如何检查字符串是否包含Ruby中的子字符串？

Question

我有一个字符串变量,内容如下:

varMessage =   
            "hi/thsid/sdfhsjdf/dfjsd/sdjfsdn\n"


            "/my/name/is/balaji.so\n"
            "call::myFunction(int const&)\n"
            "void::secondFunction(char const&)\n"
             .
             .
             .
            "this/is/last/line/liobrary.so"

在上面的字符串我必须找到一个子字符串即

"hi/thsid/sdfhsjdf/dfjsd/sdjfsdn\n"


"/my/name/is/balaji.so\n"
"call::myFunction(int const&)\n"

我该怎么找到它？我只需要确定子串是否存在.

Answer 1

您可以使用以下include?方法:

my_string = "abcdefg"
if my_string.include? "cde"
   puts "String includes 'cde'"
end

Answer 2

如果大小写无关紧要,那么不区分大小写的正则表达式是一个很好的解决方案:

'aBcDe' =~ /bcd/i  # evaluates as true

这也适用于多行字符串.

请参阅Ruby的Regexp类.

Answer 3

你也可以这样做......

my_string = "Hello world"

if my_string["Hello"]
  puts 'It has "Hello"'
else
  puts 'No "Hello" found'
end

# => 'It has "Hello"'

Answer 4

扩展Clint Pachl的答案:

当表达式不匹配时,Ruby中的正则表达式匹配返回nil.如果是,则返回匹配发生的字符的索引.例如

"foobar" =~ /bar/  # returns 3
"foobar" =~ /foo/  # returns 0
"foobar" =~ /zzz/  # returns nil

值得注意的是,在Ruby中只有nil和布尔表达式false求值为false.其他所有内容(包括空数组,空哈希或整数0)的计算结果为true.

这就是为什么上面的/ foo /示例有效,以及为什么

if "string" =~ /regex/

按预期工作.如果匹配发生,则仅输入if块的"true"部分.

Answer 5

比Rails(3.1.0及以上版本)中提供的上述接受的答案更简洁的习语是.in?.

例如:

my_string = "abcdefg"
if "cde".in? my_string
  puts "'cde' is in the String."
  puts "i.e. String includes 'cde'"
end

我也认为它更具可读性.

cf http://apidock.com/rails/v3.1.0/Object/in%3F

(请注意,它仅在Rails中可用,而不是纯Ruby.)

Answer 6

三元方式

my_string.include?('ahr') ? (puts 'String includes ahr') : (puts 'String does not include ahr')

要么

puts (my_string.include?('ahr') ? 'String includes ahr' : 'String not includes ahr')

Answer 7

您可以使用String Element Reference方法[]

在[]can 内部可以是文字子字符串,索引或正则表达式:

> s='abcdefg'
=> "abcdefg"
> s['a']
=> "a"
> s['z']
=> nil

因为nil在功能上false和返回的任何子字符串相同,[]所以true您可以像使用方法一样使用逻辑.include?:

0> if s[sub_s]
1>    puts "\"#{s}\" has \"#{sub_s}\""
1> else 
1*    puts "\"#{s}\" does not have \"#{sub_s}\""
1> end
"abcdefg" has "abc"

0> if s[sub_s]
1>    puts "\"#{s}\" has \"#{sub_s}\""
1> else 
1*    puts "\"#{s}\" does not have \"#{sub_s}\""
1> end
"abcdefg" does not have "xyz" 

只需确保不要将索引与子字符串混淆:

> '123456790'[8]    # integer is eighth element, or '0'
=> "0"              # would test as 'true' in Ruby
> '123456790'['8']  
=> nil              # correct

你也可以使用正则表达式:

> s[/A/i]
=> "a"
> s[/A/]
=> nil