如何测试有效的UUID/GUID? - Code

如何检查变量是否包含有效的UUID/GUID标识符?

我目前只对验证类型1和4感兴趣，但这不应该成为您回答的限制。

当前回答

版本1到5，当version被省略时不使用多版本正则表达式。

const uuid_patterns = { 1: /^[0-9A-F]{8}-[0-9A-F]{4}-1[0-9A-F]{3}-[0-9A-F]{4}-[0-9A-F]{12}$/i, 2: /^[0-9A-F]{8}-[0-9A-F]{4}-2[0-9A-F]{3}-[0-9A-F]{4}-[0-9A-F]{12}$/i, 3: /^[0-9A-F]{8}-[0-9A-F]{4}-3[0-9A-F]{3}-[0-9A-F]{4}-[0-9A-F]{12}$/i, 4: /^[0-9A-F]{8}-[0-9A-F]{4}-4[0-9A-F]{3}-[89AB][0-9A-F]{3}-[0-9A-F]{12}$/i, 5: /^[0-9A-F]{8}-[0-9A-F]{4}-5[0-9A-F]{3}-[89AB][0-9A-F]{3}-[0-9A-F]{12}$/i }; const isUUID = (input, version) => { if(typeof input === "string"){ if(Object.keys(uuid_patterns).includes(typeof version === "string" ? version : String(version))){ return uuid_patterns[version].test(input); } else { return Object.values(uuid_patterns).some(pattern => pattern.test(input)); } } return false; } // Testing let valid = [ 'A987FBC9-4BED-3078-CF07-9141BA07C9F3', 'A987FBC9-4BED-4078-8F07-9141BA07C9F3', 'A987FBC9-4BED-5078-AF07-9141BA07C9F3', ]; let invalid = [ '', 'xxxA987FBC9-4BED-3078-CF07-9141BA07C9F3', 'A987FBC9-4BED-3078-CF07-9141BA07C9F3xxx', 'A987FBC94BED3078CF079141BA07C9F3', '934859', '987FBC9-4BED-3078-CF07A-9141BA07C9F3', 'AAAAAAAA-1111-1111-AAAG-111111111111', ]; valid.forEach(test => console.log("Valid case, result: "+isUUID(test))); invalid.forEach(test => console.log("Invalid case, result: "+isUUID(test)));

2021-12-06 05:32:32

其他回答

如果您正在使用Node.js进行开发，建议使用名为Validator的包。它包括验证不同版本的UUID所需的所有正则表达式，还有用于验证的各种其他函数。

这里是npm链接:Validator

var a = 'd3aa88e2-c754-41e0-8ba6-4198a34aa0a2'
v.isUUID(a)
true
v.isUUID('abc')
false
v.isNull(a)
false

2014-08-06 22:43:05

感谢@usertatha做了一些修改

function isUUID ( uuid ) {
    let s = "" + uuid;

    s = s.match('^[0-9a-fA-F]{8}-[0-9a-fA-F]{4}-[0-9a-fA-F]{4}-[0-9a-fA-F]{4}-[0-9a-fA-F]{12}$');
    if (s === null) {
      return false;
    }
    return true;
}

2019-03-13 09:26:24

以上答案稍作修改，以更简洁的方式书写。这将验证任何带有连字符的GUID(但是很容易修改为可选的连字符)。这也将支持大写和小写字符，这已经成为惯例，无论规范:

/^([0-9a-fA-F]{8})-(([0-9a-fA-F]{4}\-){3})([0-9a-fA-F]{12})$/i

这里的关键是下面的重复部分

(([0-9a-fA-F]{4}\-){3})

它只是简单地重复4字符模式3次

2019-12-17 08:48:56

版本1到5，当version被省略时不使用多版本正则表达式。

const uuid_patterns = { 1: /^[0-9A-F]{8}-[0-9A-F]{4}-1[0-9A-F]{3}-[0-9A-F]{4}-[0-9A-F]{12}$/i, 2: /^[0-9A-F]{8}-[0-9A-F]{4}-2[0-9A-F]{3}-[0-9A-F]{4}-[0-9A-F]{12}$/i, 3: /^[0-9A-F]{8}-[0-9A-F]{4}-3[0-9A-F]{3}-[0-9A-F]{4}-[0-9A-F]{12}$/i, 4: /^[0-9A-F]{8}-[0-9A-F]{4}-4[0-9A-F]{3}-[89AB][0-9A-F]{3}-[0-9A-F]{12}$/i, 5: /^[0-9A-F]{8}-[0-9A-F]{4}-5[0-9A-F]{3}-[89AB][0-9A-F]{3}-[0-9A-F]{12}$/i }; const isUUID = (input, version) => { if(typeof input === "string"){ if(Object.keys(uuid_patterns).includes(typeof version === "string" ? version : String(version))){ return uuid_patterns[version].test(input); } else { return Object.values(uuid_patterns).some(pattern => pattern.test(input)); } } return false; } // Testing let valid = [ 'A987FBC9-4BED-3078-CF07-9141BA07C9F3', 'A987FBC9-4BED-4078-8F07-9141BA07C9F3', 'A987FBC9-4BED-5078-AF07-9141BA07C9F3', ]; let invalid = [ '', 'xxxA987FBC9-4BED-3078-CF07-9141BA07C9F3', 'A987FBC9-4BED-3078-CF07-9141BA07C9F3xxx', 'A987FBC94BED3078CF079141BA07C9F3', '934859', '987FBC9-4BED-3078-CF07A-9141BA07C9F3', 'AAAAAAAA-1111-1111-AAAG-111111111111', ]; valid.forEach(test => console.log("Valid case, result: "+isUUID(test))); invalid.forEach(test => console.log("Invalid case, result: "+isUUID(test)));

2021-12-06 05:32:32

我认为更好的方法是使用静态方法fromString来避免那些正则表达式。

    id = UUID.randomUUID();
    UUID uuid = UUID.fromString(id.toString());
    Assert.assertEquals(id.toString(), uuid.toString());

另一方面

   UUID uuidFalse = UUID.fromString("x");

抛出java.lang.IllegalArgumentException:无效UUID字符串:x

2019-03-25 17:05:53

aliyun