Working on the file an need to grep the line with a time stamp in the H:MM:SS format. I tried the following egrep '[0-9]\:[0-9]\:[0-9]'. Didn't work for me. What am i doing wrong in regex?
$ date -u | egrep '\d{1,2}:\d{1,2}:\d{1,2}'
Fri May 2 00:59:47 UTC 2014
Try a site like http://regexpal.com/
Here is the fix:
grep '[0-9]:[0-9][0-9]:[0-9][0-9]'
If you need get timestamp only, and your grep is gnu grep.
grep -o '[0-9]:[0-9][0-9]:[0-9][0-9]'
and if you work more harder, limit on time format only:
grep '[0-2][0-9]:[0-5][0-9]:[0-5][0-9]'
Simplest way that I know of:
grep -E '([0-9]{2}:){2}[0-9]{2}' file
If you need month and day also:
grep -E '.{3,4} .{,2} ([0-9]{2}:){2}[0-9]{2}' file
Related
I want to grep for files containing the words in different line from same log. the words are checkCredit?msisdn=766117506 and creditLimit
The log file is like this
freeMemory 103709392time Mon Mar 12 04:02:13 IST 2018
http://127.0.0.1:8087/DialogProxy/services/ServiceProxy/checkCredit?msisdn=767807544&transid=45390124
freeMemory 73117016time Mon Mar 12 04:02:14 IST 2018
statusCode200
{statusCode=200, response=outstandingAmount 0.0 creditLimit 0.0, errorResponse=, responseTime=0}
this is balnce 0.0
What is the best way to do this?
Give this a try:
grep 'checkCredit?msisdn\|creditLimit' inputfile
You can use
$ grep -e 'checkCredit?msisdn=766117506' -e 'creditLimit' <filename>
Simply try
grep creditLimit log.txt | grep checkCredit
grep 'checkCredit?msisdn=766117506\|creditLimit' inputfile catalina.out_bckp , when i run this command it also display only the creditLimit details why it doesn't show the checkCredit?msisdn=766117506 details.
The log file as shown in your question post does not contain the 766117506, so it's no wonder that grep doesn't find it. If you really have data with 766117506, add them to the question.
I have used this command grep 'checkCredit?msisdn\|this is balnce' catalina.out_bckp |awk '$4 < 10 {print ;}' it gave me good result but there is some missing values.
You haven't used creditLimit in that pattern, so it's no wonder that those lines are missing.
I'm trying to get the creation date of a file using console commands. I've managed to use this:
stat -c "%w" file
which outputs the date and time:
2015-05-01 09:33:22.503525000 -0400
But I'd like to extract only the date (without the time) from the result. Is there a way to do this using only stat, or do I have to separate the result using another command? And what might that command be?
Try this:
stat -c "%w" file | cut -d" " -f1
The output of stat will be piped into cut and the string will be cut at the first concurrence of a white space.
Assuming you are using bash:
date -d "#$(stat -c %W file)" +%F
Will do the trick. %W on stat returns the epoch date, date +%F converts it to a YYYY-MM-DD format.
I had to implement some new features on an very old awk script and now want to implement some unit tests to check if my script breaks things. I used diff to check if the script output is different from the whished output:
awk -f mygenerator.awk test.1.gen | diff - test.1.out -q
if [ $? -ne 0 ]; then
echo "test failed"
fi
But now i have some files that generate a dynamic content like a timestamp of the generation date, which causes diff to fail because obviously the timestamp will be different.
My first though was to remove the corresponding lines with grep and test the two "clean" files. then check by egrep if the line is a timestamp.
is there any better way to do this? It should all be done by common unix tools in a bash script due to compatibility reasons.
You could use sed with regular expressions.
If your output is like Fri Feb 21 22:53:54 UTC 2014 from the date command, use:
regex_timestamp="s/([A-Z]{1}[a-z]{2} [A-Z]{1}[a-z]{2} [0-9]{2} [0-9]{2}\:[0-9]{2}\:[0-9]{2} [A-Z]{3} [0-9]{4})//g";
awk -f mygenerator.awk test.1.gen | diff <(sed -r "$regex_timestamp" -) <(sed -r "$regex_timestamp" test.1.out) -q
If you're trying to filter a unix timestamp, simply use this as regex:
s/([0-9]{10})//g
Please note that the latter replaces any group of numbers the same size as a unix timestamp. What format is your timestamp?
I usually use sed to replace the timestamp with XXXXXX, so I can still compare the other information on the same line.
date | \
sed 's/\(Sun\|Mon\|Tue\|Wed\|Thu\|Fri\|Sat\) \(Jan\|Feb\|Mar\|Apr\|May\|Jun\|Jul\|Aug\|Sep\|Oct\|Nov\|Dec\) \?[0-9]\+ [0-9][0-9]:[0-9][0-9]:[0-9][0-9] [A-Z]\+ [0-9]\{4\}/XXXXXX/'
I'm trying to put together a bash/sh script that gets the UTC time of the last commit from a svn repo (the other VCSs are easy)
I understand that i can just do svn propget --revprop -rHEAD svn:date and get it rather easily, but there is no guarantee that the svn checkout will be online, so I'd prefer an offline version, if possible.
Maybe something to do with getting the UTC time from svn info? (by screwing with the timezones)
Summary: How can i get the UTC time of a svn commit, while not having access to the server?
Thanks
You can use svn log -r COMMITTED and extract date info from it. This is valid for offline copies.
svn log -r COMMITTED | sed -nE 's/^r.*\| ([0-9]{4}-[0-9]{2}-[0-9]{2} \S+ \S+).*/\1/p' | xargs -i -- date -ud '{}' '+%s'
The -u option makes date show UTC time instead.
Actually we don't need to use xargs:
date -ud "$(exec svn log -r COMMITTED | sed -nE 's/^r.*\| ([0-9]{4}-[0-9]{2}-[0-9]{2} \S+ \S+).*/\1/p')" '+%s'
UPDATE: I got the wrong command. The command above won't work offline. Here's the right one:
date -ud "$(svn info | sed -nE 's/^Last Changed Date: (\S+ \S+ \S+).*/\1/p')" '+%s'
I'm silly. As soon as i actually realised i just need to convert one timezone to another, it was obvious:
date -u +[format] -d $(svn info | <some grepping and cutting here>)
In my case, this is:
date -u +"%Y%m%d-%H%M" -d "$(svn info | grep 'Date' | cut -d' ' -f4-6)"
Of course, my solution probably isn't optimal, and if someone knows a better way, that'd be much appreciated :)
It turns out that the xml output of "svn info" has a zulu timestamp:
$ svn info --xml | grep date
<date>2015-04-30T15:38:49.371762Z</date>
So your bash command might be:
svn info --xml | grep -oP '(?<=<date>).*?(?=</date>)'
I just stumbled on this post. Ended up figuring out that svn uses env variable TZ, so for example:
TZ= svn log
will log dates in UTC
Centos / Linux
Bash
I have a log file, which has lots of text in and epoch numbers all over the place. I want to replace all epochs whereever they are into readable date/time.
I've been wanting to this via sed, as that seems the tool for the job. I can't seem to get the replacement part of sed to actually parse the variable(epoch) to it for conversion.
Sample of what I'm working with...
echo "Some stuff 1346474454 And not working" \
| sed 's/1[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]/'"`bpdbm -ctime \&`"'/g'
Some stuff 0 = Thu Jan 1 01:00:00 1970 And not working
The bpdbm part will convert a supplied epoch variable into useful date. Like this..
bpdbm -ctime 1346474454
1346474454 = Sat Sep 1 05:40:54 2012
So how do i get the "found" item to be parsed into a command. As i don't seem to be able to get it to work.
Any help would be lovely. If there is another way, that would be cool...but i suspect sed will be quickest.
Thanks for your time!
that seems the tool for the job
No, it is not. sed can use & only itself, there is no way how to make it an argument to a command. You need something more powerful, e.g. Perl:
perl -pe 'if ( ($t) = /(1[0-9]+)/ ) { s/$t/localtime($t)/e }'
You can do it with GNU sed, the input:
infile
Some stuff 1346474454 And not working
GNU sed supports /e parameter which allows for piping command output into pattern space, one way to take advantage of this with bpdbm:
sed 's/(.*)(1[0-9]{9})(.*)/echo \1 $(bpdbm -ctime \2) \3/e' infile
Or with coreutils date:
sed 's/(.*)(1[0-9]{9})(.*)/echo \1 $(date -d #\2) \3/e' infile
output with date
Some stuff Sat Sep 1 06:40:54 CEST 2012 And not working
To get the same output as with bpdbm:
sed 's/(.*)(1[0-9]{9})(.*)/echo "\1$(date -d #\2 +\"%a %b %_d %T %Y\")\3"/e' infile
output
Some stuff Sat Sep 1 06:40:54 2012 And not working
Note, this only replaces the last epoch found on a line. Re-run if there are more.