I'm using Ruby 2.0. I've currently got a string of:
我用Ruby 2.0。我现在有一连串:
str = "bar [baz] foo [with] another [one]"
str.scan(/\[.*\]/)
The output is:
的输出是:
["[baz] foo [with] another [one]"]
When I would expect it more like:
当我期望它更像:
["[baz]","[with]","[one]"]
So I basically need to put everything between "[]" into an array. Can someone please show me what I'm missing out?
所以我基本上需要把"[]"之间的所有东西都放到一个数组中。谁能告诉我我遗漏了什么吗?
2 个解决方案
#1
4
Your .* is greedy, so it doesn't stop until the final bracket.
你的。*是贪婪的，所以直到最后一个括号。
You need to use a lazy quantifier .*? or only catch non-brackets: [^\]]*
你需要使用一个懒惰的量词。*?或者只抓摘要:[^ \]]*
#2
2
Regexs are greedy by default so your regex grabbing everything from the first [ to the last ]. Make it non-greedy like so:
默认情况下，regex是贪婪的，因此您的regex将从第一个[到最后一个]获取所有内容。让它变得不贪婪:
str.scan(/\[.*?\]/)
#1
4
Your .* is greedy, so it doesn't stop until the final bracket.
你的。*是贪婪的，所以直到最后一个括号。
You need to use a lazy quantifier .*? or only catch non-brackets: [^\]]*
你需要使用一个懒惰的量词。*?或者只抓摘要:[^ \]]*
#2
2
Regexs are greedy by default so your regex grabbing everything from the first [ to the last ]. Make it non-greedy like so:
默认情况下，regex是贪婪的，因此您的regex将从第一个[到最后一个]获取所有内容。让它变得不贪婪:
str.scan(/\[.*?\]/)