unix - 找到字符串并打印出unix脚本之前和之后的行

Question

现在我有这个代码

 FNR==NR  { id[$0]; next } 
 $0 in id { f=1 }         
 f                          
 NF==0    { f=0 }   

我也想在找到字符串之前打印出该行。

我将如何修改此代码来做到这一点？

Answer 1

这是我提到的 Perl 脚本，sgrep. 今年的变化是使用#!/usr/bin/env perl而不是#!/path/to/perl/v5.16.2/bin/perl线。否则，它是在 2007 年 9 月写的。我预计还有改进的余地。

#!/usr/bin/env perl
#
# @(#)$Id: sgrep.pl,v 1.7 2013/01/28 02:07:18 jleffler Exp $
#
# Perl-based SGREP (special grep) command
# 
# Print lines around the line that matches (by default, 3 before and 3 after).
# By default, include file names if more than one file to search.
#
# Options:
# -b n1     Print n1 lines before match
# -f n2     Print n2 lines following match
# -n        Print line numbers
# -h        Do not print file names
# -H        Do     print file names

use warnings;
use strict;
use constant debug => 0;
use Getopt::Std;
my(%opts);

sub usage
{
    print STDERR "Usage: $0 [-hnH] [-b n1] [-f n2] pattern [file ...]\n";
    exit 1;
}

usage unless getopts('hnf:b:H', \%opts);
usage unless @ARGV >= 1;

if ($opts{h} && $opts{H})
{
    print STDERR "$0: mutually exclusive options -h and -H specified\n";
    exit 1;
}

my $op = shift;

print "# regex = $op\n" if debug;

# print file names if -h omitted and more than one argument
$opts{F} = (defined $opts{H} || (!defined $opts{h} and scalar @ARGV > 1)) ? 1 : 0;
$opts{n} = 0 unless defined $opts{n};

my $before = (defined $opts{b}) ? $opts{b} + 0 : 3;
my $after  = (defined $opts{f}) ? $opts{f} + 0 : 3;

print "# before = $before; after = $after\n" if debug;

my @lines = (); # Accumulated lines
my $tail  = 0;  # Line number of last line in list
my $tbp_1 = 0;  # First line to be printed
my $tbp_2 = 0;  # Last line to be printed

# Print lines from @lines in the range $tbp_1 .. $tbp_2,
# leaving $leave lines in the array for future use.
sub print_leaving
{
    my ($leave) = @_;
    while (scalar(@lines) > $leave)
    {
        my $line = shift @lines;
        my $curr = $tail - scalar(@lines);
        if ($tbp_1 <= $curr && $curr <= $tbp_2)
        {
            print "$ARGV:" if $opts{F};
            print "$curr:" if $opts{n};
            print $line;
        }
    }
}

# General logic:
# Accumulate each line at end of @lines.
# ** If current line matches, record range that needs printing
# ** When the line array contains enough lines, pop line off front and,
#    if it needs printing, print it.
# At end of file, empty line array, printing requisite accumulated lines.

while (<>)
{
    # Add this line to the accumulated lines
    push @lines, $_;
    $tail = $.;

    printf "# array: N = %d, last = $tail: %s", scalar(@lines), $_ if debug > 1;

    if (m/$op/o)
    {
        # This line matches - set range to be printed
        my $lo = $. - $before;
        $tbp_1 = $lo if ($lo > $tbp_2);
        $tbp_2 = $. + $after;
        print "# $. MATCH: print range $tbp_1 .. $tbp_2\n" if debug;
    }

    # Print out any accumulated lines that need printing
    # Leave $before lines in array.
    print_leaving($before);
}
continue
{
    if (eof)
    {
        # Print out any accumulated lines that need printing
        print_leaving(0);
        # Reset for next file
        close ARGV;
        $tbp_1 = 0;
        $tbp_2 = 0;
        $tail  = 0;
        @lines = ();
    }
}

-b 1它对匹配前的行和匹配后的行使用选项，-f 1而不是-Aand -B。

Answer 2

如果您有 gnu grep 或支持 grep 的 grep -C，您可以执行以下操作：

grep -w -f file1 -C 1 input

make只匹配整个单词，旨在模仿脚本的匹配，但不会-w做完全相同的事情（我不清楚你想要做什么匹配）你可能更愿意省略.grepawk-w

在 awk 中，您可以执行以下操作：

FNR==NR  { id[$0]; next }
$0 in id { print prev; f=1 }         
f{ f += 1; print}                          
f == 3 { f=0 }
{ prev = $0 }