找到字符串之前和離職後UNIX腳本

Question

打印出來的線現在我有這樣的代碼

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

，我想打印出前行字符串中找到爲好。

我該如何修改這段代碼呢？

Answer 1

這是我提到的Perl腳本sgrep。今年的變化是使用#!/usr/bin/env perl而不是#!/path/to/perl/v5.16.2/bin/perl行。否則，它是在2007年9月寫的。我預計還有改進的空間。

#!/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 =(); 
    } 
} 

它的比賽，而不是-A和-B後使用選項-b 1的比賽之前線和-f 1的線路。

Answer 2

如果你有GNU的grep，或支持-C一個grep的，你可以這樣做：

grep -w -f file1 -C 1 input 

的-w使得grep只匹配整個單詞，意在模仿你awk腳本的 匹配，反而會不是完全相同的事情（我不清楚你到底想要做什麼匹配）你可能更喜歡忽略-w。

在awk中，你可以這樣做：

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