年份序列的正则表达式

发布于 2024-12-11 20:04:38 字数 288 浏览 0 评论 0原文

所以我试图匹配类似

2011,2012,2013

或

2011,

或

但不是：

2011,201 或 2011,201,2012

我尝试使用 ([0-9]{ 4},?([0-9]{4},?)*) 但如果第一年匹配，则不会考虑其余年份。

原文

So I'm trying to match something like

2011,2012,2013

2011,

but NOT:

2011,201 or 2011,201,2012

I tried with ([0-9]{4},?([0-9]{4},?)*) but if the first year is matched, it does not consider the rest .

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倾｀听者〃 2024-12-18 20:04:38

你很接近。

 ^[0-9]{4}(?:,[0-9]{4})*,?$

这将匹配由 4 位数字和逗号的重复序列组成的任何字符串。

^ 和 $ 分别匹配字符串的开头和结尾。因此，仅当字符串仅由这些元素组成时，它才会匹配。

(?:) 是非捕获组。它允许您创建重复组，而无需将所有组存储到变量中。

编辑：忘记了末尾的可选逗号。添加了 ,? 来处理它。

编辑2：根据FailedDev的建议，这是我最初的想法。它也有效，但我认为它更难理解。它更聪明，但这并不总是一件好事。

^(:?[0-9]{4}(?:,|$))+$

You were close.

 ^[0-9]{4}(?:,[0-9]{4})*,?$

That will match any string consisting of a repeating sequence of 4-digit numbers and commas.

The ^ and $ match the beginning and end of the string, respectively. Thus, it will only match if the string consists of only those elements.

The (?:) is a non-capture group. It allows you to create repeating groups without storing all of them into variables.

EDIT: Forgot about the optional comma at the end. Added a ,? to take care of it.

EDIT 2: At FailedDev's advice, here was my original idea. It also works, but I think it is harder to understand. It is more clever, but that's not always a good thing.

^(:?[0-9]{4}(?:,|$))+$

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ㄟ。诗瑗 2024-12-18 20:04:38

if (subject.match(/^\d{4}(?:,?|(?:,\d{4}))+$/)) {
    // Successful match
}

这应该有效。

说明：

"^" +              // Assert position at the beginning of the string
"\\d" +             // Match a single digit 0..9
   "{4}" +            // Exactly 4 times
"(?:" +            // Match the regular expression below
   "|" +              // Match either the regular expression below (attempting the next alternative only if this one fails)
      "," +              // Match the character “,” literally
         "?" +              // Between zero and one times, as many times as possible, giving back as needed (greedy)
   "|" +              // Or match regular expression number 2 below (the entire group fails if this one fails to match)
      "(?:" +            // Match the regular expression below
         "," +              // Match the character “,” literally
         "\\d" +             // Match a single digit 0..9
            "{4}" +            // Exactly 4 times
      ")" +
")+" +             // Between one and unlimited times, as many times as possible, giving back as needed (greedy)
"$"                // Assert position at the end of the string (or before the line break at the end of the string, if any)

if (subject.match(/^\d{4}(?:,?|(?:,\d{4}))+$/)) {
    // Successful match
}

This should work.

Explanation :

"^" +              // Assert position at the beginning of the string
"\\d" +             // Match a single digit 0..9
   "{4}" +            // Exactly 4 times
"(?:" +            // Match the regular expression below
   "|" +              // Match either the regular expression below (attempting the next alternative only if this one fails)
      "," +              // Match the character “,” literally
         "?" +              // Between zero and one times, as many times as possible, giving back as needed (greedy)
   "|" +              // Or match regular expression number 2 below (the entire group fails if this one fails to match)
      "(?:" +            // Match the regular expression below
         "," +              // Match the character “,” literally
         "\\d" +             // Match a single digit 0..9
            "{4}" +            // Exactly 4 times
      ")" +
")+" +             // Between one and unlimited times, as many times as possible, giving back as needed (greedy)
"$"                // Assert position at the end of the string (or before the line break at the end of the string, if any)

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