Perl正则表达式模式

Question

我有这样一些数据：

TYPE: Travel
ADDRESS
  Barcelona
  Paris

因此，地址可以是1或很多（我需要丢弃ADDRESS并仅获取那些城市）。 由于某些原因，我的解析失败（仅打印“ ADDRESS”）无法产生正确的结果。我错过了什么吗？

elsif (/^ADDRESS/) {
    my @address_t = split /[no matter what i put,only ADDRESS is printed]+/, $_;
        shift @address_t;  #is this how i will discard ADDRESS ?

        foreach my $address (@address_t) {
                @address_names = ($address);            
    }

我认为正则表达式应该被换行，空格？

这就是我处理TYPE的方式：

elsif (/^TYPE/) {
            my @type_t = split '\s', $_;
            $type = $type_tmp[1];
                    print "$type" ; #to test, but i have a hashmap which i load them in and print at the end of the file.

谢谢

Answer 1

尝试这样的事情：

如果前一行匹配/^ADDRESS/ ，它将打印行。 让我知道您是否要停止，我可以调整...

use warnings;
use strict; 

my $current_line = "";
my $line_count = 0;

while (<IN>){
    chomp;
    my $previous_line = $current_line;
    $current_line = $_;
    if ($previous_line =~ /^ADDRESS/ or $line_count > 0 ){
    $line_count++;
    print "$current_line\n"

    }   
}

Answer 2

use warnings;
use strict;

while(<DATA>) {
    if (/^ADDRESS/) {           # if line contains ADDRESS then read addresses
        while (<DATA>) {        # ... in a loop
            last if !/^ +/;     # until we find a non-indented line
            print $_;           # here you can push $_ to a list
        }
    }   
    if ($_ && /^TYPE/) {        # a TYPE after address can be processed now
        # stuff 
    }   
}

__DATA__
TYPE: Travel
ADDRESS
  Barcelona
  Paris
TYPE: Travel
ADDRESS
  Barcelona
  Paris

生产：

  Barcelona
  Paris
  Barcelona
  Paris