http://www.unwrongest.com/projects/airport/“rel=“nofollow noreferer”>jquery Airport plugin,并试图使其接受特殊性质(如accents,“&”)。 否则,它只能与z合作。
是否有任何人发现这样做?
http://www.unwrongest.com/projects/airport/“rel=“nofollow noreferer”>jquery Airport plugin,并试图使其接受特殊性质(如accents,“&”)。 否则,它只能与z合作。
是否有任何人发现这样做?
有一个<代码>chars阵列,可以添加特性。
var chars = [ a , b , c , d , e , f , g , , h , i , j , k , l , m , n , o , p , q , r , s , t , u , v , w , x , y , z , - ];
这样做应当足够。
PS:有时看方案的来源代码,可以真正开明。 ;-
你们必须正确建立果园阵列。
var chars = [ a , ..., z , , - , & , ä , ö , ü ];
请查看这一缺陷。
这些例子使用一些高望重(äöü),但你可能包括任何其他char。
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