regex - 如何匹配重复但不相同的相似组

Question

我有许多与产品相关的字符串。这些中的每一个都有参考号，我想创建一个正则表达式，如果不止一次提到不同的参考号，它会拾取。因此，给出以下示例：

"AB01 MyProduct" >>> No match - because there is only one ID
"AB02 MyOtherProduct" >>> No match - because there is only one ID
"AB03 YetAnotherProduct" >>> No match - because there is only one ID
"AnAccessory for AB01, AB02, AB03 or AB101" >>> Matches!!
"AB02 MyOtherProduct, MyOtherProduct called the AB02" >>> No match - because the codes are the same

谁能给我一个线索？

Answer 1

如果您的正则表达式引擎支持负前瞻，这可以解决问题：

(AB\d+).*?(?!\1)AB\d+

如果有两个序列匹配AB\d+并且第二个序列与第一个序列不同（由负前瞻确保），则它匹配。

解释：

(           # start capture group 1
 AB         # match `AB` literally
 \d+        # match one or more digits
)           # end capture group one
.*?         # match any sequence of characters, non-greedy
(?!         # start negative lookahead, match this position if it does not match
 \1         # whatever was captured in capture group 1
)           # end lookahead
AB          # match `AB` literally
\d+         # match one or more digits

测试（JavaScript）：

> var pattern = /(AB\d+).*?(?!\1)AB\d+/;
> pattern.test("AB01 MyProduct")
  false
> pattern.test("AnAccessory for AB01, AB02, AB03 or AB101")
  true
> pattern.test("AB02 MyOtherProduct, MyOtherProduct called the AB02")
  false