Bash – remove 2 lines from output, grep match regular expression plus next 1

bashgrep

i have a log file from postgresql that has log entries of the format;

LOG:  execute <unnamed>: /match this here/
DETAIL:  parameters: /also want to filter this line/

I thought it might be possible with

grep -v --after-context=1 'match this here' /my/log/file.log

but it doesn't work. is there some sed/awk/grep magic, or do i have to do some perl?

Best Answer

As far as I know you cannot do this with grep alone.
awk can do it pretty simple:

awk -v skip=-1 '/match this here/ { skip = 1 } skip-- >= 0 {next } 1' /my/log/file.log

edit:
I assumed that you cannot match the second line by any regex and you really want to match line containing X and the line afterwards. If you can match against a string, it's a lot easier:

egrep -v "match this here|also want to filter this line" /my/log/file.log

Related Solutions

Bash – Don’t need the whole line, just the match from regular expression

2 Things:

As stated by @Rory, you need the -o option, so only the match are printed (instead of whole line)
In addition, you neet the -P option, to use Perl regular expressions, which include useful elements like Look ahead (?= ) and Look behind (?<= ), those look for parts, but don't actually match and print them.

If you want only the part inside the parenthesis to be matched, do the following:

grep -oP '(?<=\/\()\w(?=\).+\/)' myfile.txt

If the file contains the sting /(a)5667/, grep will print 'a', because:

/( are found by \/\(, but because they are in a look-behind (?<= ) they are not reported
a is matched by \w and is thus printed (because of -o )
)5667/ are found by \).+\/, but because they are in a look-ahead (?= ) they are not reported

Bash – Handy parsing for numbers with unit suffixes

Based on my answer at one of the questions you linked to:

awk '{
    ex = index("KMGTPEZY", substr($1, length($1)))
    val = substr($1, 0, length($1) - 1)

    prod = val * 10^(ex * 3)

    sum += prod
}
END {print sum}'

Another method that's used:

sed 's/G/ * 1000 M/;s/M/ * 1000 K/;s/K/ * 1000/; s/$/ +\\/; $a0' | bc

Best Answer

Related Solutions

Bash – Don’t need the whole line, just the match from regular expression

Bash – Handy parsing for numbers with unit suffixes

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