Question

我在这方面做了很多努力。

在MySQL中，我想查询一列以两位数字结尾的字符串。我不在乎之前的其他内容，只要最后一个字符两个数字前面至少有一个非数字

例如，这些字符串应匹配：

"Nov. 12th, 60"

"34 Bar 56"

"Foo-BAR-01"

"33-44-55"

"-------88"

"99"

当我这样做时：

SELECT <string> REGEXP <pattern>

现在，<pattern>位是我需要帮助的。

感谢。

Answer 1

SELECT * FROM mytable WHERE mycolumn REGEXP "^.*[^0-9][0-9]{2}$";

正则表达式解释：

^.*[^0-9][0-9]{2}$

Assert position at the beginning of the string «^»
Match any single character that is NOT a line break character «.*»
   Between zero and unlimited times, as few or as many times as needed to find the longest match in combination with the other quantifiers or alternatives «*»
Match any single character that is NOT present in the list below and that is NOT a line break character «[^0-9]»
   A character in the range between “0” and “9” «0-9»
Match a single character in the range between “0” and “9” «[0-9]{2}»
   Exactly 2 times «{2}»
Assert position at the very end of the string «$»

Answer 2

没有必要描述所有字符串，结尾就足够了：

SELECT 'Nov. 12th, 60' REGEXP '(^|[^[:digit:]])[[:digit:]]{2}$';
SELECT '99' REGEXP '(^|[^[:digit:]])[[:digit:]]{2}$'

简而言之，最后两位数[[:digit:]]{2}$前面是非数字字符[^[:digit:]]，或|字符串^的开头。

注意：你可以用较少的posix风格编写它，它不会改变任何东西（只是更短）：

SELECT 'Nov. 12th, 60' REGEXP '(^|[^0-9])[0-9]{2}$';

mysql REGEXP匹配以2位数字结尾的字符串，前面只有非数字

2 个答案: