ruby - 你如何在 ruby​​ 中去除子字符串？

Question

我想在两个分隔符之间替换/复制一个子字符串——例如：

"This is (the string) I want to replace" 我想去掉字符(和之间的所有内容)，并将 substr 设置为一个变量——是否有内置函数来执行此操作？

Answer 1

var = "This is (the string) I want to replace"[/(?<=\()[^)]*(?=\))/]
var # => "the string"

Answer 2

我会这样做：

my_string = "This is (the string) I want to replace"

p my_string.split(/[()]/) #=> ["This is ", "the string", " I want to replace"]

p my_string.split(/[()]/)[1] #=> "the string"

这里还有另外两种方法：

/\((?<inside_parenthesis>.*?)\)/ =~ my_string 
p inside_parenthesis #=> "the string"

my_new_var =  my_string[/\((.*?)\)/,1]
p my_new_var  #=> "the string"

编辑 - 解释最后一种方法的示例：

my_string = 'hello there'
capture = /h(e)(ll)o/ 

p my_string[capture]    #=> "hello"
p my_string[capture, 1] #=> "e"
p my_string[capture, 2] #=> "ll"

Answer 3

str = "This is (the string) I want to replace"

str.match(/\((.*)\)/)

some_var = $1  # => "the string"

Answer 4

据我了解，您想要删除或替换子字符串以及设置一个等于该子字符串的变量（不带括号）。有很多方法可以做到这一点，其中一些是其他答案的轻微变体。这是另一种方式，它也允许括号内有多个子字符串，从@sawa 的评论中获取：

def doit(str, repl)
  vars = []
  str.gsub(/\(.*?\)/) {|m| vars << m[1..-2]; repl}, vars
end

new_str, vars = doit("This is (the string) I want to replace", '')  
new_str # => => "This is  I want to replace" 
vars    # => ["the string"] 

new_str, vars = doit("This is (the string) I (really) want (to replace)", '')  
new_str # => "This is  I  want" 
vars    # => ["the string", "really, "to replace"] 

new_str, vars = doit("This (short) string is a () keeper", "hot dang")  
new_str # => "This hot dang string is a hot dang keeper" 
vars    # => ["short", ""]

在正则表达式中，?in.*?使.*“懒惰”。 gsub将每个匹配传递m给块；该块剥离括号并将其添加到vars，然后返回替换字符串。此正则表达式也适用：

/\([^\(]*\)/