正则表达式字符串取反 怎么弄?
正则表达式 正则 url
我有一个url规则如下:mm.[*|a|b|add].cn
其中 a , b , add 表示例外
例如
mm.a.cn 无法匹配
mm.x.cn 可以匹配
如何写正则???
我只知道 mm\.[a|b|add]\.cn 可以匹配所有的例外,但是我就是想要一个取反的结果?
回复讨论(解决方案)
$s1= 'mm.a.cn';$s2= 'mm.add.cn';$s3= 'mm.c.cn';$p='/mm\.(.+)(?<!a|b|add)\.cn/';$bool1=preg_match($p,$s1);$bool2=preg_match($p,$s2);$bool3=preg_match($p,$s3);var_dump($bool1,$bool2,$bool3);
int(0) int(0) int(1)
preg_match('#www\.(a|b|add)\.cn#',$str)
要不匹配的话
!preg_match('#www\.(a|b|add)\.cn#',$str)
只能对LZ说??这个正则很难写的
$s1= 'mm.a.cn';$s2= 'mm.add.cn';$s3= 'mm.c.cn';$p='/mm\.(.+)(?<!a|b|add)\.cn/';$bool1=preg_match($p,$s1);$bool2=preg_match($p,$s2);$bool3=preg_match($p,$s3);var_dump($bool1,$bool2,$bool3);
int(0) int(0) int(1)
试试 mm.aadd.cn ?
$s =<<< TXTmm.a.cn mm.x.cn mm.aadd.cnmm.c.cnTXT;preg_match_all('/mm\.(?!a|b|add).*?\.cn/s', $s, $r);print_r($r);preg_match_all('/mm\.(?!a\.|b\.|add\.).*?\.cn/s', $s, $r);print_r($r);
(
[0] => Array
(
[0] => mm.x.cn
[1] => mm.c.cn
)
)
Array
(
[0] => Array
(
[0] => mm.x.cn
[1] => mm.aadd.cn
[2] => mm.c.cn
)
)
preg_match_all('/mm\.(?!a\.|b\.|add\.).*?\.cn/s', $s, $r);
可写作
preg_match_all('/mm\.(?!(?:a|b|add)\.).*?\.cn/s', $s, $r);
preg_match_all('/mm\.(?!a\.|b\.|add\.).*?\.cn/s', $s, $r);
可写作
preg_match_all('/mm\.(?!(?:a|b|add)\.).*?\.cn/s', $s, $r);
把后一个字符也加进去,整体作为排除对象啊?受教了
四楼应该可以啊,先结贴了。
无意中看到这个贴的。。
首先。。我不懂PHP,作为正则的角度来说,这个不是很简单么。。
反义排除掉这几个字符就行了:
^mm\.[^ab(?:add)]+\.cn$
[^ab(?:add)] --> 排除a b add
preg_match_all('/mm\.(?!a\.|b\.|add\.).*?\.cn/s', $s, $r);
可写作
preg_match_all('/mm\.(?!(?:a|b|add)\.).*?\.cn/s', $s, $r);
版主这个写法是不是有点多余~

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