Finding exact match in vim

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春和景丽
春和景丽 2021-02-02 16:26

Using / or ? enables to find a match for a word in vim. But how can I find an exact match? For example, my text contains the following words: a a

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  • 2021-02-02 16:33

    You enclose the string you are looking for by \< and \> like in /\<aa\> to match exactly that string.

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  • 2021-02-02 16:34

    Bring your cursor to exact word which you want to search.

    Press * (Either Shitf+8 or from numpad of keybord).

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  • 2021-02-02 16:47

    You can search for aa[^a], which will find your two as and nothing else.

    EDIT: oh boy, y'all are after an exact match :-) So, to match exactly two aas and nothing else:

    [^a]\zsaa\ze[^a]\|^\zsaa\ze$\|^\zsaa\ze[^a]\|[^a]\zsaa\ze$
    

    And the branches expanded out:

      [^a]\zsaa\ze[^a]
    \|^\zsaa\ze$
    \|^\zsaa\ze[^a]
    \|[^a]\zsaa\ze$
    

    These cover all contingencies--the aas can be at the beginning, the middle, or the end of any line. And there can't be more than two as together.

    • \zs means the actual match starts here
    • \ze means the actual match ends here
    • the first branch finds aa in a line surrounded by other characters
    • the second branch finds aa when it makes up the whole line
    • the third branch finds aa at the beginning of a line
    • and the fourth branch finds aa at the end of a line.

    My mind boggles at fancy things like look-behind assertions, so I tried to stick to reasonably simple regex concepts.

    EDIT 2: see @benjifisher's simplified, more elegant version below for your intellectual pleasure.

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  • 2021-02-02 16:58

    To find a single, standalone aa (not a, aaa, ...), you need to assert no match of that character both before and afterwards. This is done with negative lookbehind (\@<! in Vim's regular expression syntax) and lookahead (\@!) enclosing the match itself (a\{2}):

    /\%(a\)\@<!a\{2}\%(a\)\@!/
    

    Simplification

    Those assertions are hard to type, if the border around the match is also a non-keyword / keyword border, you can use the shorter \<\a\{2}\> assertions (as in MosteM's answer), but this doesn't work in the general case, e.g. with xaax.

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