Question

我做了一个非常基本的步骤，检查一个特殊字符的存在。下一步需要一些建议，因为我希望能够在找到＃后从1个位置开始搜索另一个特殊字符。

var reg = /#/;
alert(reg.test(string))

例如：

abc#.123     //invalid - as . is immediately after #
abc#12.3     //valid  - as . is more than 1 character after #
abc#abcd.23  //valid - as . is more than 1 character after #
a#123.12     //valid - as . is more than 1 character after #
a#12123124.12 //valid - as . is more than 1 character after #
abcd#1231.09  //valid - as . is more than 1 character after #
1.23#12312.01 //invalid - as . is before #
123#122.01#12 //invalid - as there is another# after .

因此#和.之间的间距应始终为1个或多个字符，而#始终在前。

Answer 1

您可以断言字符串^的开头，不使用negated character class #匹配.或[^#.]，然后匹配{{1} }。

然后重复该部分，然后重复点，然后重复该部分直到字符串的结尾：

Regex demo

说明

^[^#.]*#[^.#]+\.[^.#]*$字符串的开头
^匹配0次以上而不是[^#.]*#或#，然后匹配.
#匹配1次以上而不是[^.#]+\.或.，然后匹配一个点
#匹配0+次，而不是[^.#]*或.
#字符串结尾

Answer 2

您可以使用/^[^\.#]*#[^\.#]+\.[^\.#]*$/。

^  beginning of line anchor
 [^\.#]*  zero or more characters other than . and #
        #  literal # character
         [^\.#]+  one or more characters other than . and #
                \.  literal . character
                  [^\.#]*  one or more characters other than . and #
                         $  EOL

通常，如果您想要一个特定大小的最小间隙（在这种情况下为5或更大），请使用/^[^\.#]*#[^\.#]{5,}\.[^#\.]*$/；如果您希望间隙恰好为5，请使用{5}。

var reg = /^[^\.#]*#[^\.#]+\.[^\.#]*$/;

[
  "abc#.123",      // invalid - as . is immediately after #
  "abc#12.3",      // valid - as . is more than 1 character after #
  "abc#abcd.23",   // valid - as . is more than 1 character after #
  "a#123.12",      // valid - as . is more than 1 character after #
  "a#12123124.12", // valid - as . is more than 1 character after #
  "abcd#1231.09",  // valid - as . is more than 1 character after #
  "1.23#12312.01", // invalid - as . is before #
  "123#122.01#12", // invalid - as there is another# after .
].forEach(test => console.log(reg.test(test)));

Answer 3

您可以使用此正则表达式确保字符串中的第一个#后跟一个点.，其中#和.之间的间隔是一个或多个字符，

^[^#]*#[^.]+\.

说明：

^-字符串的开头
[^#]*-除#以外的零个或多个字符，并且如果您希望点也不出现在#之前，则可以将其更改为[^#.]*
#-匹配第一个#字符
[^.]+-匹配点号以外的一个或多个字符
\.-匹配文字点

Regex Demo

JS代码演示

const arr = ['abc#.123','abc#12.3','abc#abcd.23','a#123.12','a#..dd']

arr.forEach(s => console.log(s + " --> " + /^[^#]*#[^.]+\./.test(s)))

RegEx用于检查字符X在另一个字符之后的位置

3 个答案: