I have a simple egrep command searching for multiple strings in a text file which outputs either null or a value. Below is the command and the output.
cat Output.txt|egrep -i "abc|def|efg"|cut -d ':' -f 2
Output is:-
xxx
(null)
yyy
Now, i am trying to prefix my search texts to the output like below.
abc:xxx
def:
efg:yyy
Any help on the code to achieve this or where to start would be appreciated.
-Abhi
Since I do not know exactly your input file content (not specified properly in the question), I will put some hypothesis in order to answer your question.
Case 1: the patterns you are looking for are always located in the same column
If it is the case, the answer is quite straightforward:
$ cat grep_file.in
abc:xxx:uvw
def:::
efg:yyy:toto
xyz:lol:hey
$ egrep -i "abc|def|efg" grep_file.in | cut -d':' -f1,2
abc:xxx
def:
efg:yyy
After the grep just use the cut with the 2 columns that you are looking for (here it is 1 and 2)
REMARK:
Do not cat the file, pipe it and then grep it, since this is doing the work twice!!! Your grep command will already read the file so do not read it twice, it might not be that important on small files but you will feel the difference on 10GB files for example!
Case 2: the patterns you are looking for are NOT located in the same column
In this case it is a bit more tricky, but not impossible. There are many ways of doing, here I will detail the awk way:
$ cat grep_file2.in
abc:xxx:uvw
::def:
efg:yyy:toto
xyz:lol:hey
If your input file is in this format; with your pattern that could be located anywhere:
$ awk 'BEGIN{FS=":";ORS=FS}{tmp=0;for(i=1;i<=NF;i++){tmp=match($i,/abc|def|efg/);if(tmp){print $i;break}}if(tmp){printf "%s\n", $2}}' grep_file
2.in
abc:xxx
def:
efg:yyy
Explanations:
FS=":";ORS=FS define your input/output field separator at : Then on each line you define a test variable that will become true when you reach your pattern, you loop on all the fields of the line until you reach it if it is the case you print it, break the loop and print the second field + an EOL char.
If you do not meet your pattern you do nothing.
If you prefer the sed way, you can use the following command:
$ sed -n '/abc\|def\|efg/{h;s/.*\(abc\|def\|efg\).*/\1:/;x;s/^[^:]*:\([^:]*\):.*/\1/;H;x;s/\n//p}' grep_file2.in
abc:xxx
def:
efg:yyy
Explanations:
/abc\|def\|efg/{} is used to filter the lines that contain only one of the patterns provided, then you execute the instructions in the block. h;s/.*\(abc\|def\|efg\).*/\1:/; save the line in the hold space and replace the line with one of the 3 patterns, x;s/^[^:]*:\([^:]*\):.*/\1/; is used to exchange the pattern and hold space and extract the 2nd column element. Last but not least, H;x;s/\n//p is used to regroup both extracted elements on 1 line and print it.
try this
$ egrep -io "(abc|def|efg):[^:]*" file
will print the match and the next token after delimiter.
If we can assume that there are only two fields, that abc etc will always match in the first field, and that getting the last match on a line which contains multiple matches is acceptable, a very simple sed script could work.
sed -n 's/^[^:]*\(abc\|def\|efg\)[^:]*:\([^:]*\)/\1:\2/p' file
If other but similar conditions apply (e.g. there are three fields or more but we don't care about matches in the first two) the required modifications are trivial. If not, you really need to clarify your question.
Related
I was wondering if it was possible to use the sed command to find a range between 2 patterns (in this case, dates) and output these lines in the range to a new file.
Right now, I am just looking at one file and getting lines within my time range of the file FileMoverTransfer.log. However, after a certain time period, these logs are moved to new log files with a suffix such as FileMoverTransfer.log-20180404-xxxxxx.gz. Here is my current code:
sed -n '/^'$start_date'/,/^'$end_date'/p;/^'$end_date'/q' FileMoverTransfer.log >> /public/FileMoverRoot/logs/intervalFMT.log
While this doesn't work, as sed isn't able to look through all of the files in the directory starting with FileMoverTransfer.log?
sed -n '/^'$start_date'/,/^'$end_date'/p;/^'$end_date'/q' FileMoverTransfer.log* >> /public/FileMoverRoot/logs/intervalFMT.log
Any help would be greatly appreciated. Thanks!
The range operator only operates within a single file, so you can't use it if the start is in one file and the end is in another file.
You can use cat to concatenate all the files, and pipe this to sed:
cat FileMoverTransfer.log* | sed -n "/^$start_date/,/^$end_date/p;/^$end_date/q" >> /public/FileMoverRoot/logs/intervalFMT.log
And instead of quoting and unquoting the sed command, you can use double quotes so that the variables will be expanded inside it. This will also prevent problems if the variables contain whitespace.
awk solution
As the OP confirmed that an awk solution would be acceptable, I post it.
(gunzip -c FileMoverTransfer.log-*.gz; cat FileMoverTransfer.log ) \
|awk -v st="$start_date" -v en="$end_date" '$1>=st&&$1<=en{print;next}$1>en{exit}'\
>/public/FileMoverRoot/logs/intervalFMT.log
This solution is functionally almost identical to Barmar’s sed solution, with the difference that his solution, like the OP’s, will print and quit at the first record matching the end date, while mine will print all lines matching the end date and quit at the first record past the end date, without printing it.
Some remarks:
The OP didn't specify the date format. I suppose it is a format compatible with ordinary string order, otherwise some conversion function should be used.
The files FileMoverTransfer.log-*.gz must be named in such a way that their alphabetical ordering corresponds to the chronological order (which is probably the case.)
I suppose that the dates are separated from the rest of the line by whitespace. If they aren’t, you have to supply the -F option to awk. E.g., if the dates are separated by -, you must write awk -F- ...
awk is much faster than sed in this case, because awk simply looks for the separator (whitespace or whatever was supplied with -F) while sed performs a regexp match.
There is no concept of range in my code, only date comparison. The only place where I suppose that the lines are ordered is when I say $1>en{exit}, that is exit when a line is newer than the end date. If you remove that final pattern and its action, the code will run through the whole input, but you could drop the requirement that the files be ordered.
I'm having a hard time getting a grasp of using grep for a class i am in was hoping someone could help guide me in this assignment. The Assignment is as follows.
Using grep print all 5 letter lower case words from the linux dictionary that have a single letter duplicated one time (aabbe or ababe not valid because both a and b are in the word twice). Next to that print the duplicated letter followed buy the non-duplicated letters in alphabetically ascending order.
The Teacher noted that we will need to use several (6) grep statements (piping the results to the next grep) and a sed statement (String Editor) to reformat the final set of words, then pipe them into a read loop where you tear apart the three non-dup letters and sort them.
Sample Output:
aback a bck
abaft a bft
abase a bes
abash a bhs
abask a bks
abate a bet
I haven't figured out how to do more then printing 5 character words,
grep "^.....$" /usr/share/dict/words |
Didn't check it thoroughly, but this might work
tr '[:upper:]' '[:lower:]' | egrep -x '[a-z]{5}' | sed -r 's/^(.*)(.)(.*)\2(.*)$/\2 \1\3\4/' | grep " " | egrep -v "(.).*\1"
But do your way because someone might see it here.
All in one sed
sed -n '
# filter 5 letter word
/[a-zA-Z]\{5\}/ {
# lower letters
y/ABCDEFGHIJKLMNOPQRSTUVWXYZ/abcdefghijklmnopqrstuvwxya/
# filter non single double letter
/\(.\).*\1/ !b
/\(.\).*\(.\).*\1.*\1/ b
/\(.\).*\(.\).*\1.*\2/ b
/\(.\).*\(.\).*\2.*\1/ b
# extract peer and single
s/\(.\)*\(.\)\(.*\)\2\(.*\)/a & \2:\1\3\4/
# sort singles
:sort
s/:\([^a]*\)a\(.*\)$/:\1\2a/
y/abcdefghijklmnopqrstuvwxyz/zabcdefghijklmnopqrstuvwxy/
/^a/ !b sort
# clean and print
s/..//
s/:/ /p
}' YourFile
posix sed so --posix on GNU sed
The first bit, obviously, is to use grep to get it down to just the words that have a single duplication in. I will give you some clues on how to do that.
The key is to use backreferences, which allow you to specify that something that matched a previous expression should appear again. So if you write
grep -E "^(.)...\1...\1$"
then you'll get all the words that have the starting letter reappearing in fifth and ninth positions. The point of the brackets is to allow you to refer later to whatever matched the thing in brackets; you do that with a \1 (to match the thing in the first lot of brackets).
You want to say that there should be a duplicate anywhere in the word, which is slightly more complicated, but not much. You want a character in brackets, then any number of characters, then the repeated character (with no ^ or $ specified).
That will also include ones where there are two or more duplicates, so the next stage is to filter them out. You can do that by a grep -v invocation. Once you've got your list of 5-character words that have at least one duplicate, pipe them through a grep -v call that strips out anything with two (or more) duplicates in. That'll have a (.), and another (.), and a \1, and a \2, and these might appear in several different orders.
You'll also need to strip out anything that has a (.) and a \1 and another \1, since that will have a letter with three occurrences.
That should be enough to get you started, at any rate.
Your next step should be to find the 5-letter words containing a duplicate letter. To do that, you will need to use back-references. Example:
grep "[a-z]*\([a-z]\)[a-z]*\$1[a-z]*"
The $1 picks up the contents of the first parenthesized group and expects to match that group again. In this case, it matches a single letter. See: http://www.thegeekstuff.com/2011/01/advanced-regular-expressions-in-grep-command-with-10-examples--part-ii/ for more description of this capability.
You will next need to filter out those cases that have either a letter repeated 3 times or a word with 2 letters repeated. You will need to use the same sort of back-reference trick, but you can use grep -v to filter the results.
sed can be used for the final display. Grep will merely allow you to construct the correct lines to consider.
Note that the dictionary contains capital letters and also non-letter characters, plus that strange characters used in Southern Europe. say "è".
If you want to distinguish "A" and "a", it's automatic, on the other hand if "A" and "a" are the same letter, in ALL grep invocations you must use the -i option, to instruct grep to ignore case.
Next, you always want to pass the -E option, to avoid the so called backslashitis gravis in the regexp that you want to pass to grep.
Further, if you want to exclude the lines matching a regexp from the output, the correct option is -v.
Eventually, if you want to specify many different regexes to a single grep invocation, this is the way (just an example btw)
grep -E -i -v -e 'regexp_1' -e 'regexp_2' ... -e 'regexp_n'
The preliminaries are after us, let's look forward, use the answer from chiastic-security as a reference to understand the procedings
There are only these possibilities to find a duplicate in a 5 character string
(.)\1
(.).\1
(.)..\1
(.)...\1
grep -E -i -e 'regexp_1' ...
Now you have all the doubles, but this doesn't exclude triples etc that are identified by the following patterns (Edit added a cople of additional matching triples patterns)
(.)\1\1
(.).\1\1
(.)\1.\1
(.)..\1\1
(.).\1.\1
(.)\1\1\1
(.).\1\1\1
(.)\1\1\1\1\
you want to exclude these patterns, so grep -E -i -v -e 'regexp_1' ...
at his point, you have a list of words with at least a couple of the same character, and no triples, etc and you want to drop double doubles, these are the regexes that match double doubles
(.)(.)\1\2
(.)(.)\2\1
(.).(.)\1\2
(.).(.)\2\1
(.)(.).\1\2
(.)(.).\2\1
(.)(.)\1.\2
(.)(.)\2.\1
and you want to exclude the lines with these patterns, so its grep -E -i -v ...
A final hint, to play with my answer copy a few hundred lines of the dictionary in your working directory, head -n 3000 /usr/share/dict/words | tail -n 300 > ./300words so that you can really understand what you're doing, avoiding to be overwhelmed by the volume of the output.
And yes, this is not a complete answer, but it is maybe too much, isn't it?
I have thousand of files in a directory and each file contains numbers of defined variables starting with keyword DEFINE and ending with a semicolon (;), I want to copy all the occurrences of the data between this keyword(Inclusive) into a target file.
Example: Below is the content of the text file:
/* This code is for lookup */
DEFINE variable as a1 expr= extract (n123f1 using brach, code);
END.
Now from the above content i just want to copy the section starting with DEFINE and ending with ; into a target file i.e. the output should be:
DEFINE variable as a1 expr= extract (n123f1 using brach, code);
this needs to done for thousands of scripts and multiple occurences, Please help out.
Thanks a lot , the provided code works, but to a limited extent only when the whole sentence is in a single line but the data is not supposed to be in one single line it is spread in multiple line like below:
/* This code is for lookup */
DEFINE variable as a1 expr= if branchno > 55
then
extract (n123f1 using brach, code)
else
branchno = null
;
END.
The code is also in the above fashion i need to capture all the data between DEFINE and semicolon (;) after every define there will be an ending semicolon ;, this is the pattern.
It sounds like you want grep(1):
grep '^DEFINE.*;$' input > output
Try using grep. Let's say you have files with extension .txt in present directory,
grep -ho 'DEFINE.*;' *.txt > outfile
Output:
DEFINE variable as a1 expr= extract (n123f1 using brach, code);
Short Description
-o will give you only matching string rather than whole line, if line also contains something else and want to ommit it.
-h will suppress file names before matching result
Read man page of grep by typing man grep on your terminal
EDIT
If you want capability to search in multiple lines, you can use pcregrep with -M option
pcregrep -M 'DEFINE.*?(\n|.)*?;' *.txt > outfile
Works fine on my system. Check man pcregrep for more details
Reference : SO Question
One can make a simple solution using sed with version :
sed -n -e '/^DEFINE/{:a p;/;$/!{n;ba}}' your-file
Option -n prevents sed from printing every line; then each time a line begins with DEFINE, print the line (command p) then enter a loop: until you find a line ending with ;, grab the next line and loop to the print command. When exiting the loop, you do nothing.
It looks a bit dirty; it seems that the version sed15 has a shorter (and more straightforward) way to achieve this in one line:
sed -n -e '/^DEFINE/,/;$/p' your-file
Indeed, only for this version of sed, both patterns are treated; for other versions of sed like mine under cygwin, the range patterns must be on separate lines to work properly.
One last thing to remember: it does not treat inclusive patterned ranges, i.e. it stops printing after the first encountered end-pattern even if multiple start patterns have been matched. Prefer something with awk if this is a feature you are looking for.
I am using the following command in order to extract domain names & the full domain extension from a file. Ex: www.abc.yahoo.com, www.efg.yahoo.com.us.
[a-z0-9\-]+\.com(\.[a-z]{2})?' source.txt | sort | uniq | sed -e 's/www.//'
> dest.txt
The command write correctly when I specify small maximum parameter -m 100 after the source.txt. The problem if I didn't specify, or if I specified a huge number. Although, I could write to files with grep (not egrep) before with huge numbers similar to what I'm trying now and that was successful. I also check the last modified date and time during the command being executed, and it seems there is no modification happening in the destination file. What could be the problem ?
As I mentioned in your earlier question, it's probably not an issue with egrep, but that your file is too big and that sort won't output anything (to uniq) until egrep is done. I suggested that you split the files into manageable chucks using the split command. Something like this:
split -l 10000000 source.txt split_source.
This will split the source.txt file into 10 million line chunks called split_source.a, split_source.b, split_source.c etc. And you can run the entire command on each one of those files (and maybe changing the pipe to append at the end: >> dest.txt).
The problem here is that you can get duplicates across multiple files, so at the end you may need to run
sort dest.txt | uniq > dest_uniq.txt
Your question is missing information.
That aside, a few thoughts. First, to debug and isolate your problem:
Run the egrep <params> | less so you can see what egreps doing, and eliminate any problem from sort, uniq, or sed (my bets on sort).
How big is your input? Any chance sort is dying from too much input?
Gonna need to see the full command to make further comments.
Second, to improve your script:
You may want to sort | uniq AFTER sed, otherwise you could end up with duplicates in your result set, AND an unsorted result set. Maybe that's what you want.
Consider wrapping your regular expressions with "^...$", if it's appropriate to establish beginning of line (^) and end of line ($) anchors. Otherwise you'll be matching portions in the middle of a line.
I'm having trouble with data manipulation in a txt file. My file currently looks like this:
HG02239 -23.42333333
NA06985NA06985 -20.125
NA06991NA06991 -20.92
This shows some of my tab-delimited data. Half the entries are in the correct seven-characters (letterletternumbernumbernumbernumbernumber) format, but some are doubled up. I want to go into the second column (first column is empty for a reason!) and remove the repeats in the string so it would read
HG02239 -23.42333333
NA06985 -20.125
NA06991 -20.92
I can't work out how to do this with sed/awk on a per column basis. I feel like I should be able to write a regex, but because the data is a repeat, I don't want to lose the first half of the string; and I can't work out how to cut on a specific column, or I would just delete the 7th character. Any help much appreciated!
Solution
You can solve this with a backreference. For example, using GNU sed:
$ cat << EOF | sed --regexp-extended 's/(.{7})\1/\1/'
HG02239 -23.42333333
NA06985NA06985 -20.125
NA06991NA06991 -20.92
EOF
HG02239 -23.42333333
NA06985 -20.125
NA06991 -20.92
If you aren't using GNU sed, you may need to escape the capture groups. In addition, you can tune the regular expression if you need a more accurate character match.
Explanation
The cat pipeline is just a here-document to make it easy to display and test the code. You can call sed directly on your file, or use the -i flag to perform an in-place edit when you're comfortable with the results.
The sed script does the following:
It stores any group of 7 consecutive characters in a capture group using an "interval expression" (the number in the curly braces).
The \1 is a backreference that matches the first capture group.
The match looks for "a capture group followed by a copy of the capture group."
The substitution replaces the match with a single copy of the capture group.
One way, using awk:
awk '{ print substr($1, 1, 7), $2 }' file.txt
Output:
HG02239 -23.42333333
NA06985 -20.125
NA06991 -20.92
You could use something like that:
sed -i 's|\([A-Z]\{2\}[0-9]\{5\}\)[A-Z0-9]*\s*\(.*\)|\1 \2|g' <your-file>