extracting part of a text file

Shell Programming and Scripting
1

Hi guys

So I have a very large log file where each event is logged along with the time that it occurred.

So for e.g. The contents of the file look like:

...
12:00:07 event 0 happened.
12:01:01 event 1 happened.
12:01:05 event 2 happened.
12:01:30 event 3 happened.
12:02:01 event 4 happened.
12:02:40 event 5 happened.
12:03:40 event 6 happened.
12:04:06 event 6 happened.
...

I want to extract every thing in that file listed between a given start time and an end time.

so some thing like all events between
the first entry of 12:01:00 and the last entry of 12:03:59
and store them in a random text file. Note time stamps can be repeated and therefore I mentioned the first and the last entries.

And not all lines in the log file start with a time stamp. Some lines are just random information about the server etc. But I want to extract those lines as well.

Hope I have clarified my question.

And thanks in advance every one :slight_smile:

ali

Use CODE tags when posting code, data or logs for better readability and to preserve formatting like indention etc., ty.

2

Besides adding code tags moving it to shell scripting area - nothing AIX particular in here.

3

You can start from this example:

awk '{a="file."srand()}/12:01/,/12:03/{print >> a}'  file
4 
sed  -n '/^12:00/,/^12:03/p' event.log > new_event.log
5

Or a nice simple grep would do it:

grep -E "12:01|12:02|12:03" <input_text_file_name> > <output_text_file_name>

So here we use the grep command to search for multiple "arguments", delimited by the pipe (|) symbol in the <input_text_file_name> and then redirect with the > to the <output_text_file_name> and then you should have the results you want in the <output_text_file_name> file.

pSeries and AIX Information Center

HTH

6

This may not meet all the requirements.
Remember the requirement says:
"""And not all lines in the log file start with a time stamp. Some lines are just random information about the server etc. But I want to extract those lines as well."""

7

yeah it wouldn't and this is more simpler

grep 12:0[1-3] event.log
8

hi guys
Thanks for your replies so far

This pattern doesnt do anything :
sed -n '/^12:00/,/^12:03/p' event.log > new_event.log

And the grep ones I believe will only extract lines that contain the pattern that I have entered. But they would miss the lines in between. I do need the lines without the pattern as well.

I again summarise my requirement:
In a file extract every thing from the first occurance of pattern: 12:01:00 till the last occurance of pattern: 12:03:00.

many thanks
ali

9

I have no issues.
Is it possible your input file is DOS formatted?

Here is my working.

cat event.log
12:00:07 event 0 happened.
some line at 2
12:01:01 event 1 happened.
some line at 4
12:01:05 event 2 happened.
12:01:30 event 3 happened.
some line at 7
some line at 8
12:02:01 event 4 happened.
12:02:40 event 5 happened.
12:03:40 event 6 happened.
12:04:06 event 6 happened.

Command:

sed -n '/^12:00/,/^12:03/p' event.log

Output:

12:00:07 event 0 happened.
some line at 2
12:01:01 event 1 happened.
some line at 4
12:01:05 event 2 happened.
12:01:30 event 3 happened.
some line at 7
some line at 8
12:02:01 event 4 happened.
12:02:40 event 5 happened.
12:03:40 event 6 happened.

May be version issues?
Remove the "> new_event.log" part and check.

10

Did you ever try my previous solution

11

Try with a gap before the "p" like this.

sed -n '/^12:00/,/^12:03/ p' event.log > new_event.log