我正在尝试使用以下要求验证文本输入：
允许使用字符＆amp;长度/^\w{8,15}$/
必须包含/[a-z]+/
必须包含/[A-Z]+/
必须包含/[0-9]+/
不得包含重复的字符（例如。aba=pass和aab=fail）
与.test()一起使用时，每个测试都会返回true。
熟悉程度不高，我能够编写前4个测试，尽管是单独的。第五次测试没有成功，否定前瞻（这是我认为我需要使用的）是具有挑战性的。
以下是一些价值/结果示例：
re.test("Fail1");//returns false, too short
re.test("StringFailsRule1");//returns false, too long
re.test("Fail!");//returns false, invalid !
re.test("FAILRULE2");//returns false, missing [a-z]+
re.test("failrule3");//returns false, missing [A-Z]+
re.test("failRuleFour");//returns false, missing [0-9]+
re.test("failRule55");//returns false, repeat of "5"
re.test("TestValue1");//returns true
最后，理想的是用于强制执行所有要求的单一组合测试。
答案 0 :(得分：0)
这为测试使用负向和正向前导零长度断言，.{8,15}位验证长度。
^(?!.*(.)\1)(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9])\w{8,15}$
对于你的第五条规则，我使用了一个负面的先行，以确保任何角色的捕获组永远不会被它自己跟随。
NODE EXPLANATION
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^ the beginning of the string
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(?! look ahead to see if there is not:
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.* any character except \n (0 or more times
(matching the most amount possible))
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( group and capture to \1:
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. any character except \n
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) end of \1
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\1 what was matched by capture \1
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) end of look-ahead
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(?= look ahead to see if there is:
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.* any character except \n (0 or more times
(matching the most amount possible))
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[a-z] any character of: 'a' to 'z'
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) end of look-ahead
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(?= look ahead to see if there is:
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.* any character except \n (0 or more times
(matching the most amount possible))
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[A-Z] any character of: 'A' to 'Z'
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) end of look-ahead
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(?= look ahead to see if there is:
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.* any character except \n (0 or more times
(matching the most amount possible))
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[0-9] any character of: '0' to '9'
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) end of look-ahead
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\w{8,15} word characters (a-z, A-Z, 0-9, _)
(between 8 and 15 times (matching the most
amount possible))
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$ before an optional \n, and the end of the
string