[英]Regex - Every spaces without back on line
I would like to accept any spaces with [\\s] between to words, but not returns on the line -> [^\\n].我想接受单词之间带有 [\\s] 的任何空格,但不返回行 -> [^\\n]。
I'm using我正在使用
([\w\-]+)([\s[^\n]])([\w\-]+)
but i doesn't work.但我不工作。
How can I do that?我怎样才能做到这一点?
A character class within another character class will not work.另一个字符类中的字符类将不起作用。 It would be better to specify what you want to match.
最好指定您要匹配的内容。 So
[ \\t]+
which means one or more spaces or tabs
.所以
[ \\t]+
表示one or more spaces or tabs
。
Two more points:还有两点:
1. -
does not need to be escaped in a character class if it is the first or the last character, so [\\w-]
or [-\\w]
will work fine. 1.
-
如果它是第一个或最后一个字符,则不需要在字符类中进行转义,因此[\\w-]
或[-\\w]
可以正常工作。
2. Unless you want to use the white space characters between the words you do not need to use brackets. 2. 除非您想在单词之间使用空格字符,否则您不需要使用括号。
In summary, the following should work fine:总之,以下应该可以正常工作:
([-\\w]+)[ \\t]+([-\\w]+)
try the following regex:尝试以下正则表达式:
([\w\-]+)([\s^[\n]])([\w\-]+)
That should add the "not" you want for new lines.这应该为新行添加您想要的“不”。
cheers!干杯!
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