In javascript, I would like to get a regular expression of the sentence ending with "." in the part that is not surrounded by "to" from the text below.
Ayeo
[
AIUEO。
kakikukeko。
]
Hi, everyone.
goodbye。
hello
The method I tried was to exclude the regular expression pattern that matches in [ ] with a negative look-ahead, and the result was
in [ ].
Oh, my. Kakikakeko.
It also matches the .
/^(?!\[[\s\S]*?\].*?$/gm
How do I write to satisfy my purpose?
javascript regular-expression
The flag is m, so ^$ matches the beginning and end of the line.
So the question /^(?!\[[\s\S]*?\).*?$/gm hits all lines that do not have at the end of the line, regardless of the negative look-ahead..
For example, if you rewrite to , the line will also be a hit.
It is difficult to determine . at each end of each line while excluding [].
Like @cubick's comment, I think the proper way to write is to first exclude parentheses without using the m flag and then use the m flag to get matches for each line.
vars="Aiueo\r\n"+
US>"[\r\n" +
US>"Aye.\r\n" +
US>"Scratch.\r\n" +
US>" ]\r\n" +
"Hello,\r\n"+
US>"Goodbye\r\n"
"Hello";
r=s.replace(/^([\s\S])*)\[[\s\S]*\]([\s\S]*)$/, "$1$2");
m=r.match(/^.+.$/gm);
console.log(m);
Is it like this?
vars=
Nice to meet you.\n"+
US>"Aye\n"+
US>"[\n"+
US>"Aye.\n"
US>"Kakikakeko\n"+
US>"]\n" +
"Hello,\n"+
US>"[\n"+
"Stand up.\n" +
US>"]\n" +
US>"Goodbye\n"
"Hello";
var regex=/(?<![^\]]*)^[^\[\]]*.$/gm;
var match=s.match(regex);
console.log(match);"After [, ] must not appear" is a negative look-ahead.
In a simple example, I think it works (although Safari didn't do a negative look-ahead…), but I think it would be easier to understand and better to say [ and ].
© 2024 OneMinuteCode. All rights reserved.