regexp_replace 替换字符串的错误部分

Question

运行时:

select regexp_replace('( (test :Name (x) :Table (y) )','\s+\:Name \(.*?\)',' avner ');

我得到:

"( (test avner "

但是如果我运行:

select regexp_replace('( (test :Name (x) :Table (y) )','\:Name \(.*?\)',' avner ');

我得到:

"( (test  avner  :Table (y) )"

为什么开头的\s+会匹配到字符串的结尾？

Answer 1

原因是(per documentation) :

A branch — that is, an RE that has no top-level | operator — has the same greediness as the first quantified atom in it that has a greediness attribute.

大胆强调我的。将您的问题简化为:

SELECT substring('( (test :Name (x) :Table (y) )', '\s+\:Name \(.*?\)')
      ,substring('( (test :Name (x) :Table (y) )',    '\:Name \(.*?\)')

如果您希望第二个量词是非贪婪的，请将第一个量词也更改为非贪婪。特别是，因为这不会改变任何东西:

SELECT substring('( (test :Name (x) :Table (y) )', '\s+<b>?</b>:Name \(.*?\)')

而且不需要转义冒号(:)。

SQL Fiddle.