While converting Oracle to Postgresql I came to know the following Oracle query need to be converted in Postgres.
Oracle Query: Find pattern and replace with null
select regexp_replace('1', '[^0-9]', null) from dual;
select regexp_replace('a', '[^0-9]', null) from dual;
select regexp_replace('1a1', '[^0-9]', null) from dual;
My try:
As per the postgres document we need to use REGEXP_REPLACE with [[:alpha:]] pattern.
But the statement is replacing with empty string if match found. I'm looking for null instead.
PostgreSQL Query:
select REGEXP_REPLACE('1','[[:alpha:]]','','g') --Correct
select REGEXP_REPLACE('a','[[:alpha:]]','','g') --Wrong: output should be NULL
select REGEXP_REPLACE('1a1','[[:alpha:]]','','g') --Correct
select REGEXP_REPLACE(' ','[[:alpha:]]','','g') --Wrong: output should be NULL
Definitely we can use case statement like following but I want the solution in single line without using case condition.
SELECT case when REGEXP_REPLACE('1a','[[:alpha:]]','','g') = ''
then
null
else
REGEXP_REPLACE('1a','[[:alpha:]]','','g')
end;
Related
I have varchar field in the database that contains text. I need to replace every occurrence of a any 2 letter + 8 digits string to a link, such as VA12345678 will return /cs/page.asp?id=VA12345678
I have a regex that replaces the string but how can I replace it with a string where part of it is the string itself?
SELECT REGEXP_REPLACE ('test PI20099742', '[A-Z]{2}[0-9]{8}$', 'link to replace with')
FROM dual;
I can have more than one of these strings in one varchar field and ideally I would like to have them replaced in one statement instead of a loop.
As mathguy had said, you can use backreferences for your use case. Try a query like this one.
SELECT REGEXP_REPLACE ('test PI20099742', '([A-Z]{2}[0-9]{8})', '/cs/page.asp?id=\1')
FROM DUAL;
For such cases, you may want to keep the "text to add" somewhere at the top of the query, so that if you ever need to change it, you don't have to hunt for it.
You can do that with a with clause, as shown below. I also put some input data for testing in the with clause, but you should remove that and reference your actual table in your query.
I used the [:alpha:] character class, to match all letters - upper or lower case, accented or not, etc. [A-Z] will work until it doesn't.
with
text_to_add (link) as (
select '/cs/page.asp?id=' from dual
)
, sample_strings (str) as (
select 'test VA12398403 and PI83048203 to PT3904' from dual
)
select regexp_replace(str, '([[:alpha:]]{2}\d{8})', link || '\1')
as str_with_links
from sample_strings cross join text_to_add
;
STR_WITH_LINKS
------------------------------------------------------------------------
test /cs/page.asp?id=VA12398403 and /cs/page.asp?id=PI83048203 to PT3904
I want to write a query in a hive Table using CASE WHEN, LIKE and a regular expression. I have used regexp and rlike, but I do not get the desired results. My attempts so far are the following
select distinct ending from
(select date, ending, name, count(distinct id)
from (select CONCAT_WS("/",year,month,day,hour) as date, id, name,
case when type = 'TRAN' then 'tran'
when events regexp '%[:]no_reply[:]%[^o][^n][:]incomplete[:]%' and type rlike '%HUP' then 'con'
when events not regexp '%[:]no_reply[:]%[^o][^n][:]incomplete[:]%' and type rlike '%HUP' then 'aban'
else 'other'
end as ending
from data_struct1) tmp
group by date, ending, name) tmp2;
and also
select distinct ending from
(select date, ending, name, count(distinct id)
from (select CONCAT_WS("/",year,month,day,hour) as date, id, name,
case when type = 'TRAN' then 'tran'
when events rlike '%[:]no_reply[:]%[^o][^n][:]incomplete[:]%' and type rlike '%HUP' then 'con'
when events not rlike '%[:]no_reply[:]%[^o][^n][:]incomplete[:]%' and type rlike '%HUP' then 'aban'
else 'other'
end as ending
from data_struct1) tmp
group by date, ending, name) tmp2;
Both queries return incorrect results (not bad syntax, just not the correct results).
There are a lot of docs on regex quantifiers, for example this one: https://learn.microsoft.com/en-us/dotnet/standard/base-types/quantifiers-in-regular-expressions
select 'opencase_2,initial_state:inquiry,inquiry:no_reply:initial_state:incomplete::,inquiry:reask:secondary_state:complete::' regexp 'no_reply:[^:]+:incomplete';
OK
true
Also this is wrong: rlike '%HUP'. It should be like this '.*HUP$' (in the end of the string) or simply 'HUP' if it does not matter where the HUP is located: in the middle or in the end or in the beginning of the string
rlike and regexp in your query work the same, better use the same operator: regexp or rlike only. These two are synonyms.
Test: https://regex101.com/r/ksG67v/1
select REGEXP_REPLACE (' if function_text() hi how do you do text_jon','(test\S.*)','',1,0,'m') from dual;
Output for this:
if function_
I don't want this "if function_" too i.e null output should be there
How to achieve the above?
I would like to eliminate all duplicate words in a comma separated list.
I've tried with:
SELECT
REGEXP_REPLACE(
'1234,234,1234,1234,928,1234,123,1234,Abcd,1234,1234',
'([^,\w]+)(,[ ]*[\1])+') AS r
FROM dual
It should return
1234,234,928,123,Abcd
But in fact it returns
1234,234,234,234
Also tried with ([^,\w]+)(,[ ]*\1)+ but with '1234,1234,1234' it returns (null)
Also tried with
SELECT
REGEXP_REPLACE(
'1234,234,1234,1234,928,1234,123,1234,Abcd,1234,1234',
'([^,\w]+)(,[ ]*[\1])+', '\1') AS r
FROM dual
and following replacements, even '\1\2' but none of them is giving the desired result.
Please, any ideas?
I know this isn't exactly the method you were asking for, but it still achieves the same result:
WITH DATA AS
( SELECT '1234,234,1234,1234,928,1234,123,1234,Abcd,1234,1234' str FROM dual)
SELECT DISTINCT trim(regexp_substr(str, '[^,]+', 1, LEVEL)) str
FROM DATA
CONNECT BY instr(str, ',', 1, LEVEL - 1) > 0
I have a problem in matching regular expression in Oracle PL/SQL.
To be more specific, the problem is that regex doesn't want to match any zero occurrence.
For example, I have something like:
select * from dual where regexp_like('', '[[:alpha:]]*');
and this doesn't work. But if I put space in this statement:
select * from dual where regexp_like(' ', '[[:alpha:]]*');
it works.
I want to have the first example running, so that person doesn't have to put 'space' for it to work.
Any help is appreciated, and thank you for your time.
T
For better or worse, empty strings in Oracle are treated as NULL:
SQL> select * from dual where '' like '%';
DUMMY
-----
Take that into account when querying with Oracle:
SQL> SELECT *
2 FROM dual
3 WHERE regexp_like('', '[[:alpha:]]*')
4 OR '' IS NULL;
DUMMY
-----
X
Does Oracle still treat empty strings as NULLs? And if so, does regexp_like with a NULL input source string return UNKNOWN? Both of these would be semi-reasonable, and a reason why your test does not work as you expected.