REGEXP_LIKE Confusion

Hi I want to create a function using regexp_like.

In seeking,

SQL> WITH t1 AS (SELECT '[email protected]' email FROM DUAL 2 UNION 3 SELECT '[email protected]' email FROM DUAL 4 UNION 5 SELECT 'test3@[email protected]' email FROM DUAL) 6 SELECT email 7 FROM t1 8 WHERE REGEXP_LIKE (EMAIL, 9 '^[A-Za-z0-9._%+-]+@[A-Za-z0-9.-]+\.[A-Za-z]{2,4}$'); EMAIL --------------------- [email protected] [email protected]

Now i trying to create function on that.

declare

p BOOLEAN

BEGIN

P := REGEXP_LIKE ('test3@[email protected]', '^[A-Za-z0-9._%+-]+@[A-Za-z0-9.-]+\.[A-Za-z]{2,4}$');

IF TRUE THEN

DBMS_OUTPUT.PUT_LINE('T');

ELSE

DBMS_OUTPUT.PUT_LINE('F');

END IF;

END;

But always i get answer 'T'.....but if u see this email is invalid so i have to get 'F' insted of 'T'

So may i know why i get 'T' always ??

Thanks

Swapnil

First of all, you can not assign regexp_like<...> to a variable.

Secondly you always compare IF (TRUE) which means its always true forever, he penetrates inside and Don 't' on the screen.

Maybe something like that is what you are looking for.

set serveroutput on

DECLARE
   p   NUMBER;
BEGIN
   SELECT 1
     INTO p
     FROM DUAL
    WHERE REGEXP_LIKE ('test3@[email protected]',
                       '^[A-Za-z0-9._%+-]+@[A-Za-z0-9.-]+\.[A-Za-z]{2,4}$');
   IF (p = 1)
   THEN
      DBMS_OUTPUT.PUT_LINE ('T');
   END IF;
EXCEPTION
   WHEN NO_DATA_FOUND
   THEN
      DBMS_OUTPUT.PUT_LINE ('F');
END;

See you soon,.

Manik.

Tags: Database

Similar Questions

Work with REGEXP_LIKE
Hello
I am confused with the following example of REGEXP_LIKE
```
create table test_like (nm1 varchar2(11))

insert into test_like values ('AUS' ) ;
insert into test_like values ('ATS' ) ;

select * from test_like where regexp_like(nm1,'*AT*');
```
The statement SELECT abvoe gives 2records. I don't expect that one folder must be returned by the query above
Please give me a hint on this

Thank you
You don't need *.
```
select * from test_like where regexp_like(nm1,'AT');
```
or
```
SQL> select * from test_like where regexp_like(nm1,'.*AT.*');

NM1
-----------
ATS
```
Published by: JAC on May 22, 2013 16:19

Your model (* AT *) means, whatever it is - A then - then T zero or more times.

Help REGEXP_LIKE
Hello

Browsed the web looking for assistance using the syntax REGEXP_LIKE. Did not have a precise answer, but I think that I wrote my own correct and just wanted to get validated by other users.

Running Oracle 10.2

Sample data:
```
create table test(data varchar2(10));
    insert into test values('01W00012');
    insert into test values('50321070');
    insert into test values('A1391010');
    insert into test values('ZZ260010');
    insert into test values('29374C01');
    insert into test values('A938523A');
    insert into test values('23W48AC3');
    insert into test values('DKFUGHCK');
    insert into test values('09W00032');
    insert into test values('94857283');
    insert into test values('AB-29348');
    insert into test values('98W-8923');
    insert into test values('0AW00005');
    insert into test values('04300W05');
    commit;
```
What I'm doing:
I have a table that contains millions of work orders. They are all of length 8. I want to find all the good work where the first 2 characters are a number only, the third character is a 'W' and the last 5 characters are only numbers. Another thing I want to throw. I think that I came up with an expression of work... but I'm always hesitant when he runs against millions of rows. That's what I did:
```
select * from test
    where regexp_like(data, '(^[0-9]{2}W{1}[[:digit:]]{5})');
```
There are has exactly 2 occurrences of room above that match the criteria that I try to answer.
This expression is correctly written?

Any help would be appreciated... reg expressions always make my turn head :(
Hello

dvsoukup wrote:
Running Oracle 10.2

Samples:...

Thanks for posting this; It is really useful!
What I'm doing:
I have a table that contains millions of work orders. They are all of length 8. I want to find all the good work where the first 2 characters are a number only, the third character is a 'W' and the last 5 characters are only numbers. Another thing I want to throw. I think that I came up with an expression of work... but I'm always hesitant when he runs against millions of rows. That's what I did:
```
select * from test
where regexp_like(data, '(^[0-9]{2}W{1}[[:digit:]]{5})');
```
There are has exactly 2 occurrences of room above that match the criteria that I try to answer.
This expression is correctly written?
Yes, it's a good regular expression. However, one very important thing about regular expressions is that you should only use the when you actually need. In this case, you can get the results you want more effectively and at least as easily, without regular expressions. Here's one way:
```
SELECT  *
FROM     test
WHERE   TRANSLATE ( data
            , '012345678'
            , '999999999'
            )          = '99W99999'
;
```
It will be more effective than the use of REGEXP_LIKE, and it seems that efficiency is important in this case.

If you really want to use a regular expression, there are a few things you can do to make it more clear. (With regular expressions, nothing overlooked.)
(a) If you are sure that the data is always exactly 8 characters, then you don't need the ^ anchor at the beginning. If you want the regular expression to verify that the length is 8 characters, then keep the ^ anchor at the beginning, but also, to use an anchor $ at the end, as in the example below. As you posted it, the expression is TRUE when data are greater of 8 characters, as long the first 8 characters fit the pattern.
(b) [0-9] is equivalent to [[: digit:]]. Either it's good, but I find it confusing to use both in the same expression. Use one or the other in both places.
(c) {1} is the default value. I would say just 'W' instead of 'W {1} '.
(d) you don't need parentheses in this expression. If parentheses are not required, but they improve the clarity, then it's OK to include them. In this case, I don't think they add anything.

There are has exactly 2 occurrences of room above that match the criteria that I try to answer.

It helps if you post exactly what results you want. For example:
```
DATA
----------
01W00012
09W00032
```
In this case, I found your description very clear, but it does not hurt to post actual results anyway.

This expression is correctly written?

You can consider writing this way, especially if you think you'll need to debug:
```
SELECT  *
FROM      test
WHERE      REGEXP_LIKE ( data
              , '^'          || -- beginning of string
                '[0-9]{2}'     || -- 2 digits
                'W'          || -- a capital W
                '[0-9]{5}'     || -- 5 more digits
                '$'             -- end of string
              )
;
```
As previously mentioned, you only need the ^ and $ if you are not sure that all the strings are exactly 8 characters.

... reg expressions always make my head spin

All the more reason to use TRANSLATE.

Published by: Frank Kulash, 26 July 2012 14:42

Address with REGEXP_LIKE

Hey all,.
I am a newbie to REGEXP and have a problem where I have 2 spatial tables, for each object in table1 and do a SDO_NN to find the nearest 2 objects in table2. I also have an address for each spatial table column. I'm trying to compare Address1 (from table1) to the two addresses (address2 from table2) to find the most likely candidate to match.
After reading REGEXP it seemed the REGEXP_LIKE was probably best way to reach my matches.
What I came up with works many times, but not always and I don't quite understand.

Here's an example where I do not understand what is happening when I used REGEXP_LIKE like this
REGEXP_LIKE (Address1, replace (address2,' ',' |'))

SELECT 
  id,  
  address1,
  address2,
  CASE
    WHEN address1=address2 THEN 'EXACT MATCH'
    WHEN REGEXP_LIKE(address1,replace(address2,' ','|')) THEN 'PROBABLE MATCH'
    ELSE 'NON MATCH'
  END AS TEST
FROM
  (
  SELECT
    1 ID, '930 OLD STEESE HWY' address1, '930 OLD STEESE HWY' address2
  FROM
    DUAL  
  UNION SELECT
    3 ID, '930 OLD STEESE HWY' address1, '920 OLD STEESE HWY' address2
  FROM
    DUAL    
  UNION SELECT
    3 ID, '930 OLD STEESE HWY' address1, '99 COLLEGE RD' address2
  FROM
    DUAL    
  UNION SELECT
    4 ID, '930 OLD STEESE HWY' address1, '80 5TH ST' address2
  FROM
    DUAL      
  );

RESULTS
ID ADDRESS1            ADDRESS2            TEST
-- ------------------  ------------------  --------------
1  930 OLD STEESE HWY  930 OLD STEESE HWY  EXACT MATCH
2  930 OLD STEESE HWY  920 OLD STEESE HWY  PROBABLE MATCH
3  930 OLD STEESE HWY  99 COLLEGE RD       NON MATCH
4  930 OLD STEESE HWY  80 5TH ST           PROBABLE MATCH

I am really confused about the 4th line in the results?
Thoughts and or suggestions?
Enjoy it!
See you soon,.
Eric

Second and fourth row match because there is a word in address2 who is present (at least as a substring) in Address1: for fourth ST is STEESE substring.

If you want a PROBABLE MATCH when there is a Word now as a word in Address1 address2, you can change this way:

SQL> SELECT
  2    id,
  3    address1,
  4    address2,
  5    CASE
  6      WHEN address1=address2 THEN 'EXACT MATCH'
  7      WHEN REGEXP_LIKE(address1,'(^| )'||replace(address2,' ','|')||'($| )') THEN 'PROBABLE MATCH'
  8      ELSE 'NON MATCH'
  9    END AS TEST
 10  FROM
 11    (
 12    SELECT
 13      1 ID, '930 OLD STEESE HWY' address1, '930 OLD STEESE HWY' address2
 14    FROM
 15      DUAL
 16    UNION SELECT
 17      3 ID, '930 OLD STEESE HWY' address1, '920 OLD STEESE HWY' address2
 18    FROM
 19      DUAL
 20    UNION SELECT
 21      3 ID, '930 OLD STEESE HWY' address1, '99 COLLEGE RD' address2
 22    FROM
 23      DUAL
 24    UNION SELECT
 25      4 ID, '930 OLD STEESE HWY' address1, '80 5TH ST' address2
 26    FROM
 27      DUAL
 28    );

        ID ADDRESS1           ADDRESS2           TEST
---------- ------------------ ------------------ --------------
         1 930 OLD STEESE HWY 930 OLD STEESE HWY EXACT MATCH
         3 930 OLD STEESE HWY 920 OLD STEESE HWY PROBABLE MATCH
         3 930 OLD STEESE HWY 99 COLLEGE RD      NON MATCH
         4 930 OLD STEESE HWY 80 5TH ST          NON MATCH

In Oracle 11 g, you can use REGEXP_COUNT to count the common words in the address1 and address2:

  2    id,
  3    address1,
  4    address2,
  5    CASE
  6      WHEN address1=address2 THEN 'EXACT MATCH'
  7      WHEN REGEXP_LIKE(address1,'(^| )'||replace(address2,' ','|')||'($| )') THEN 'PROBABLE MATCH'
  8      ELSE 'NON MATCH'
  9    END AS TEST,
 10    regexp_count(address1,'(^| )'||replace(address2,' ','|')||'($| )') common_words
 11  FROM
 12    (
 13    SELECT
 14      1 ID, '930 OLD STEESE HWY' address1, '930 OLD STEESE HWY' address2
 15    FROM
 16      DUAL
 17    UNION SELECT
 18      3 ID, '930 OLD STEESE HWY' address1, '920 OLD STEESE HWY' address2
 19    FROM
 20      DUAL
 21    UNION SELECT
 22      3 ID, '930 OLD STEESE HWY' address1, '99 COLLEGE RD' address2
 23    FROM
 24      DUAL
 25    UNION SELECT
 26      4 ID, '930 OLD STEESE HWY' address1, '80 5TH ST' address2
 27    FROM
 28      DUAL
 29    );

        ID ADDRESS1           ADDRESS2           TEST           COMMON_WORDS
---------- ------------------ ------------------ -------------- ------------
         1 930 OLD STEESE HWY 930 OLD STEESE HWY EXACT MATCH               4
         3 930 OLD STEESE HWY 920 OLD STEESE HWY PROBABLE MATCH            3
         3 930 OLD STEESE HWY 99 COLLEGE RD      NON MATCH                 0
         4 930 OLD STEESE HWY 80 5TH ST          NON MATCH                 0

Max
[My Italian blog Oracle | http://oracleitalia.wordpress.com/2009/12/29/estrarre-i-dati-in-formato-xml-da-sql/]

Published by: Massimo Ruocchio on December 29, 2009 22:48
Added example Regexp_count.

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