javascript 从范围中排除某些字符 - javascript正则表达式
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Excluding some character from a range - javascript regular expression
提问by Saif
To validate only word simplest regex would be (I think)
仅验证单词最简单的正则表达式是（我认为）
/^\w+$/
I want to exclude digits and _from this (as it accept aa10aaand aa_aa now, I want to reject them)
我想从中排除数字_（因为它aa10aa现在接受和 aa_aa，我想拒绝它们）
I think it can be gained by
我认为它可以通过
/^[a-zA-z]+$/
which means I have to take a different approach other than the previous one.
这意味着我必须采取不同于前一种方法的方法。
but what if I want to exclude any character from this range
suppose I will not allow k,K,p,Por more.
但是如果我想从这个范围中排除任何字符，假设我不允许k, K, p,P或更多怎么办。
Is there a way to add an excluding list in the range without changing the range.?
有没有办法在不改变范围的情况下在范围内添加排除列表。？
回答by Avinash Raj
To exclude kor pfrom [a-zA-Z]you need to use a negative lookahead assertion.
要排除k或p从[a-zA-Z]您需要使用否定前瞻断言。
(?![kpKP])[a-zA-Z]+
Use anchors if necessary.
必要时使用锚点。
^(?:(?![kpKP])[a-zA-Z])+$
It checks for not of kor pbefore matching each character.
它检查不是k或p每个字符匹配之前。
OR
或者
^(?!.*[kpKP])[a-zA-Z]+$
It just excludes the lines which contains kor pand matches only those lines which contains only alphabets other than kor p.
它只是排除其中包含线k或p与只匹配那些只包含字母以外的其他线路k或p。
回答by vks
^(?!.*(?:p|k))[a-zA-Z]+$
This should do it.See demo.The negative lookahead will assert that matching word has no por k.Use imodifier as well.
这应该可以做到。参见演示。否定前瞻将断言匹配的单词也没有p或k.Usei修饰符。
https://regex101.com/r/vD5iH9/31
https://regex101.com/r/vD5iH9/31
var re = /^(?!.*(?:p|k))[a-zA-Z]+$/gmi;
var str = 'One\nTwo\n\nFour\nFourp\nFourk';
var m;
while ((m = re.exec(str)) != null) {
if (m.index === re.lastIndex) {
re.lastIndex++;
}
// View your result using the m-variable.
// eg m[0] etc.
}
