I have a file with phrases such as "Canyon St / 27th Way" that I am trying to turn into "Canyon St and 27th Way" with Ruby regular expressions.
我有一个带有诸如“Canyon St / 27th Way”之类的短语的文件，我试图用Ruby正则表达式变成“Canyon St和27th Way”。
I used file = file.gsub(/(\b) \/ (\b)/, "#{$1} and #{$2}") to make the match, but I am a little stumped about what \b really means and why $1 contains all of the characters before the word boundary that precedes the slash and why $2 contains all of the characters after the word boundary starting the next word.
我使用file = file.gsub（/（\ b）\ /（\ b）/，“＃{$ 1}和＃{$ 2}”来进行匹配，但我对于什么\ b真正意味着什么感到困惑为什么$ 1包含在斜线之前的单词边界之前的所有字符以及为什么$ 2包含单词边界开始下一个单词之后的所有字符。
Usually, I expect that whatever is in parentheses in a regular expression would be in $1 and $2, but I am not sure what parentheses around a word boundary would really mean because there really is nothing between the transition from a word character to a white space character.
通常，我希望正则表达式中括号中的任何内容都是1美元和2美元，但我不确定单词边界周围的括号是什么意思，因为从单词字符到空格的转换之间确实没有任何内容字符。
2 个解决方案
#1
6
The $1 and $2 are not actually related to your regex match: a method's arguments are evaluated before the method is called, so
$ 1和$ 2实际上与你的正则表达式匹配无关：在调用方法之前评估方法的参数，所以
"#{$1} and #{$2}"
Is evaluated before the regex is matched against your string. If you haven't done earlier regex matches then these variables will be nil, so you're actually doing
在正则表达式与字符串匹配之前进行评估。如果你没有做过早期的正则表达式匹配，那么这些变量将是零，所以你实际上在做
file = file.gsub(/(\b) \/ (\b)/, " and ")
that is you are replacing a slash surrounded by spaces by "and", also surrounded by spaces. $1 and $2 will be updated to be empty strings, and so you'll see the same behaviour when you process the next string.
那就是你用“和”替换空格包围的斜线，也用空格包围。 $ 1和$ 2将更新为空字符串，因此当您处理下一个字符串时，您将看到相同的行为。
#2
8
The parentheses aren't doing anything in this context. You could get the same result using /\b \/ \b/.
括号在这种情况下没有做任何事情。您可以使用/ \ b \ / \ b /获得相同的结果。
I think you are getting a little confused by $1 and $2. Those aren't actually doing anything either. They are nil because they are matching nothing (just a word boundry). What you have written is the logical equivalent of .gsub(/\b \/ \b/, " and ")
我认为你会因为1美元和2美元而感到困惑。那些实际上也没有做任何事情。它们是零，因为它们没有匹配（只是一个单词边界）。你写的是.gsub（/ \ b \ / \ b /，“和”）的逻辑等价物
#1
6
The $1 and $2 are not actually related to your regex match: a method's arguments are evaluated before the method is called, so
$ 1和$ 2实际上与你的正则表达式匹配无关：在调用方法之前评估方法的参数，所以
"#{$1} and #{$2}"
Is evaluated before the regex is matched against your string. If you haven't done earlier regex matches then these variables will be nil, so you're actually doing
在正则表达式与字符串匹配之前进行评估。如果你没有做过早期的正则表达式匹配，那么这些变量将是零，所以你实际上在做
file = file.gsub(/(\b) \/ (\b)/, " and ")
that is you are replacing a slash surrounded by spaces by "and", also surrounded by spaces. $1 and $2 will be updated to be empty strings, and so you'll see the same behaviour when you process the next string.
那就是你用“和”替换空格包围的斜线，也用空格包围。 $ 1和$ 2将更新为空字符串，因此当您处理下一个字符串时，您将看到相同的行为。
#2
8
The parentheses aren't doing anything in this context. You could get the same result using /\b \/ \b/.
括号在这种情况下没有做任何事情。您可以使用/ \ b \ / \ b /获得相同的结果。
I think you are getting a little confused by $1 and $2. Those aren't actually doing anything either. They are nil because they are matching nothing (just a word boundry). What you have written is the logical equivalent of .gsub(/\b \/ \b/, " and ")
我认为你会因为1美元和2美元而感到困惑。那些实际上也没有做任何事情。它们是零，因为它们没有匹配（只是一个单词边界）。你写的是.gsub（/ \ b \ / \ b /，“和”）的逻辑等价物