除了括号中的空格之外的所有空格拆分字符串 [英] Split string by all spaces except those in brackets

问题描述

可能重复：

根据正则表达式拆分字符串

我'从来没有成为正规表达大师，所以我需要你的帮助！我有一个这样的字符串：

I've never been a regular expression guru, so I need your help! I have a string like this:

String s = "a [b c] d [e f g]";

我想用空格分隔这个字符串作为分隔符 - 但我不想拆分出现在 [] 括号内的空格。所以，从上面的例子中，我想要这个数组：

I want to split this string using spaces as delimiters -- but I don't want to split on spaces that appear within the [] brackets. So, from the example above, I would like this array:

{"a", "[b c]", "d", "[e f g]"}

有关正则表达式可以与<$一起使用的任何建议c $ c>拆分以实现这个目标？

Any advice on what regex could be used in conjunction with split in order to achieve this?

这是另一个例子：

"[a b] c [[d e] f g]"

变为

{"[a b]", "c", "[[d e] f g]"}

推荐答案

我认为这是应该工作，使用否定前瞻 - 它不匹配没有开口的关闭括号之前的空格括号：

I think this should work, using negative lookahead - it matches no whitespace that comes before closing bracket without an opening bracket:

"a [b c] d [e f g]".split("\\s+(?![^\\[]*\\])");

对于嵌套括号，你需要编写一个解析器，正则表达式无法承受无限级别并得到超过一个或两个级别太复杂了。例如，我的表达式失败

For nested brackets you will need to write a parser, regexes can't afford an infinite level and get too complicated for more than one or two levels. My expression for example fails for

"[a b [c d] e] f g"

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