匹配 UID 的正则表达式

Question

我正在尝试验证具有以下规则的 UID

它必须包含至少 2 个大写英文字母字符。
它必须包含至少 3 个数字 (0-9)。
它应该只包含字母数字字符。
任何字符都不应重复。
有效的 UID 中必须恰好有 10 个字符。

我试过用正则表达式来做，但我无法创建一个可以满足所有规则的模式。

所以我没有使用正则表达式。

#import re
#pattern = r'^[A-Z0-9]{10}$'

testCases = int(input())

for _ in range(testCases):
    uid = input()
    if len(uid) == 10 and len(uid) == len(set(uid)):
       countChr = countNum = 0
       for i in uid:
           if i.isdigit():
               countNum += 1
           elif i.isalpha():
               countChr +=1
       if countChr >= 2 and countNum >= 3:
           print("Valid")
       else:
           print("Invalid")
   else:
       print("Invalid")

上面的程序完美运行，但我想用正则表达式验证 UID，那么有什么模式可以满足所有给定的规则吗？

Answer 1

let regex = /^(?!.*?(.).*?)(?=(?:.*?[A-Z]){2,})(?=(?:.*?\d){3,})[a-zA-Z\d]{10}$/;

console.log(regex.test("A0ieu5Wsl2"));
console.log(regex.test("A0ieuiWsl2"));
console.log(regex.test("A0ieu5Wsl2l"));

Answer 2

试试这个正则表达式：

^(?=(?:.*[A-Z]){2})(?=(?:.*[0-9]){3})(?:([a-zA-Z0-9])(?!.*)){10}$

^                           beginning of the text

    (?=                     positive lookahead
        (?:                 non-capturing group
            .*[A-Z]         anything followed by an uppercased letter
        ){2}                match 2 of them
    )                       end of the lookahead

    (?=                     positive lookahead
        (?:                 non-capturing group
            .*[0-9]         anything followed by a digit
        ){3}                match 3 of them
    )                       end of the lookahead

    (?:                     non-capturing group
        ([a-zA-Z0-9])       any alphanumeric character and put it in group 1
        (?!.*)            negative lookahead, make sure there's no repeated character ahead
    ){10}                   match 10 of them
    
$                           end of the text

勾选test cases

使用python

import re

pattern = r'^(?=(?:.*[A-Z]){2})(?=(?:.*[0-9]){3})(?:([a-zA-Z0-9])(?!.*)){10}$'
testCases = int(input())

for _ in range(testCases):
    uid = input()
    if re.match(pattern, uid):
        print("Valid")
    else:
        print("Invalid")

匹配 UID 的正则表达式

Regular expression to match an UID

regex

validation

python-3.x

python-re