Perl：修剪多行記錄

Question

我需要讀取多行記錄並將它們修剪成完全40行。然後 填充它們爲45行。它們可能高達70多條線。這些記錄需要 最終爲45行。

記錄分隔符是以模式/ ^＃matchee /開頭的行。

我假設你將$ /設置爲#matchee。

{ 
    $/ = "#matchee"; 

    while (<>) { 
     # I need to print first 40 
     # lines of each record then 
     # pad to 45 with delimiter as 
     # last line. 
    } 
} 

樣本記錄

REDUNDANCY DEPARTMENT 
Anonymous Ave 

Item 1 
Item 2 



<bunch of blank lines> 
#matchee 

Answer 1

這裏是我的解決方案...

#! /usr/bin/env perl 
use strict; 
use warnings; 

{ 
    $/ = "#matchee"; 

    while (my @line = split "\n", <>) { 

    # print first 40 lines of record 
     for my $counter (0..39) { 
      print($line[$counter] . "\n"); 
     } 

     # pad record with four extra blank lines 
     # (last record already ends with a newline) 
     print "\n" x 4; 
    } 
} 

+1使用$/ = "#matchee";

這是不完全正確......第一條記錄有45行，第二行有44行。

Answer 2

指定「記錄分隔符是以模式/ ^＃matchee /開頭的行」。這使記錄分離有點複雜，因爲$/是a special string, but not a regex。您沒有指定您的輸出是否使用相同的記錄分隔符，但我假設如此。這是一種似乎可行的方法。

#!/usr/bin/env perl 
use strict; 
use warnings; 

sub take_and_pad_lines { 
    my ($str, $take, $pad) = @_; 

    my @lines = (split(/\n/, $str))[0..$take-1]; 
    return join "\n", @lines, ('') x ($pad - $take); 
} 


{ 
    $/ = "#matchee"; 

    while (my $record = <>) { 
    # because RS is really begins-with we must clean up first line 
    # and double check last record 
    unless (1 == $.) { 
     $record =~ s/\A.*\n//m; 
     last if eof() && $record eq ''; 
    } 

    print take_and_pad_lines($record, 40, 45), "\n"; 
    print "$/\n" unless eof(); 
    } 
}