正则表达式Oracle中的子表达式

Question

Why do we need a subexpression in regular expressions in Oracle?

It is new feature of Oracle 11g, we can specify which subexpression from pattern we want to find. We can use this parameter in REGEXP_SUBSTR and REGEXP_INSTR .

Here is an example from docs:

SELECT REGEXP_INSTR('1234567890', '(123)(4(56)(78))', 1, 1, 0, 'i', 2) 
"REGEXP_INSTR" FROM DUAL;

REGEXP_INSTR
-------------------
4

But it isn't clear to me, why do we really need this parameter (in the above example we could just use 4(56)(78) pattern). Does anyone have real-world example?

Answer 1

Oracle regex does not support lookaround . Just like ^ and $ anchors can be used to specify start and end of the string, lookarounds(lookbehind/lookahead) can be used to match(or not) any pattern either preceding or succeeding the pattern that you are interested in.

For example consider the following values in a column and you need to select pickup date only.

event_dte
----------------------
pickup_dte 2015-04-03
shipped_dte 2015-03-02
PU_dte 2015-03-11
pickup_date 2014-05-02
delivery_dte 2015-07-11

There are all possible dates in this and the wording of pickup date is also not consistent.

You can write a regex like (pickup|PU)_d(a?)te (\\d{4}-]d{2}-\\d{2}) . This will match the entire string and when used in regexp_substr will return entire string. If you use subexpressions you can extract only the date part. For the above example, it will be third sub expression.

Answer 2

Well, I figured that out, so if someone interested in that is my answer:

We use subexpression when we want to find a specific string, which follows some string, which also could follows some string etc. (or otherwise, find a string, which followed by some strings)

So for upper example I modify source string:

SELECT REGEXP_INSTR('456781234567890', '(123)(4(56)(78))', 1, 1, 0, 'i', 2) 
"REGEXP_INSTR" FROM DUAL;

REGEXP_INSTR
-------------------
9

Here we will find position of string 45678 only if it follows 123 . And we get 9 instead of 1