將周圍的單詞與匹配一起提取

Question

我正在尋找搜索詞的每一次出現，例如， ddd，在一個文件中並輸出的環境中，這樣的：

FILE.TXT

aaa bbb ccc ddd eee fff 
ttt uuu iii eee ddd 
ddd 
ggg jjj kkk ddd lll 

輸出

ccc ddd eee 
eee ddd 
ddd 
kkk ddd lll 

作爲一個起點，我使用的這一段代碼

#!/usr/bin/perl -w  
while(<>) { 
    while (/ddd(\d{1,3}))/g) {  
    print "$1\n" 
    } 
} 

Answer 1

#!/usr/bin/perl 

while (<>) { 
    chomp; 
    @F = split /\s+/; 
    if (/^ddd$/) {print $_."\n";next}; 
    for ($i=0; $i<=$#F;$i++) { 
     if ($F[$i] eq 'ddd') { 
      print "$F[$i-1] $F[$i] $F[$i + 1]\n"; 
     } 
    } 
} 

Answer 2

#!/usr/bin/perl -w 
while (<>) { 
    if (/((?:[a-z]{3})?ddd(?: [a-z]{3})?)/) 
     print "$1\n"; 
} 

Answer 3

你可以嘗試following..it給你想要的輸出：

while(<>) { 
     if(/((?:\w+)?ddd(?: \w+)?)/) { 
       print "$1\n"; 
     } 
} 

正則表達式中使用：

(  # open the grouping. 
(?:\w+)? # an optional word of at least one char followed by a space. 
ddd  # 'ddd' 
(?: \w+)? # an optional space followed by a word of at least one char. 
)   # close the grouping. 

Answer 4

while (<>) { 
    chomp; 
    my @words = split; 
    for my $i (0..$#words) { 
     if ($words[$i] eq 'ddd') { 
      print join ' ', $i > 0 ? $words[$i-1] :(), $words[$i], $i < $#words ? $words[$i+1] :(); 
      print "\n"; 
     } 
    } 

}