the stuff in between. I want to match one field followed by a ,. How can you say you want to match a string without a , ? You use the negatioin ^ in the character class [,], so [^,] will match any character except the ,. Only once.
Using [^,]* will match a string without a ,.
The second , in \([^,]*,\) is ... just a plain ,.
The complete match is a the first field followed by a ,.
Now match the first 12 csv fields with {12}, but do not forget the backslashes.
In input string has slashes, so use another character like the | (you already found that):
sed 's|from|to|'
# or everything filled in
sed -e 's|\([^,]*,\)\{12\}\([0-9][0-9]\)/\([0-9][0-9]\)/\([0-9]\{4\}\)|year \4 month \2 day \3|'
^^^^^^^^^^ ^^^^ ^^^^^^^ ^^^^^^^^ ^^^^^^ ^^^ ^^^ ^^
field+, repeat month day year recall recall recall
Sed is simple, but I switch over to Perl when the regular expressions are more complicated. In the example above, converting 01/03/2015 to 01/02/2015:
echo 01/03/2015 | perl -pe 's/([0-9][0-9])\/([0-9][0-9])\/([0-9]{4})/\1\/02\/\3/'
There are three regular expression backreferences: month, day and year. I added back month and year, and changed the day to 02. The '{4}' in the regular expression means that there must be four matches.
If you just want to change the year, it would be:
echo 01/03/2015 | perl -pe 's/([0-9][0-9])\/([0-9][0-9])\/([0-9]{4})/\1\/\2\/2016/'