Include null in oracle regex - regex

I've been trying the following but it does not return any (null) values. How can I include null values in using a regular expression with oracle an in fact return everything including null. (It is necessary to use a regex as the value is replaced with a variable in python and so the value will not always be null. Otherwise i could just leave out the expression.)
SELECT * from table WHERE
REGEXP_LIKE (column1,'.*|NULL$')

Two methods.
Explicit comparison:
WHERE (REGEXP_LIKE(column1, '.*') OR column1 IS NULL)
Or replacement:
WHERE REGEXP_LIKE(COALESCE(column1, '<NULL>'), '.*|<NULL>')
The second method is more dangerous, because you could already have the replacement string in the column.

Related

Having difficulty in pattern matching Postal Codes for an oracle regexp_like command

The Problem:
All I'm trying to do is come up with a pattern matching string for my regular expression that lets me select Canadian postal codes in this format: 'A1A-2B2' (for example).
The types of data I am trying to insert:
Insert Into Table
(Table_Number, Person_Name, EMail_Address, Street_Address, City, Province, Postal_Code, Hire_Date)
Values
(87, 'Tommy', 'mobster#gmail.com', '123 Street', 'location', 'ZY', 'T4X-1S2', To_Date('30-Aug-2020 08:50:56');
This is a slightly modified/generic version to protect some of the data. All of the other columns enter just fine/no complaints. But the postal code it does not seem to like when I try to run a load data script.
The Column & Constraint in question:
Postal_Code varchar2(7) Constraint Table_Postal_Code Null
Constraint CK_Postal_Code Check ((Regexp_like (Postal_Code, '^\[[:upper:]]{1}[[:digit:]]{1}[[:upper:]][[:punct:]]{1}[[:digit:]]{1}[[:upper:]](1}[[:digit:]]{1}$')),
My logic here: following the regular expression documentation:
I have:
an open quote
a exponent sign to indicate start of string
Backslash (I think to interpet a string literal)
-1 upper case letter, 1 digit, 1 uppercase , 1 :punct: to account for the hypen, 1 digit, 1 upper case letter, 1 digit
$ to indicate end of string
Close quote
In my mind, something like this should work, it accounts for every single letter/character and the ranges they have to be in. But something is off regarding my formatting of this pattern matching string.
The error I get is:
ORA-02290: check constraint (user.CK_POSTAL_CODE) violated
(slightly modified once more to protect my identity)
Which tells me that the data insert statement is tripping off my check constraint and thats about it. So its as issue with the condition of the constraint itself - ie string I'm using to match it. My instructor has told me that insert data is valid, and doesn't need any fix-up so I'm at a loss.
Limits/Rules: The Hyphen has to be there/matched to my understanding of the problem. They are all uppercase in the dataset, so I don't have to worry about lowercase for this example.
I have tried countless variations of this regexp statement to see if anything at all would work, including:
changing all those uppers to :alpha: , then using 'i' to not check for case sensitivity for the time being
removing the {1} in case that was redudant
using - (backslash hyphen) , to turn into a string literal maybe
using only Hyphen by itself
even removing regexp altogether and trying a LIKE [A-Z][0-9][A-Z]-[0-9][A-Z][0-9] etc
keeping the uppers , turning :digit:'s to [0-9] to see if that would maybe work
The only logical thing I can think of now is: the check constraint is actually working fine and tripping off when it matches my syntax. But I didn't write it clearly enough to say "IGNORE these cases and only get tripped/activated if it doesn't meet these conditions"
But I'm at my wits end and asking here as a last resort. I wouldn't if I could see my mistake eventually - but everything I can think of, I probably tried. I'm sure its some tiny formatting rule I just can't see (I can feel it).Thank you kindly to anyone who would know how to format a pattern matching string like this properly.
It looks like you may have been overcomplicating the regex a bit. The regex below matches your description based on the first set of bullets you lined out:
REGEXP_LIKE (postal_code, '^[A-Z]\d[A-Z]-\d[A-Z]\d$')
I see two problems with that regexp.
Firstly, you have a spurious \ at the start. It serves you no purpose, get rid of it.
Secondly, the second-from last {1} appears in your code with mismatched brackets as (1}. I get the error ORA-12725: unmatched parentheses in regular expression because of this.
To be honest, you don't need the {1}s at all: they just tell the regular expression that you want one of the previous item, which is exactly what you'd get without them.
So you can fix the regexp in your constraint by getting rid of the \ and removing the {1}s, including the one with mismatched parentheses.
Here's a demo of the fixed constraint in action:
SQL> CREATE TABLE postal_code_test (
2 Postal_Code varchar2(7) Constraint Table_Postal_Code Null
3 Constraint CK_Postal_Code Check ((Regexp_like (Postal_Code, '^[[:upper:]][[:digit:]][[:upper:]][[:punct:]][[:digit:]][[:upper:]][[:digit:]]$'))));
Table created.
SQL> INSERT INTO postal_code_test (postal_code) VALUES ('T4X-1S2');
1 row created.
SQL> INSERT INTO postal_code_test (postal_code) VALUES ('invalid');
INSERT INTO postal_code_test (postal_code) VALUES ('invalid')
*
ERROR at line 1:
ORA-02290: check constraint (user.CK_POSTAL_CODE) violated
You do not need the backslash and you have (1} instead of {1}.
You can simplify the expression to:
Postal_Code varchar2(7)
Constraint Table_Postal_Code Null
Constraint CK_Postal_Code Check (
REGEXP_LIKE(Postal_Code, '^[A-Z]\d[A-Z][[:punct:]]\d[A-Z]\d$')
)
or:
Constraint CK_Postal_Code Check (
REGEXP_LIKE(
Postal_Code,
'^[A-Z][0-9][A-Z][[:punct:]][0-9][A-Z][0-9]$'
)
)
or:
Constraint CK_Postal_Code Check (
REGEXP_LIKE(
Postal_Code,
'^[[:upper:]][[:digit:]][[:upper:]][[:punct:]][[:digit:]][[:upper:]][[:digit:]]$'
)
)
or (although the {1} syntax is redundant here):
Constraint CK_Postal_Code Check (
REGEXP_LIKE(
Postal_Code,
'^[[:upper:]]{1}[[:digit:]]{1}[[:upper:]]{1}[[:punct:]]{1}[[:digit:]]{1}[[:upper:]]{1}[[:digit:]]{1}$'
)
)
fiddle
removing regexp altogether and trying a LIKE [A-Z][0-9][A-Z]-[0-9][A-Z][0-9] etc
That will not work as the LIKE operator does not match regular expression patterns.

How can I replace multiple words "globally" using regexp_replace in Oracle?

I need to replace multiple words such as (dog|cat|bird) with nothing in a string where there may be multiple consecutive occurrences of a word. The actual code is to remove salutations and suffixes from a name. Unfortunately the garbage data I get sometimes contains "SNERD JR JR."
I was able to create a regular expression pattern that accomplishes my goal but only for the first occurrence. I implemented a stupid hack to get rid of the second occurrence, but I believe there has to be a better way. I just can't figure it out.
Here is my "hacked" code;
FUNCTION REMOVE_SALUTATIONS(IN_STRING VARCHAR2) RETURN VARCHAR2 DETERMINISTIC
AS
REGEX_SALUTATIONS VARCHAR2(4000) := '(^|\s)(MR|MS|MISS|MRS|DR|MD|M D|SR|SIR|PHD|P H D|II|III|IV|JR)(\.?)(\s|$)';
BEGIN
RETURN TRIM(REGEXP_REPLACE(REGEXP_REPLACE(IN_STRING,REGEX_SALUTATIONS,' '),REGEX_SALUTATIONS,''));
END REMOVE_SALUTATIONS;
I was actually proud that I was able to get this far, as regular expression are not very regular to me. All help is appreciated.
EDIT:
The default for regexp_replace based on my understanding is to do a global replace. But on the outside chance my DB is configured different I did try;
select REGEXP_REPLACE('SNERD JR JR','(^|\s)(MR|MS|MISS|MRS|DR|MD|M D|SR|SIR|PHD|P H D|II|III|IV|JR)(\.?)(\s|$)',' ',1,0) from dual;
and the results are;
SNERD JR
Use occurrence parameter of REGEXP_REPLACE function. The docs says:
occurrence is a nonnegative integer indicating the occurrence of the replace operation:
If you specify 0, then Oracle replaces all occurrences of the match.
If you specify a positive integer n, then Oracle replaces the nth occurrenc
https://docs.oracle.com/cd/B28359_01/server.111/b28286/functions137.htm#SQLRF06302
It should look like:
...
REGEXP_REPLACE(IN_STRING,REGEX_SALUTATIONS,' ', 1,0 )
...

OpenRefine custom text faceting

I have a column of names like:
Quaglia, Pietro Paolo
Bernard, of Clairvaux, Saint, or
.E., Calvin F.
Swingle, M Abate, Agostino, Assereto
Abati, Antonio
10-NA)\u, Ferraro, Giuseppe, ed, Biblioteca comunale ariostea. Mss. (Esteri
I want to make a Custom text facet with openrefine that mark as "true" the names with one comma and "false" all the others, so that I can work with those last (".E., Calvin F." is not a problem, I'll work with that later).
I'm trying using "Custom text facet" and this expression:
if(value.match(/([^,]+),([^,]+)/), "true", "false")
But the result is all false. What's the wrong part?
The expression you are using:
if(value.match(/([^,]+),([^,]+)/), "true", "false")
will always evaluate to false because the output of the 'match' function is either an array, or null. When evaluated by 'if' neither an array nor 'null' evaluate to true.
You can wrap the match function in a 'isNonBlank' or similar to get a boolean true/false, which would then cause the 'if' function to work as you want. However, once you have a boolean true/false result the 'if' becomes redundant as its only function is to turn the boolean true/false into string "true" or "false" - which won't make any difference to the values function of the custom text facet.
So:
isNonBlank(value.match(/([^,]+),([^,]+)/))
should give you the desired result using match
Instead of using 'match' you could use 'split' to split the string into an array using the comma as a split character. If you measure the length of the resulting array, it will give you the number of commas in the string (i.e. number of commas = length-1).
So your custom text facet expression becomes:
value.split(",").length()==2
This will give you true/false
If you want to break down the data based on the number of commas that appear, you could leave off the '==2' to get a facet which just gives you the length of the resulting array.
I would go with lookahead assertion to check if only 1 "," can find from the beginning until the end of line.
^(?=[^\,]+,[^\,]+$).*
https://regex101.com/r/iG4hX6/2

DB2: find field value where first character is a lower case letter

I am trying to pick out a value in a field where the first character is a lower case letter. This is difficult since DB2 does not permit regular expressions. My current attempt is:
select * from mytable
where field1 like lcase('_%')
where I was hoping the underscore followed by percent wildcard would find any character in the first position, and then wrap the lcase() around that to ensure it is lower case. the result is that any and every value gets selected, so the lcase() is not performing what I want it to do, and in hindsight is used to cast to lowercase.
With that in mind, how to I ensure that the result of
('_%')
is lowercase only?
Thanks very much
i would use something like:
... where substr(field1,1,1) <> upper(substr(field1,1,1))
solution with 'a'...'z' will not work with characters different from latin characterset (e.g. cyrilic etc)
Why not
where field1 >= 'a' and field1 < '{'
This will even make use of an appropriate index, if any.
Be warned, however, that this won't work when your DB instance does lexicongraphic ordering. I am not sure if the latter is a DB attribute or a session attribute, however.
Another, more general way (especially when considering non ASCII letters) would be to check if the length of the field is > 0 and the lowercased substring consisting of the first character equals the substring consisting of the first character while the uppercased first character does not equal the first character. (Look up the functions in the DB2 reference, I have mine not ready at the moment.)
DB2 DOES allow Regular expressions with xQuery. For example:
with cteGender(VALUE) as
(
values
('M'),('F'),('U'),('S'),(' M'),('f')
),
cteResult(VALUE,RESULT_BOOLEAN) as
(
select '"' || VALUE || ‘"',
xmlquery('fn:matches($VALUE,''^[MFU]{1}$'')') from cteGender
)
select VALUE, RESULT_BOOLEAN,
xmlcast(RESULT_BOOLEAN as integer) RESULT_INTEGER from cteResult;
I took this example from: http://www.idug.org/p/bl/et/blogid=278&blogaid=187 That article explain very well how to use xQuery.
DB2 does not have SQL functions for Regular Expressions, but with xQuery you can do that. But if you really want SQL functions for RegEx, please visit this site: https://www.ibm.com/developerworks/jp/data/library/db2/j_d-regularexpression/ (In Japanese, but the code can be understood)
For more information about RegEx in DB2 please visit: http://pic.dhe.ibm.com/infocenter/db2luw/v10r5/topic/com.ibm.db2.luw.xml.doc/doc/xqrregexp.html

Postgres set varchar field to regular expression of itself

I'm trying to normalise a data field by removing a fairly common postfix. I've got as far as using the substring() function in postgres, but can't quite get it to work. For example, if I want to strip the postfix 'xyz' from any values that have it;
UPDATE my_table SET my_field=substring(my_field from '#"%#"xyz' for '#');
But this is having some weird effects that I cant pin down. Any thoughts? Many thanks as always.
update my_table
set my_field = regexp_replace(my_field, 'xyz$', '')
where my_field ~ 'xyz$';
This will also change the value 'xyz' into an empty string. I don't know if you want that (or if the suffix can exists "on it's own".
The where clause is not strictly necessary but will make the update more efficient because only those rows are updated that actually meet the criteria.
UPDATE my_table
SET my_field = left(my_field, -3)
WHERE my_field LIKE '%xyz';
For several reasons:
If you don't want to change every single row, always add a WHERE clause to your UPDATE. Even if only some rows are actually changed by the expression. An UPDATE from the same value to the same value is still an UPDATE and will produce dead rows and table bloat and trigger triggers ...
Use left() in combination with LIKE.
left() with a negative second parameter effectively trims the number of character from the end of the string. left() was introduced with PostgreSQL 9.1. I quote the manual here:
When n is negative, return all but last |n| characters.
Always pick LIKE over a regular expression (~) if you can. LIKE is not as versatile, but much faster. (SIMILAR TO is rewritten as regular expression internally). Details in this related answer on dba.SE.
If you want to make sure that a minimum of characters remains:
WHERE my_field LIKE '_%xyz'; -- prepend as many _ as you want chars left
substring() would work like this (one possibility):
substring(my_field, '^(.*)xyz$');