I have a directory containing log files.
We are interested in a particular log line which goes like 'xxxxxxxxx|platform=SUN|.......|orderId=ABCDEG|........'
We have to extract all similar lines from the log files in this directory,and print out the token 'ABCDEG'.
Duplication is acceptable.
How do we achieve this with a single unix command operation?
sed -r '/platform=.*orderId=/s/.*orderId=([^|]+).*/\1/g' *
From all lines containing platform= && orderId= (/platform=.*orderId=/), take the non-| sequence of characters (([^|]+))after orderId=.
awk -F'|' '$2=="platform=SUN"{sub(/orderId=/,"", $4); print $4}' logFile*
output
ABCDEG
IHTH
grep -rP "\|platform=SUN\|.*(?<=\|orderId=)" | sed s/.*platform=SUN.*orderId=// | sed s/\|.*//
$ str='xxxxxxxxx|platform=SUN|.......|orderId=ABCDEG|........'
$ grep -Po 'platform=SUN.*orderId=\K[^|]*' <<< "$str"
ABCDEG
This requires Perl compatible regular expressions (-P); -o retains just the match. \K is variable length look-behind: "match the stuff to the left of it, but don't include it in the matched string".
From the logs directory you could run the following command:
sed -n /platform=SUN/p * | sed 's#.*orderId=\(.*\)|.*$#\1#'
Related
i'm trying to extract a specific string from a grep output
uci show minidlna
produces a large list
.
.
.
minidlna.config.enabled='1'
minidlna.config.db_dir='/mnt/sda1/usb/db'
minidlna.config.enable_tivo='1'
minidlna.config.wide_links='1'
.
.
.
so i tried to narrow down what i wanted by running
uci show minidlna | grep -oE '\bdb_dir=\S+'
this narrows the output to
db_dir='/mnt/sda1/usb/db'
what i want is to output only
/mnt/sda1/usb/db
without the quotes and without the starting "db_dir" so i can run rm /mnt/sda1/usb/db/file.db
i've used the answers found here
How to extract string following a pattern with grep, regex or perl
and that's as close as i got.
EDIT: after using Ed Morton's awk command i needed to pass the output to rm command.
i used:
| ( read DB; (rm $DB/files.db) .
read DB passes the output into the vairable DB.
(...) combines commands.
rm $DB/files.db deletes the the file files.db.
Is this what you're trying to do?
$ awk -F"'" '/db_dir/{print $2}' file
/mnt/sda1/usb/db
That will work in any awk in any shell on every UNIX box.
If that's not what you want then edit your question to clarify your requirements and post more truly representative sample input/output.
Using sed with some effort to avoid single quotes:
sed -n 's/^minidlna.config.db_dir=\s*\S\(\S*\)\S\s*$/\1/p' input
Well, so you end up having a string like db_dir='/mnt/sda1/usb/db'.
I would first remove the quotes by piping this to
.... | tr -d "'"
Now you end up with a string like db_dir=/mnt/sda1/usb/db.
Say you have this string stored in a variable named confstr, then
${confstr##*=}
gives you just /mnt/sda1/usb/db, since *= denotes everything from the start to the equal sign, and ## denotes removal.
I would do this:
Once you either extracted your line about into file.txt (or pipe it into this command), split the fields using the quote character. Use printf to generate the rm command and pass this into bash to execute.
$ awk -F"'" '{printf "rm %s.db/file.db\n", $2}' file.txt | bash
rm: /mnt/sda1/usb/db.db/file.db: No such file or directory
With your original command:
$ uci show minidlna | grep -oE '\bdb_dir=\S+' | \
awk -F"'" '{printf "rm %s.db/file.db\n", $2}' | bash
I am working on shell script and new to it. I want to extract the string between double $$ characters, for example:
input:
$$extractabc$$
output
extractabc
I used grep and sed but not working out. Any suggestions are welcome!
You could do
awk -F"$" '{print $3}' file.txt
assuming the file contained input:$$extractabc$$ output:extractabc. awk splits your data into pieces using $ as a delimiter. First item will be input:, next will be empty, next will be extractabc.
You could use sed like so to get the same info.
sed -e 's/.*$$\(.*\)$$.*/\1/' file.txt
sed looks for information between $$s and outputs that. The goal is to type something like this .*$$(.*)$$.*. It's greedy but just stay with me.
looks for .* - i.e. any character zero or more times before $$
then the string should have $$
after $$ there'll be any character zero or more times
then the string should have another $$
and some more characters to follow
between the 2 $$ is (.*). String found between $$s is given a placeholder \1
sed finds such information and publishes it
Using grep PCRE (where available) and look-around:
$ echo '$$extractabc$$' | grep -oP "(?<=\\$\\$).*(?=\\$\\$)"
extractabc
echo '$$extractabc$$' | awk '{gsub(/\$\$/,"")}1'
extractabc
Here is an other variation:
echo "$$extractabc$$" | awk -F"$$" 'NF==3 {print $2}'
It does test of there are two set of $$ and only then prints whats between $$
Does also work for input like blabla$$some_data$$moreblabla
How about remove all the $ in the input?
$ echo '$$extractabc$$' | sed 's/\$//g'
extractabc
Same with tr
$ echo '$$extractabc$$' | tr -d '$'
extractabc
How would I use sed to delete all lines in a text file that contain a specific string?
To remove the line and print the output to standard out:
sed '/pattern to match/d' ./infile
To directly modify the file – does not work with BSD sed:
sed -i '/pattern to match/d' ./infile
Same, but for BSD sed (Mac OS X and FreeBSD) – does not work with GNU sed:
sed -i '' '/pattern to match/d' ./infile
To directly modify the file (and create a backup) – works with BSD and GNU sed:
sed -i.bak '/pattern to match/d' ./infile
There are many other ways to delete lines with specific string besides sed:
AWK
awk '!/pattern/' file > temp && mv temp file
Ruby (1.9+)
ruby -i.bak -ne 'print if not /test/' file
Perl
perl -ni.bak -e "print unless /pattern/" file
Shell (bash 3.2 and later)
while read -r line
do
[[ ! $line =~ pattern ]] && echo "$line"
done <file > o
mv o file
GNU grep
grep -v "pattern" file > temp && mv temp file
And of course sed (printing the inverse is faster than actual deletion):
sed -n '/pattern/!p' file
You can use sed to replace lines in place in a file. However, it seems to be much slower than using grep for the inverse into a second file and then moving the second file over the original.
e.g.
sed -i '/pattern/d' filename
or
grep -v "pattern" filename > filename2; mv filename2 filename
The first command takes 3 times longer on my machine anyway.
The easy way to do it, with GNU sed:
sed --in-place '/some string here/d' yourfile
You may consider using ex (which is a standard Unix command-based editor):
ex +g/match/d -cwq file
where:
+ executes given Ex command (man ex), same as -c which executes wq (write and quit)
g/match/d - Ex command to delete lines with given match, see: Power of g
The above example is a POSIX-compliant method for in-place editing a file as per this post at Unix.SE and POSIX specifications for ex.
The difference with sed is that:
sed is a Stream EDitor, not a file editor.BashFAQ
Unless you enjoy unportable code, I/O overhead and some other bad side effects. So basically some parameters (such as in-place/-i) are non-standard FreeBSD extensions and may not be available on other operating systems.
I was struggling with this on Mac. Plus, I needed to do it using variable replacement.
So I used:
sed -i '' "/$pattern/d" $file
where $file is the file where deletion is needed and $pattern is the pattern to be matched for deletion.
I picked the '' from this comment.
The thing to note here is use of double quotes in "/$pattern/d". Variable won't work when we use single quotes.
You can also use this:
grep -v 'pattern' filename
Here -v will print only other than your pattern (that means invert match).
To get a inplace like result with grep you can do this:
echo "$(grep -v "pattern" filename)" >filename
I have made a small benchmark with a file which contains approximately 345 000 lines. The way with grep seems to be around 15 times faster than the sed method in this case.
I have tried both with and without the setting LC_ALL=C, it does not seem change the timings significantly. The search string (CDGA_00004.pdbqt.gz.tar) is somewhere in the middle of the file.
Here are the commands and the timings:
time sed -i "/CDGA_00004.pdbqt.gz.tar/d" /tmp/input.txt
real 0m0.711s
user 0m0.179s
sys 0m0.530s
time perl -ni -e 'print unless /CDGA_00004.pdbqt.gz.tar/' /tmp/input.txt
real 0m0.105s
user 0m0.088s
sys 0m0.016s
time (grep -v CDGA_00004.pdbqt.gz.tar /tmp/input.txt > /tmp/input.tmp; mv /tmp/input.tmp /tmp/input.txt )
real 0m0.046s
user 0m0.014s
sys 0m0.019s
Delete lines from all files that match the match
grep -rl 'text_to_search' . | xargs sed -i '/text_to_search/d'
SED:
'/James\|John/d'
-n '/James\|John/!p'
AWK:
'!/James|John/'
/James|John/ {next;} {print}
GREP:
-v 'James\|John'
perl -i -nle'/regexp/||print' file1 file2 file3
perl -i.bk -nle'/regexp/||print' file1 file2 file3
The first command edits the file(s) inplace (-i).
The second command does the same thing but keeps a copy or backup of the original file(s) by adding .bk to the file names (.bk can be changed to anything).
You can also delete a range of lines in a file.
For example to delete stored procedures in a SQL file.
sed '/CREATE PROCEDURE.*/,/END ;/d' sqllines.sql
This will remove all lines between CREATE PROCEDURE and END ;.
I have cleaned up many sql files withe this sed command.
echo -e "/thing_to_delete\ndd\033:x\n" | vim file_to_edit.txt
Just in case someone wants to do it for exact matches of strings, you can use the -w flag in grep - w for whole. That is, for example if you want to delete the lines that have number 11, but keep the lines with number 111:
-bash-4.1$ head file
1
11
111
-bash-4.1$ grep -v "11" file
1
-bash-4.1$ grep -w -v "11" file
1
111
It also works with the -f flag if you want to exclude several exact patterns at once. If "blacklist" is a file with several patterns on each line that you want to delete from "file":
grep -w -v -f blacklist file
to show the treated text in console
cat filename | sed '/text to remove/d'
to save treated text into a file
cat filename | sed '/text to remove/d' > newfile
to append treated text info an existing file
cat filename | sed '/text to remove/d' >> newfile
to treat already treated text, in this case remove more lines of what has been removed
cat filename | sed '/text to remove/d' | sed '/remove this too/d' | more
the | more will show text in chunks of one page at a time.
Curiously enough, the accepted answer does not actually answer the question directly. The question asks about using sed to replace a string, but the answer seems to presuppose knowledge of how to convert an arbitrary string into a regex.
Many programming language libraries have a function to perform such a transformation, e.g.
python: re.escape(STRING)
ruby: Regexp.escape(STRING)
java: Pattern.quote(STRING)
But how to do it on the command line?
Since this is a sed-oriented question, one approach would be to use sed itself:
sed 's/\([\[/({.*+^$?]\)/\\\1/g'
So given an arbitrary string $STRING we could write something like:
re=$(sed 's/\([\[({.*+^$?]\)/\\\1/g' <<< "$STRING")
sed "/$re/d" FILE
or as a one-liner:
sed "/$(sed 's/\([\[/({.*+^$?]\)/\\\1/g' <<< "$STRING")/d"
with variations as described elsewhere on this page.
cat filename | grep -v "pattern" > filename.1
mv filename.1 filename
You can use good old ed to edit a file in a similar fashion to the answer that uses ex. The big difference in this case is that ed takes its commands via standard input, not as command line arguments like ex can. When using it in a script, the usual way to accomodate this is to use printf to pipe commands to it:
printf "%s\n" "g/pattern/d" w | ed -s filename
or with a heredoc:
ed -s filename <<EOF
g/pattern/d
w
EOF
This solution is for doing the same operation on multiple file.
for file in *.txt; do grep -v "Matching Text" $file > temp_file.txt; mv temp_file.txt $file; done
I found most of the answers not useful for me, If you use vim I found this very easy and straightforward:
:g/<pattern>/d
Source
I get the following result in my script when I run it against my services. The result differs depending on the service but the text pattern showing below is similar. The result of my script is assigned to var1. I need to extract data from this variable
$var1=HOST1*prod*gem.dot*serviceList : svc1 HOST1*prod*kem.dot*serviceList : svc3, svc4 HOST1*prod*fen.dot*serviceList : svc5, svc6
I need to strip the name of the service list from $var1. So the end result should be printed on separate line as follow:
svc1
svc2
svc3
svc4
svc5
svc6
Can you please help with this?
Regards
Using sed and grep:
sed 's/[^ ]* :\|,\|//g' <<< "$var1" | grep -o '[^ ]*'
sed deletes every non-whitespace before a colon and commas. Grep just outputs the resulting services one per line.
Using gnu grep and gnu sed:
grep -oP ': *\K\w+(, \w+)?' <<< "$var1" | sed 's/, /\n/'
svc1
svc3
svc4
svc5
svc6
grep is the perfect tool for the job.
From man grep:
-o, --only-matching
Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.
Sounds perfect!
As far as I'm aware this will work on any grep:
echo "$var1" | grep -o 'svc[0-9]\+'
Matches "svc" followed by one or more digits. You can also enable the "highly experimental" Perl regexp mode with -P, which means you can use the \d digit character class and don't have to escape the + any more:
grep -Po 'svc\d+' <<<"$var1"
In bash you can use <<< (a Here String) which supplies "$var1" to grep on the standard input.
By the way, if your data was originally on separate lines, like:
HOST1*prod*gem.dot*serviceList : svc1
HOST1*prod*kem.dot*serviceList : svc3, svc4
HOST1*prod*fen.dot*serviceList : svc5, svc6
This would be a good job for awk:
awk -F': ' '{split($2,a,", "); for (i in a) print a[i]}'
I have a file as show below
1.2.3.4.ask
sanma.nam.sam
c.d.b.test
I want to remove the last field from each line, the delimiter is . and the number of fields are not constant.
Can anybody help me with an awk or sed to find out the solution. I can't use perl here.
Both these sed and awk solutions work independent of the number of fields.
Using sed:
$ sed -r 's/(.*)\..*/\1/' file
1.2.3.4
sanma.nam
c.d.b
Note: -r is the flag for extended regexp, it could be -E so check with man sed. If your version of sed doesn't have a flag for this then just escape the brackets:
sed 's/\(.*\)\..*/\1/' file
1.2.3.4
sanma.nam
c.d.b
The sed solution is doing a greedy match up to the last . and capturing everything before it, it replaces the whole line with only the matched part (n-1 fields). Use the -i option if you want the changes to be stored back to the files.
Using awk:
$ awk 'BEGIN{FS=OFS="."}{NF--; print}' file
1.2.3.4
sanma.nam
c.d.b
The awk solution just simply prints n-1 fields, to store the changes back to the file use redirection:
$ awk 'BEGIN{FS=OFS="."}{NF--; print}' file > tmp && mv tmp file
Reverse, cut, reverse back.
rev file | cut -d. -f2- | rev >newfile
Or, replace from last dot to end with nothing:
sed 's/\.[^.]*$//' file >newfile
The regex [^.] matches one character which is not dot (or newline). You need to exclude the dot because the repetition operator * is "greedy"; it will select the leftmost, longest possible match.
With cut on the reversed string
cat youFile | rev |cut -d "." -f 2- | rev
If you want to keep the "." use below:
awk '{gsub(/[^\.]*$/,"");print}' your_file