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Hi All, Can you please let me know how to grep for a
particular pattern in unix. I want to print the dates from
the file exp.txt. the date pattern is DD:MM:YYYY, I just
want to print all the dates from the file exp.txt.
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Answer / thiru
Above answers allow 09:2332:19811981 input also to display.
Following may use to avoid that issue
grep -o '[0-9][0-9]:[0-9][0-9]:[0-9][0-9][0-9][0-9]' infile.txt
| Is This Answer Correct ? | 44 Yes | 6 No |
Answer / uttamjit
You can use the below command:
grep -o '[[0-9]*:[0-9]*:[0-9]*' infile
where infile is the input file to grep command.
| Is This Answer Correct ? | 16 Yes | 12 No |
Answer / sona shaw
grep, egrep, fgrep - print lines matching a pattern
SYNOPSIS
grep [options] PATTERN [FILE...]
grep [options] [-e PATTERN | -f FILE] [FILE...]
e.g
grep "24:02:2010" date.txt
o/p is :hi today is 24:02:2010
here we should keep the matching patten in double
qoute"xxxxxxxx" and date.txt is inputfile in which u have
to search the pattern. for more details see manual page of
grep.type man grep and see various option
| Is This Answer Correct ? | 4 Yes | 2 No |
Answer / mo
#1: grep '[0-9][0-9]-[0-9][0-9]-[0-9][0-9][0-9][0-9]'
<filename> | more
from #1 output see which column is returned for date.
use next this command.
#1: grep '[0-9][0-9]-[0-9][0-9]-[0-9][0-9][0-9][0-9]'
<filename> | cut -f<from the first command> -d"<Delimeter
of file>
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / venkat
grep -o -P '(([0-2][0-9]|3[01]):([0][1-9]|[1][012]):[12][09][0-9][0-9])'
use Perl regex to validate the limits of date and month.
Pitfall: it might miss 31st of feb
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / mad
Small modification required
grep -o '[0-9]*:[0-9]*:[0-9]*' infile
| Is This Answer Correct ? | 5 Yes | 12 No |
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