Question

假设：

my $s1='A "sentence" ... is a concatination of nondelimiters followed by delimiters, 
     followed by nondelimiters, followed by delimiters, ...'; 

或

my $s2='"A "sentence" ... is a concatination of delimiters followed by nondelimiters, 
     followed by delimiters, followed by nondelimiters, ..."'; 

我试图通过正则表达式来分割到一个列表@a使用的分隔符标量$ d设置：

my $d = ' ,."'; # delimiter set 
my @a = $s1 =~ /([^$d]+)([$d]+)/g; # gives |A| "|sentence|" ... |is| |a| |... 

在第二种情况下，句子可以使用

my @a = $s2 =~ /([$d]+)([^$d]+)/g; # gives |"|A| "|sentence|" ... |is| |a|... 

如何处理这两个案件与一个正则表达式？需要帮助 - 请。

Answer 1

您需要连续的分隔符或非限定符。只需使用

$s =~ /([$d]+|[^$d]+)/g;