Getting specific part of output in Linux

Question

I have an output from a shell script like this:

aaa.sh output

     Tue Mar 04 01:00:53 2014
     Time drift detected. Please check VKTM trace file for more details.
     Tue Mar 04 07:21:52 2014
     Time drift detected. Please check VKTM trace file for more details.
     Tue Mar 04 13:17:16 2014
     Time drift detected. Please check VKTM trace file for more details.
     Tue Mar 04 16:56:01 2014
     SQL> ALTER DISKGROUP fra  ADD DISK  '/dev/rhdisk20' 
     Wed Mar 05 00:03:42 2014
     Time drift detected. Please check VKTM trace file for more details.
     Wed Mar 05 04:13:39 2014
     Time drift detected. Please check VKTM trace file for more details.
     Tue Mar 05 05:56:07 2014
     GMON querying group 3 at 10 for pid 18, osid 27590856
     GMON querying group 3 at 11 for pid 18, osid 27590856

I need to get the part, beginning from today's date:

     Wed Mar 05 00:03:42 2014
     Time drift detected. Please check VKTM trace file for more details.
     Wed Mar 05 04:13:39 2014
     Time drift detected. Please check VKTM trace file for more details.
     Tue Mar 05 05:56:07 2014
     GMON querying group 3 at 10 for pid 18, osid 27590856
     GMON querying group 3 at 11 for pid 18, osid 27590856

@AloneInTheDark man awk or man sed has all the information you need to find a solution. Also man date. If you still can't manage to do it out you should add your attempt to the question. — DBedrenko
– DBedrenko, Commented Mar 5, 2014 at 9:01
The solution is to become a little professional on bash. That does need several hours of reading Bash Guide for Beginners, Advanced Bash Scripting Guide, gawk manual — Basile Starynkevitch
– Basile Starynkevitch, Commented Mar 5, 2014 at 9:15

Mark Setchell · Accepted Answer · 2014-03-05 09:49:03Z

3

You can get the date in the correct format like this:

today=$(date +'%a %b %d')

and then search for it like this:

grep "$today" aaa.sh

If there are lines from today without a date, such as your GMON lines, you could add -A to say how many lines after the match you want and use a big number:

grep -A 999999 "$today" aaa.sh

If you are on AIX and there is no -A option, use sed like this:

today=$(date +'%a %b %d')
sed -n "/${today}/,$ p" aaa.sh

Explanation:

That says store today's date in the variable today in the format "Wed Mar 05". Then search, without printing anything (-n) till you find that date, From that point on, till the end of file ($) print all lines (p).

edited Mar 5, 2014 at 9:49

answered Mar 5, 2014 at 9:34

Mark Setchell

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3 Comments

AloneInTheDark Over a year ago

How to do that in AIX? Because there is no -A parameter in that.

AloneInTheDark Over a year ago

I got an error: aaa.sh[10]: Wed: not found. sed: 0602-410 The first regular expression cannot be null. My date format is like that("Wed Mar 05")

Mark Setchell Over a year ago

Try again, I added braces.

DBedrenko · Accepted Answer · 2014-03-05 09:22:47Z

1

I think I have an easy solution:

Get date to output the date in a format that would match the date in the file (check man date on formatting options). Since we don't want to match the hours/minutes/seconds we have to call date twice: once for the weekday/month/day half and once for the year half on the end of the full date. Between these two halves we match the horus/minutes/seconds with .* regex.

Then do:

aaa.sh | grep -E '`date --only-weekday-month-day`.*`date --only-year`' -A 999999

edited Mar 5, 2014 at 9:22

answered Mar 5, 2014 at 9:14

DBedrenko

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Comments

Nachiket Kate · Accepted Answer · 2014-03-05 09:23:04Z

1

though I am using answer by NewWorld it can be modified as,

convert output of date similar to your file format

suppose in variable 'D'you get that output

sed '1,/${D}/d' aaa.sh

that will output all lines after match date match.

example: suppose you get D="Wed Mar 05 00:03:42 2014"

output will be as expected.

answered Mar 5, 2014 at 9:23

Nachiket Kate

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3 Comments

AloneInTheDark Over a year ago

Thanks for the solution. I have another question, how to do the same thing in AIX? Because syntax is different.

Harshitha K V Over a year ago

@AloneInTheDark the above solution prints all lines after the pattern match, it excludes the first line. Just FYI

Mark Setchell Over a year ago

You can use sed -n "/${D}/,$ p" aaa.sh to avoid that.

tusharmakkar08 · Accepted Answer · 2014-03-05 09:32:24Z

1

You can use

  tail -n 7 filename

for getting the desired output . It will basically give you the last seven lines of the text file named filename .

For getting solution from today's date you can use :

 k=$(date +"%a %b %d")
 g=$(grep -nr "$k" in|cut -f1 -d:|head -1)
 total=$(wc -l<in)
 l=`expr $total - $g + 1
 tail -n$l in

edited Mar 5, 2014 at 9:32

answered Mar 5, 2014 at 9:05

tusharmakkar08

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1 Comment

AloneInTheDark Over a year ago

I need the part from beginning of today's date. There may be much more lines.

Harshitha K V · Accepted Answer · 2014-03-05 10:07:58Z

1

Try

sed -n '/Wed Mar 05/,$p' aaa.sh

Here -n means "don't print anything unless specified to". First appearance of a line that matches the expression /Wed\ Mar\ 05/ till the end of the file, will be printed(p)"

edited Mar 5, 2014 at 10:07

answered Mar 5, 2014 at 9:28

Harshitha K V

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1 Comment

Mark Setchell Over a year ago

Just a friendly word of advice... you do not need to escape the spaces with slashes. And you do not need to specify -e when there is only one piece of code to execute since sed knows what it is and what to do with it.

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