Question

我搜索了这些问题，却找不到答案。我需要一个与php preg_match函数一起使用的模式，只匹配不包含3个或更多连续数字的字符串，例如：

rrfefzef        => TRUE
rrfef12ze1      => TRUE
rrfef1zef1      => TRUE
rrf12efzef231   => FALSE
rrf2341efzef231 => FALSE

到目前为止，我已经写了以下正则表达式：

@^\D*(\d{0,2})?\D*$@

它只匹配只出现一次\d{0,2}

的字符串

如果其他人有时间帮助我，我会很感激：）

此致

Answer 1

如果字符串有两个或多个连续数字，则拒绝该字符串：\d{2,}

或者只有在没有连续数字时才使用否定前瞻匹配：^(?!.*\d{2}).*$

Answer 2

/^(.(?!\d\d\d))+$/

匹配所有未跟随三位数的字符。 Example

Answer 3

您可以搜索\d\d，它将匹配所有错误的字符串。然后你可以调整你的进一步程序逻辑以正确地做出反应。

如果你真的需要对包含相邻数字的字符串进行“肯定”匹配，那么这也应该有效：

^\D?(\d?\D)*$

Answer 4

有没有什么可以阻止你简单地在preg_match()函数前加上“！”，从而反转布尔结果？

!preg_match( '/\d{2,}/' , $subject );

更容易......

Answer 5

如果我正确解释您的要求，则跟随正则表达式匹配您的有效输入而不匹配无效输入。

^\D*\d*\D*\d?(?!\d+)$

解释如下

> # ^\D*\d*\D*\d?(?!\d+)$
> # 
> # Options: case insensitive; ^ and $ match at line breaks
> # 
> # Assert position at the beginning of a line (at beginning of the string or
> after a line break character) «^»
> # Match a single character that is not a digit 0..9 «\D*»
> #    Between zero and unlimited times, as many times as possible, giving back
> as needed (greedy) «*»
> # Match a single digit 0..9 «\d*»
> #    Between zero and unlimited times, as many times as possible, giving back
> as needed (greedy) «*»
> # Match a single character that is not a digit 0..9 «\D*»
> #    Between zero and unlimited times, as many times as possible, giving back
> as needed (greedy) «*»
> # Match a single digit 0..9 «\d?»
> #    Between zero and one times, as many times as possible, giving back as
> needed (greedy) «?»
> # Assert that it is impossible to match the regex below starting at this
> position (negative lookahead)
> «(?!\d+)»
> #    Match a single digit 0..9 «\d+»
> #       Between one and unlimited times, as many times as possible,
> giving back as needed (greedy) «+»
> # Assert position at the end of a line (at the end of the string or before a
> line break character) «$»

正则表达式PCRE：验证不包含3个或更多连续数字的字符串

5 个答案: