javascript - 如何检查给定的字符串是否包含某些指定的特殊字符在javascript中是否存在？

Question

当用户在字符串中输入数据时，我有一个输入字符串，然后找出该字符串是否包含某些指定的(!@#$%)特殊字符。以下字符串值输出如

 string str="Mn@";    --> true
 string str="m@*";    --> false
 string str="@Mn";    --> true
 string str="Mn&";    --> false
 string str="@";      --> true
 string str="M";      --> false
 string str="*";      --> false
 string str=" ";      --> false
 string str=" Mn";    --> false
 string str="M *";    --> false
 string str="m@ ";    --> false    

Answer 1

使用正则表达式，将要搜索的所有字符放入一个字符集中:

const pattern = /[!@#$%]/;
console.log(pattern.test('Mn@'));
console.log(pattern.test('@Mn'));
console.log(pattern.test('Mn&'));
console.log(pattern.test('@'));
console.log(pattern.test('M'));
console.log(pattern.test('*'));

对于您的新问题，如果您想确保文本不包含空格，请为空格添加否定前瞻:

const pattern = /^(?!.* )[!@#$%]/;
console.log(pattern.test('Mn@'));
console.log(pattern.test('@Mn'));
console.log(pattern.test('Mn&'));
console.log(pattern.test('@'));
console.log(pattern.test('M'));
console.log(pattern.test('*'));
console.log(pattern.test(' '));
console.log(pattern.test(' Mn'));
console.log(pattern.test('M *'));
console.log(pattern.test('m@ *'));