I'm trying to regexp_replace() all the values of a column ending without "/", by adding "/".
I can get the correct values by using this statement (the pattern was tested with a PCRE checker):
SELECT * FROM `table` WHERE `column` REGEXP("(?<=[^\/])$");
And the non-matching ones with:
SELECT * FROM `table` WHERE `column` REGEXP("(?<![^\/])$");
But when the statement is:
UPDATE `table` SET `column` = REGEXP_REPLACE(`column`, "(?<=[^\/])$", "/");
Then, there is no change, whatever value I put into the third parameter:
Query OK, 0 rows affected (0.00 sec)
Rows matched: 1031 Changed: 0 Warnings: 0
You could do this easily without regex:
UPDATE `table` SET `column` = `column` + '/'
WHERE RIGHT(`column`, 1) <> '/'
trying to understand why it does not work
As I rationalize the problem, you are asking REGEXP_REPLACE to do two things:
Discover that something is missing, and
Point to a location in the string.
Your regexp says that it is missing, but I question whether it points to a specific substring (even an empty one) for replacing. It's easy to point to a found substring (or substrings). It is hard to point to a missing substring. And such a 'pointer' is needed to do the replacement.
Hence, Michal's approach (even if some regexp were needed) is the "right" way to solve the problem.
Related
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.
we have ranges like this
"0,5, 0,5"
"0,112, 0,118"
and want to split by the second comma.
Any idea?
You can update the regex you split by with comma then a space after.
select regexp_split_to_array('0,112, 0,118', ', ')
demo:db<>fiddle
Supposing, there ist always at least one space after the second comma and none after the others, you could use this for the split regex:
SELECT
regexp_split_to_array(ranges, ',\s+')
FROM
t
This returns an array like {"0,5","0,5"}.
You can split both ranges into columns using a subquery:
SELECT
r[1],
r[2]
FROM (
SELECT
regexp_split_to_array(ranges, ',\s+') as r
FROM
t
) s
Edit:
TO wants to get everything after the second comma. So you need a regex for splitting, which finds the nth (here n = 2) occurrence of a comma:
(?:(^.*?,.*)),
This can be used to query the required data:
demo:db<>fiddle
SELECT
(regexp_split_to_array(ranges, '(?:(^.*?,.*)),'))[2]
FROM
t
Use regexp_replace:
select regexp_replace('0,112, 0,118', '.*,\s+', '') as foo;
Output:
foo
-------
0,118
(1 row)
Thank you all for the quick answers. It finally worked by using this
regexp_matches(your_string_value, '\d+[,|.]\d+|\d+','g'))[1]
This helped me getting rid of all unnecessary characters within the values + delivered me back the second value in the range.
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 )
...
please, I have in Oracle table this texts (as 2 records)
"Sample text with replace parameter %1%"
"You reached 90% of your limit"
I need replace %1% with specific text from input parameter in Oracle Function. In fact, I can have more than just one replace parameters. I have also record with "Replace this %12% with real value"
This functionality I have programmed:
IF poc > 0 THEN
FOR i in 1 .. poc LOOP
p := get_param(mString => mbody);
mbody := replace(mbody,
'%' || p || '%', parameters(to_number(p, '99')));
END LOOP;
END IF;
But in this case I have problem with text number 2. This functionality trying replace "90%" also and I then I get this error:
ORA-06502: PL/SQL: numeric or value error: NULL index table key value
It's a possible to avoid try replace "90%"? Many thanks for advice.
Best regards
PS: Oracle version: 10g (OCI Version: 10.2)
Regular expressions can work here. Try the following and build them into your script.
SELECT REGEXP_REPLACE( 'Sample text with replace parameter %1%',
'\%[0-9]+\%',
'db_size' )
FROM DUAL
and
SELECT REGEXP_REPLACE( 'Sample text with replace parameter 1%',
'\%[0-9]+\%',
'db_size' )
FROM DUAL
The pattern is pretty simple; look for patterns where a '%' is followed by 1 or more numbers followed by a '%'.
The only issue here will be if you have more than one replacement to make in each string and each replacement is different. In that case you will need to loop round the string each time replacing the next parameter. To do this add the position and occurrence parameters to REGEXP_REPLACE after the replacement string, e.g.
REGEXP_REPLACE( 'Sample text with replace parameter %88888888888%','\%[0-9]+\%','db_size',0,1 )
You are getting the error because at parameters(to_number(p, '99')). Can you please check the value of p?
Also, if the p=90 then then REPLACE will not try to replace "90%". It will replace "%90%". How have you been sure that it's trying to replace "90%"?
I have a table with a column po_number of type varchar in Postgres 8.4. It stores alphanumeric values with some special characters. I want to ignore the characters [/alpha/?/$/encoding/.] and check if the column contains a number or not. If its a number then it needs to typecast as number or else pass null, as my output field po_number_new is a number field.
Below is the example:
SQL Fiddle.
I tired this statement:
select
(case when regexp_replace(po_number,'[^\w],.-+\?/','') then po_number::numeric
else null
end) as po_number_new from test
But I got an error for explicit cast:
Simply:
SELECT NULLIF(regexp_replace(po_number, '\D','','g'), '')::numeric AS result
FROM tbl;
\D being the class shorthand for "not a digit".
And you need the 4th parameter 'g' (for "globally") to replace all occurrences.
Details in the manual.
For a known, limited set of characters to replace, plain string manipulation functions like replace() or translate() are substantially cheaper. Regular expressions are just more versatile, and we want to eliminate everything but digits in this case. Related:
Regex remove all occurrences of multiple characters in a string
PostgreSQL SELECT only alpha characters on a row
Is there a regexp_replace equivalent for postgresql 7.4?
But why Postgres 8.4? Consider upgrading to a modern version.
Consider pitfalls for outdated versions:
Order varchar string as numeric
WARNING: nonstandard use of escape in a string literal
I think you want something like this:
select (case when regexp_replace(po_number, '[^\w],.-+\?/', '') ~ '^[0-9]+$'
then regexp_replace(po_number, '[^\w],.-+\?/', '')::numeric
end) as po_number_new
from test;
That is, you need to do the conversion on the string after replacement.
Note: This assumes that the "number" is just a string of digits.
The logic I would use to determine if the po_number field contains numeric digits is that its length should decrease when attempting to remove numeric digits.
If so, then all non numeric digits ([^\d]) should be removed from the po_number column. Otherwise, NULL should be returned.
select case when char_length(regexp_replace(po_number, '\d', '', 'g')) < char_length(po_number)
then regexp_replace(po_number, '[^0-9]', '', 'g')
else null
end as po_number_new
from test
If you want to extract floating numbers try to use this:
SELECT NULLIF(regexp_replace(po_number, '[^\.\d]','','g'), '')::numeric AS result FROM tbl;
It's the same as Erwin Brandstetter answer but with different expression:
[^...] - match any character except a list of excluded characters, put the excluded charaters instead of ...
\. - point character (also you can change it to , char)
\d - digit character
Since version 12 - that's 2 years + 4 months ago at the time of writing (but after the last edit that I can see on the accepted answer), you could use a GENERATED FIELD to do this quite easily on a one-time basis rather than having to calculate it each time you wish to SELECT a new po_number.
Furthermore, you can use the TRANSLATE function to extract your digits which is less expensive than the REGEXP_REPLACE solution proposed by #ErwinBrandstetter!
I would do this as follows (all of the code below is available on the fiddle here):
CREATE TABLE s
(
num TEXT,
new_num INTEGER GENERATED ALWAYS AS
(NULLIF(TRANSLATE(num, 'ABCDEFGHIJKLMNOPQRSTUVWXYZ. ', ''), '')::INTEGER) STORED
);
You can add to the 'ABCDEFG... string in the TRANSLATE function as appropriate - I have decimal point (.) and a space ( ) at the end - you may wish to have more characters there depending on your input!
And checking:
INSERT INTO s VALUES ('2'), (''), (NULL), (' ');
INSERT INTO t VALUES ('2'), (''), (NULL), (' ');
SELECT * FROM s;
SELECT * FROM t;
Result (same for both):
num new_num
2 2
NULL
NULL
NULL
So, I wanted to check how efficient my solution was, so I ran the following test inserting 10,000 records into both tables s and t as follows (from here):
EXPLAIN (ANALYZE, BUFFERS, VERBOSE)
INSERT INTO t
with symbols(characters) as
(
VALUES ('ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789')
)
select string_agg(substr(characters, (random() * length(characters) + 1) :: INTEGER, 1), '')
from symbols
join generate_series(1,10) as word(chr_idx) on 1 = 1 -- word length
join generate_series(1,10000) as words(idx) on 1 = 1 -- # of words
group by idx;
The differences weren't that huge but the regex solution was consistently slower by about 25% - even changing the order of the tables undergoing the INSERTs.
However, where the TRANSLATE solution really shines is when doing a "raw" SELECT as follows:
EXPLAIN (ANALYZE, BUFFERS, VERBOSE)
SELECT
NULLIF(TRANSLATE(num, 'ABCDEFGHIJKLMNOPQRSTUVWXYZ. ', ''), '')::INTEGER
FROM s;
and the same for the REGEXP_REPLACE solution.
The differences were very marked, the TRANSLATE taking approx. 25% of the time of the other function. Finally, in the interests of fairness, I also did this for both tables:
EXPLAIN (ANALYZE, BUFFERS, VERBOSE)
SELECT
num, new_num
FROM t;
Both extremely quick and identical!