问题描述
Vim 中什么正则表达式会匹配所有在下面的刺中最多no.的字符?
What regular expression in Vim will match allcharacters up to no. in the below sting?
foo, bar, and baz, no. 13, qux
换句话说，匹配必须是foo, bar, and baz,
我是正则表达式的新手，在 Stack 上做了我的研究溢出，一个建议是 .+?(?= no.).但它似乎有效仅具有正则表达式的正则(perl?)风味，而不是
I'm new with regular expressions, did my research here on StackOverflow, and one suggestion is .+?(?= no.). But it seems to workonly with the regular (perl?) flavour of regular expressions, not inVim.
请帮忙.
对于我想要实现的目标，也许有一个更简单的解决方案.我的最终目标是将 foo、bar 和 baz 放在大括号中，并且我计划使用包含正则表达式的全局替换命令.
Maybe there's a simpler solution to what I'm trying to achieve. My ultimate goal is to put foo, bar, and baz in curly brackets, and I planned to to a global substitution command involving regular expressions.
推荐答案
你可以试试
:s/\v(.*)(, no\.)/{\1}\2/
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