一道关于REGEXP_REPLACE的题目

   The following example examines phone_number, looking for the pattern xxx.xxx.xxxx. Oracle reformats this pattern with (xxx)xxx-xxxx.

SELECT 

       REGEXP_REPLACE(phone_number,

                                        '([[:digit:]]{3})\.([[:digit:]]{3})\.([[:digit:]]{4})' ,              

                                         '(\1) \2-\3') "REGEXP_REPLACE" 

FROM employees; 

REGEXP_REPLACE

 -------------------------------------------------------------------------------- 

[phone] 

[phone]

 [phone] 

[phone]

这一题主要考察了正则表达式函数中元字符“\n"（n是一个正整数）的用法。

Oracle® Database Application Developer's Guide - Fundamentals 10g Release 2 (10.2)对\n的解释如下：

\n

Backreference

Matches the nth preceding subexpression, that is, whatever is grouped within parentheses, where n is an integer from 1 to 9. The parentheses cause an expression to be remembered; a backreference refers to it. A backreference counts subexpressions from left to right, starting with the opening parenthesis of each preceding subexpression. The expression is invalid if the source string contains fewer than n subexpressions preceding the \n. Oracle supports the backreference expression in the regular expression pattern and the replacement string of the REGEXP_REPLACE function.

The expression (abc|def)xy\1 matches the stringsabcxyabc and defxydef, but does not matchabcxydef or abcxy. A backreference enables you to search for a repeated string without knowing the actual string ahead of time. For example, the expression^(.*)\1$ matches a line consisting of two adjacent instances of the same string.

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