[英]Match anything but this group of characters?
I'm looking for a pattern that will match anything but this: 我正在寻找一个可以匹配任何东西的模式,但是:
"/[a-z0-9]/i"
How can i do that? 我怎样才能做到这一点?
With ^
, which negates a character set. 使用
^
,它否定字符集。
You want /[^a-z0-9]/i
. 您想要
/[^a-z0-9]/i
。
[az]
matches a..z. [az]
匹配a..z。 [^az]
matches anything not a..z. [^az]
匹配非 a..z的任何内容。
/[^a-z0-9]/i
The ^
at the start inside a [ ]
means "none of these" rather than the conventional "any of these" [ ]
内开头的^
表示“这些都不是”,而不是常规的“其中任何一个”
If he wants something matches not a-z0-9 one time, this is it. 如果他希望某件事与a-z0-9一次不匹配,那就是。 this matches spaces, everything that isn't a-z0-9, one time
这会匹配空格,不是a-z0-9的所有内容,一次
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.