How to replace character accented characters in Snowflake? - sql

I'm trying to replace accented characters from a column to "normal" characters.
select 'áááããã'
I'd like some operation which would return 'aaaaaa'.

There is a more general way that uses a built-in JavaScript function to replace them:
Remove Diacritics from string in Snowflake
create or replace function REPLACE_DIACRITICS("str" string)
returns string
language javascript
strict immutable
as
$$
return str.normalize("NFD").replace(/\p{Diacritic}/gu, "");
$$;
select REPLACE_DIACRITICS('ö, é, č => a, o e, c');

Just found a solution with one of my colleagues.
select translate('áááããã','áéíóúãõâêôàç','aeiouaoaeoac')
We can also add a lower() to make it generalized for more cases
select translate(lower('ÁÁÁÃÃÃ'),'áéíóúãõâêôàç','aeiouaoaeoac')

Related

Remove template text on regexp_replace in Oracle's SQL

I am trying to remove template text like &#x; or &#xx; or &#xxx; from long string
Note: x / xx / xxx - is number, The length of the number is unknown, The cell type is CLOB
for example:
SELECT 'H'ello wor±ld' FROM dual
A desirable result:
Hello world
I know that regexp_replace should be used, But how do you use this function to remove this text?
You can use
SELECT REGEXP_REPLACE(col,'&&#\d+;')
FROM t
where
& is put twice to provide escaping for the substitution character
\d represents digits and the following + provides the multiple occurrences of them
ending the pattern with ;
or just use a single ampersand ('&#\d+;') for the pattern as in the case of Demo , since an ampersand has a special meaning for Oracle, a usage is a bit problematic.
In case you wanted to remove the entities because you don't know how to replace them by their character values, here is a solution:
UTL_I18N.UNESCAPE_REFERENCE( xmlquery( 'the_double_quoted_original_string' RETURNING content).getStringVal() )
In other words, the original 'H'ello wor±ld' should be passed to XMLQUERY as '"H'ello wor±ld"'.
And the result will be 'H'ello wo±ld'

Regular Expression PL/SQL

I am trying to use some regular expressions in PL/SQL.
I try to check the following pattern: LOKATIONS_ID => /LAND/ORT/GEBÄUDE/
I've tried this one:
create or replace function check_lok_id(lok_id in varchar2) return boolean
is
begin
if regexp_like (lok_id, '^(/[A-Z]+?){3}/$)')
then
return true;
else
return false;
end if;
end;
But unfortunately this one and several other regular expressions i've tested so far doesn't work.
Any suggestions?
Your example won't match because the umlauted character Ä is not in the range A-Z.
Try this regex:
^(/\w+?){3}/$
Or if you want to match only uppercase letters, but from all languages:
^(/[[:upper:]]+?){3}/$
See live demo

How to remove part of the string in oracle

Input data:
abcdef_fhj_viji.dvc
Expected output:
fhj_viji.dvc
The part to be trimmed is not constant.
Use the REPLACE method
Select REPLACE('abcdef_fhj_viji.dvc','abcde','')
If you want this query for your table :
Select REPLACE(column,'abcde','') from myTable
For update :
UPDATE TABLE
SET column = REPLACE(column,'abcde','')
select substr('abcdef_fhj_viji.dvc',instr('abcdef_fhj_viji.dvc','_')+1) from dual
So, Its all depends on INSTR function, define from which position and which occurrence, you will get the index and pass that index to SUBSTR to get your string.
Since you didn't give a lot of information I'm gonna assume some.
Let's assume you want a prefix of some string to be deleted. A good way to do that is by using Regular Expressions. There's a function called regexp_replace, that can find a substring of a string, depending on a pattern, and replace it with a different string. In PL/SQL you could write yourself a function using regexp_replace, like this:
function deletePrefix(stringName in varchar2) return varchar2 is
begin
return regexp_replace(stringName, '^[a-zA-Z]+_', '');
end;
or just use this in plain sql like:
regexp_replace(stringName, '^[a-zA-Z]+_', '');
stringName being the string you want to process, and the ^[a-zA-Z]+_ part depending on what characters the prefix includes. Here I only included upper- and lowercase letters.

simple parameter substitution in regexp_matches postgreSQL function

I have a table with a structure like this...
the_geom data
geom1 data1+3000||data2+1000||data3+222
geom2 data1+500||data2+900||data3+22232
I want to create a function that returns the records by user request.
Example: for data2, retrieve geom1,1000 and geom2, 900
Till now I created this function (see below) which works quite good but I am facing a parameter substitution problem... (you can see I am not able to substitute 'data2' for $1 in... BUT yes I can use $1 later
regexp_matches(t::text, E'(data2[\+])([0-9]+)'::text)::text)[2]::integer
MY FUNCTION
create or replace function get_counts(taxa varchar(100))
returns setof record
as $$
SELECT t2.counter,t2.the_geom
FROM (
SELECT (regexp_matches(t.data::text, E'(data2[\+])([0-9]+)'::text)::text)[2]::integer as counter,the_geom
from (select the_geom,data from simple_inpn2 where data ~ $1::text) as t
) t2
$$
language sql;
SELECT get_counts('data2') will work **but we should be able to make this substitution**:
regexp_matches(t::text, E'($1... instead of E'(data2....
I think its more a syntaxis issue, as the function execution gives no error, just interprets $1 as a string and gives no result.
thanks in advance,
A E'$1' is a string literal (using the escape string syntax) containing a dollar sign followed by a one. An unquoted $1 is the first parameter to your function. So this:
regexp_matches(t, E'($1[\+])([0-9]+)'))[2]::integer
as you've found, won't interpolate the $1 with the function's first parameter.
The regex is just a string, a string with an internal structure but still just a string. If you know that $1 will be a normal word then you could say:
regexp_matches(t, E'(' || $1 || E'[\+])([0-9]+)'))[2]::integer
to paste your strings together into a suitable regex. However, it is better to be a little paranoid, sooner or later someone is going to call your function with a string like 'ha ha (' so you should be prepared for it. The easiest way that I can think of to add an arbitrary string to a regex is to escape all the non-word characters:
-- Don't forget to escape the escaped escapes! Hence all the backslashes.
str := regexp_replace($1, E'(\\W)', E'\\\\\\1', 'g');
and then paste str into the regex as above:
regexp_matches(t, E'(' || str || E'[\+])([0-9]+)'))[2]::integer
or better, build the regex outside the regexp_matches to cut down on the nested parentheses:
re := E'(' || str || E'[\+])([0-9]+)';
-- ...
select regexp_matches(t, re)[2]::integer ...
PostgreSQL doesn't have Perl's \Q...\E and the (?q) metasyntax applies until the end of the regex so I can't think of any better way to paste an arbitrary string into the middle of a regex as a non-regex literal value than to escape everything and let PostgreSQL sort it out.
Using this technique, we can do things like:
=> do $$
declare
m text[];
s text;
r text;
begin
s = E'''{ha)?';
r = regexp_replace(s, E'(\\W)', E'\\\\\\1', 'g');
r = '(ha' || r || ')';
raise notice '%', r;
select regexp_matches(E'ha''{ha)?', r) into m;
raise notice '%', m[1];
end$$;
and get the expected
NOTICE: ha'{ha)?
output. But if you leave out the regexp_replace escaping step, you'll just get an
invalid regular expression: parentheses () not balanced
error.
As an aside, I don't think you need all that casting so I removed it. The regexes and escaping are noisy enough, there's no need to throw a bunch of colons into the mix. Also, I don't know what your standard_conforming_strings is set to or which version of PostgreSQL you're using so I've gone with E'' strings everywhere. You'll also want to switch your procedure to PL/pgSQL (language plpgsql) to make the escaping easier.

urlencode with only built-in functions

Without using plpgsql, I'm trying to urlencode a given text within a pgsql SELECT statement.
The problem with this approach:
select regexp_replace('héllo there','([^A-Za-z0-9])','%' || encode(E'\\1','hex'),'g')
...is that the encode function is not passed the regexp parameter, unless there's another way to call functions from within the replacement expression that actually works. So I'm wondering if there's a replacement expression that, by itself, can encode matches into hex values.
There may be other combinations of functions. I thought there would be a clever regex (and that may still be the answer) out there, but I'm having trouble finding it.
select regexp_replace(encode('héllo there','hex'),'(..)',E'%\\1','g');
This doesn't leave the alphanumeric characters human-readable, though.
Here is pretty short version, and it's even "pure SQL" function, not plpgsql. Multibyte chars (including 3- and 4-bytes emoji) are supported.
create or replace function urlencode(in_str text, OUT _result text) returns text as $$
select
string_agg(
case
when ol>1 or ch !~ '[0-9a-za-z:/#._?#-]+'
then regexp_replace(upper(substring(ch::bytea::text, 3)), '(..)', E'%\\1', 'g')
else ch
end,
''
)
from (
select ch, octet_length(ch) as ol
from regexp_split_to_table($1, '') as ch
) as s;
$$ language sql immutable strict;
Here's a function I wrote that handles encoding using built in functions while preserving the readability of the URL.
Regex matches to capture pairs of (optional) safe characters and (at most one) non-safe character. Nested selects allow those pairs to be encoded and re-combined returning a fully encoded string.
I've run through a test suite with all sorts of permutations (leading/trailing/only/repeated encoded characters and thus far it seems to encode correctly.
The safe special characters are _ ~ . - and /. My inclusion of "/" on that list is probably non-standard, but fits the use case I have where the input text may be a path and I want that to remain.
CREATE OR REPLACE FUNCTION oseberg.encode_uri(input text)
RETURNS text
LANGUAGE plpgsql
IMMUTABLE STRICT
AS $function$
DECLARE
parsed text;
safePattern text;
BEGIN
safePattern = 'a-zA-Z0-9_~/\-\.';
IF input ~ ('[^' || safePattern || ']') THEN
SELECT STRING_AGG(fragment, '')
INTO parsed
FROM (
SELECT prefix || encoded AS fragment
FROM (
SELECT COALESCE(match[1], '') AS prefix,
COALESCE('%' || encode(match[2]::bytea, 'hex'), '') AS encoded
FROM (
SELECT regexp_matches(
input,
'([' || safePattern || ']*)([^' || safePattern || '])?',
'g') AS match
) matches
) parsed
) fragments;
RETURN parsed;
ELSE
RETURN input;
END IF;
END;
$function$
You can use CLR and import the namespace or use the function shown in this link , this creates a T-SQL function that does the encoding.
http://www.sqljunkies.com/WebLog/peter_debetta/archive/2007/03/09/28987.aspx