perl 正则表达式匹配“"(字符串)语法

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我是 Perl 和 regex 的新手，我需要从文本文件中提取所有字符串.字符串由双引号包裹的任何内容标识.

I am new to Perl and regex and I need to extract all the strings from a text file. A string is identified by anything that is wrapped by double quotes.

字符串示例:

"This is string" "1!=2" "This is \"string\"" "string1"."string2" "S t r i n g"

代码:

my $fh; open($fh,'<','text.txt') or die "$!"; undef $/; my $text = <$fh>; my @strings = m/".*"/g; # this returns the most out "" in example 4 my @strings2 = m/"[^"]*"/g #fixed the above issue but does not take in example 3

我想得到(1)一个双引号，然后是(2)零次或多次出现的非双引号非反斜杠或反斜杠后跟任何字符，然后是(3)双引号.(2) 可以是

Edited : I want to get (1) a double quote, followed by (2) zero or more occurrences of either a non-double-quote-non-backslash or a backslash followed by any character, followed by (3) a double quote. (2) can be anything but "

下面提供的正则表达式 m/"(?:\.|[^"])*"/g 但是当有一行带有 "string1".string2."string2" 时将返回 "string1" string2 "string3"

The regex provided below m/"(?:\.|[^"])*"/g however when the there is a line with "string1".string2."string2" it will return "string1" string2 "string3"

有什么地方可以跳过之前匹配的单词吗?

Is there any wher to skip the previously matched word?

有人可以帮忙吗?

推荐答案

一种可能的方法:

/"(?:\\.|[^"])*"/

... 读作:

• 匹配双引号，

• 后跟任意数量的...

• match double quotation mark,

• followed by any number of...

--- 任何转义字符(任何以 \ 开头的符号)

--- either any escaped character (any symbol prepended by \)

--- 或任何不是双引号的字符

--- or any character that's not a double quotation mark

这里的关键技巧是使用交替使用任何转义符号 - 包括转义双引号.

The key trick here is using alternation that'll eat any escaped symbol - including escaped double quotation mark.

演示.