在常规的前pression限制字符长度 [英] Restricting Character length in Regular expression

问题描述

我使用下面的正前pression没有任何限制字符长度

I am using the following regular expression without restricting any character length

var test =  /^(a-z|A-Z|0-9)*[^$%^&*;:,<>?()\""\']*$/ //Works Fine

在当我试图将字符长度的限制为15，如下上面，它抛出一个错误。

In above when I am trying to restrict the characters length to 15 as below, it throws an error.

var test =  /^(a-z|A-Z|0-9)*[^$%^&*;:,<>?()\""\']*${1,15}/    //**Uncaught SyntaxError: Invalid regular expression**

请帮我作出上述正则表达式的工作与字符限制为15。

Please help me to make the above regEx work with the characters limit to 15.

推荐答案

您不能量词适用于锚。相反，以限制输入字符串的长度，使用开头挂靠一个前瞻：

You cannot apply quantifiers to anchors. Instead, to restrict the length of the input string, use a lookahead anchored at the beginning:

^(?=.{1,15}$)[a-zA-Z0-9]*[^$%^&*;:,<>?()\"']*$
  ^^^^^^^^^^^

另外，我假设你想匹配0个或多个字母或数字与（A-Z | A-Z | 0-9）* 。它应该看起来像 [A-ZA-Z0-9] * （即这里使用一个字符类）。

Also, I assume you wanted to match 0 or more letters or digits with (a-z|A-Z|0-9)*. It should look like [a-zA-Z0-9]* (i.e. use a character class here).

为什么不使用限制量词，如 {1,15} ，在结束了吗？

Why not use a limiting quantifier, like {1,15}, at the end?

量词仅施加于子模式的左侧，是它的一组或一个字符类，或者一个文字符号。因此， ^ [A-ZA-Z0-9] * [^ $％^＆放大器; *;：？，＆LT;＆GT;（）\\'] {1,15} $ 将有效地限制第二个字符长度类 [^ $％^＆放大器; *;：？，＆LT;＆GT;（）\\'] 来1到15个字符。在 ^（[A-ZA-Z0-9] *？[^ $％^＆放大器; *;：？，＆LT;＆GT;（）\\'] *）{} 1.15 $ 将限制的2子模式无限长度的序列（如 * （和 + ，太）可以匹配的字符数量不受限制），以1〜15倍，我们仍然不限制长度的整个输入字符串的

Quantifiers are only applied to the subpattern to the left, be it a group or a character class, or a literal symbol. Thus, ^[a-zA-Z0-9]*[^$%^&*;:,<>?()\"']{1,15}$ will effectively restrict the length of the second character class [^$%^&*;:,<>?()\"'] to 1 to 15 characters. The ^(?:[a-zA-Z0-9]*[^$%^&*;:,<>?()\"']*){1,15}$ will "restrict" the sequence of 2 subpatterns of unlimited length (as the * (and +, too) can match unlimited number of characters) to 1 to 15 times, and we still do not restrict the length of the whole input string.

如何超前限制工作？

的 的 正向前查找 出现 ^ 启动的字符串的 锚 。这是一个的零宽度断言的，只有返回如果它的子模式后续字符匹配检查之后真或假。所以，这个先行尝试匹配任何1到15（因 <强>限制量词 {1,15} ）字符，但就在字​​符串的结尾换行符（由于 $ 锚）。如果我们从先行删除 $ 锚，超前只需要该字符串的包含的1到15个字符，但总字符串的长度是任何

The (?=.{1,15}$) positive lookahead appears right after ^ start-of-string anchor. It is a zero-width assertion that only returns true or false after checking if its subpattern matches the subsequent characters. So, this lookahead tries to match any 1 to 15 (due to the limiting quantifier {1,15}) characters but a newline right at the end of the string (due to the $ anchor). If we remove the $ anchor from the lookahead, the lookahead will only require the string to contain 1 to 15 characters, but the total string length can be any.

如果输入的字符串可以包含一个新行序列，你应该使用 [\\ S \\ S] 便携式任何字符的正则表达式结构（这将在JS和其他共同工作正则表达式口味）：

If the input string can contain a newline sequence, you should use [\s\S] portable any-character regex construct (it will work in JS and other common regex flavors):

^(?=[\s\S]{1,15}$)[a-zA-Z0-9]*[^$%^&*;:,<>?()\"']*$
 ^^^^^^^^^^^^^^^^^

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