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，#和之间的差距.应始终为 1 个或多个字符，并且#始终在前。
您可以断言字符串的开头^，不匹配 a#或.使用否定字符类 [^#.]，然后匹配#。
然后重复该部分，然后重复该部分，然后重复该部分直到字符串的末尾：
^[^#.]*#[^.#]+\.[^.#]*$
解释
^ 字符串的开始[^#.]*#匹配 0+ 次不匹配#或.匹配#[^.#]+\.不匹配 1+ 次.或#匹配一个点[^.#]*匹配 0+ 次不.或#$ 字符串结束let pattern = /^[^#.]*#[^.#]+\.[^.#]*$/;
[
"abc#.123",
"abc#12.3",
"abc#abcd.23",
"a#123.12",
"a#12123124.12",
"abcd#1231.09",
"1.23#12312.01",
"123#122.01#12"
].forEach(s => console.log(s + ": " + pattern.test(s)))本文收集自互联网，转载请注明来源。
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我来说两句