Javascript reg ex仅匹配整个单词，仅由空格绑定 [英] Javascript reg ex to match whole word only, bound only by whitespace

问题描述

所以我知道\ bblah \ b将匹配整个Blah，但它也会匹配Blah的Blah.jpg我不想要它。我想只匹配整个单词和两边的空格。

so I know \bBlah\b will match a whole Blah, however it will also match Blah in "Blah.jpg" I don't want it to. I want to match only whole words with a space on either side.

推荐答案

你可以尝试： \\ \\ sBlah \s 。

或者，如果您允许开始和结束锚点，（^ | \ s） Blah（\ s | $）

Or if you allow beginning and end anchors, (^|\s)Blah(\s|$)

这将匹配Blah by本身，或每个 Blah in Blah and Blah

This will match "Blah" by itself, or each Blah in "Blah and Blah"

• regular-expressions.info/Character classes 和 Anchors
  
  
  
  
  • \s 代表空白字符。
  
  
  • 插入符 ^ 匹配字符串中第一个字符前的位置
  
  
  • 同样， $ 在字符串中的最后一个字符后面匹配
  
  
  • regular-expressions.info/Character classes and Anchors
    • \s stands for "whitespace character".
    • The caret ^ matches the position before the first character in the string
    • Similarly, $ matches right after the last character in the string

    如果你想在Blah Blah中匹配 Blah ，然后由于在两次出现之间共享一个空格，您必须使用断言。类似于：

    If you want to match both Blah in "Blah Blah", then since the one space is "shared" between the two occurrences, you must use assertions. Something like:

    (^|\s)Blah(?=\s|$)
    

    
    
    

    参见

    
    
    
    
    
    • regular-expressions.info/Lookarounds
    
    

    See also

    • regular-expressions.info/Lookarounds

    上面的正则表达式也会匹配前导空格。

    The above regex would also match the leading whitespace.

    如果你只想要 Blah ，理想情况下，lookbehind将会很好：

    If you want only Blah, ideally, lookbehind would've been nice:

    (?<=^|\s)Blah(?=\s|$)
    

    但是由于Javascript不支持它，你可以改写：

    But since Javascript doesn't support it, you can instead write:

    (?:^|\s)(Blah)(?=\s|$)
    

    现在 Blah 将在 \1 中捕获，没有领先的空白。

    Now Blah would be captured in \1, with no leading whitespace.

    
    
    • regular-expressions.info/Grouping 和风味比较
    
    
    • regular-expressions.info/Grouping and flavor comparison

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