Cellular number valid in pl sql oracle - regex

I want to check cellular number if it current and if true - get it in fromat
for example
the numbers correct :
0521234567
521234567 - need only to 0 in the start
052-1234567
(052)1234567
052-123-456-7
numbers not correct:
052123
0871234567
how I do it??
i tried to write:
SELECT REGEXP_REPLACE('0521234567', '^0?(5[0-9])(\-)?\d{7}$', '') FROM dual;
but it's return '' ;
thank.

SELECT CASE WHEN REGEXP_LIKE('0521234567', '^0?(5[0-9])(\-)?\d{7}$')
THEN '0521234567'
ELSE NULL END
FROM dual;
If a string satisfies the format return the string, otherwise return NULL

Related

Casting not working correctly in Amazon Athena (Presto)?

I have a doctor license registry dataset which includes the total_submitted_charge_amount for each doctor as well as the number of entitlements with medicare & medicaid . I used the query from the answer suggested below:
with datamart AS
(SELECT npi,
provider_last_name,
provider_first_name,
provider_mid_initial,
provider_address_1,
provider_address_2,
provider_city,
provider_zipcode,
provider_state_code,
provider_country_code,
provider_type,
number_of_services,
CASE
WHEN REPLACE(num_entitlement_medicare_medicaid,',', '') ='' THEN
null
ELSE CAST(REPLACE(num_entitlement_medicare_medicaid,',', '') AS DECIMAL)
END AS medicare_medicaid_entitlement,
CASE
WHEN REPLACE(total_submitted_charge_amount,',', '') ='' THEN
null
ELSE CAST(REPLACE(num_entitlement_medicare_medicaid,',', '') AS DECIMAL)
END AS total_submitted_charge_amount
FROM cmsaggregatepayment2017)
SELECT *
FROM datamart
ORDER BY total_submitted_charge_amount DESC
Unfortunately I get the error
INVALID_CAST_ARGUMENT: Cannot cast VARCHAR '' to DECIMAL(38, 0)
This query ran against the aggregatepayment_data_2017 database, unless qualified by the query. Please post the error message on our forum or contact customer support with Query Id: be01d1e8-dc4d-4c75-a648-428dcb6be3a5." I have tried Decimal, Real, Big int and nothing works for casting num_entitlement_medicare_medicaid. Below is a screenshot of how the data looks like:
Can someone please suggest how to rephrase this query?
Instead of putting cast/replace in your queries, you could convert the data into a new table with 'clean' data:
CREATE TABLE clean_table
WITH (format='Parquet', external_location='s3://my_bucket/clean_data/')
AS
SELECT
npi,
provider_last_name,
provider_first_name,
...
CASE WHEN REPLACE(num_entitlement_medicare_medicaid,',', '') ='' THEN null
ELSE CAST(REPLACE(num_entitlement_medicare_medicaid,',', '') AS DECIMAL)
END AS medicare_medicaid_entitlement,
CASE WHEN REPLACE(total_submitted_charge_amount,',', '') ='' THEN null
ELSE CAST(REPLACE(num_entitlement_medicare_medicaid,',', '') AS DECIMAL)
END AS total_submitted_charge_amount
FROM cmsaggregatepayment2017
You can the SELECT ... FROM clean_table without having to do any conversions.
In data warehousing, this type of process is known as ETL (Extract, Transform, Load). The cleaning process is the 'transform' to convert the data into a more useful format.
See: CREATE TABLE AS - Amazon Athena
You might want to try try_cast() in presto. This version works on coercion. If there is any error it avoids it and moves to the next item.
Documention: https://prestodb.io/docs/current/functions/conversion.html
The reason you are getting error is you have blank value(but it is not null) in the column and we cannot cast varchar '' as decimal. You can probably use case statement. Also as per the data set column num_entitlement_medicare_medicaid has comma ',' in it which you are not replacing.
SELECT npi,
case
when REPLACE(num_entitlement_medicare_medicaid,'[^A-Za-z0-9.]', '') ='' then null
else CAST(REPLACE(num_entitlement_medicare_medicaid,'[^0-9.]', '') AS DECIMAL)
end as medicare_medicaid_entitlement,
case
when REPLACE(total_submitted_charge_amount,'[^A-Za-z0-9.]', '') ='' then null
else CAST(REPLACE(total_submitted_charge_amount,'[^0-9.]', '') AS DECIMAL)
end as total_submitted_charge_amount
FROM cmsaggregatepayment2017

Date format substitution in PL/SQL. Example: from 5y 6m 20d to 050620

I am writing a query where I need to perform a date format transformation to meet the specified requirements.
In the database which I have to search, the date format looks like the one in the example: 5y 6m 10d with spaces in between and with optional digits (10y 30d; 1m 23d; 6m are also valid) and they are always ordered (first years, then month and then days).
The format transformation should be the following:
10y 6m 10d => 100610
1y 10m 1d => 011001
6m 2d => 000602
So that the output is always a 6-digit number.
I tried writing regular expressions within REGEX_SUBSTR to isolate the tokens and then concatenate them together in the type of SELECT REGEXP_SUBSTR(text_source, '(\d+)*y') FROM database and I also tried using the REGEX_REPLACE function. Nevertheless, I am not able to perform the transformation to two digits per token without spaces, nor replace one pattern by another, I can only replace the pattern by another string.
Although I am able to output the token separation without spaces by writing the function above. I am not able to get the whole transformation. Is there any possibility of writing a RegEx and combining it with any of the PL/SQL functions in order to transform the dates stated on the list above ? I am also open to hear any other solutions not involving RegEx, I just thought it was sensible to make a proper use of them here.
Here is a simple solution in SQL.
you get the values for year, month and day e.g. with regexp_substr.
with nvl you set the value to 0 if there it is null.
lpad it with 0
with tab as(
select '10y 6m 10d' as str from dual union all
select '1y 10m 1d ' as str from dual union all
select '6m 2d ' as str from dual
)
select lpad(nvl(y,0), 2,'0') ||lpad(nvl(m,0), 2,'0')|| lpad(nvl(d,0), 2,'0')
from (
select rtrim(regexp_substr(str, '[0-9]{1,2}y', 1),'y') as y
,rtrim(regexp_substr(str, '[0-9]{1,2}m', 1),'m') as m
,rtrim(regexp_substr(str, '[0-9]{1,2}d', 1),'d') as d
from tab
)
;
LPAD(N
------
100610
011001
000602
I hope it works
declare
myDate_ varchar2(50) := REPLACE('1y 10m 81d',' ','');
year_ varchar2(50);
month_ varchar2(50);
day_ varchar2(50);
begin
if instrb(myDate_,'y',1,1)>0 then
year_ := lpad(regexp_substr(substr(myDate_,0,instrb(myDate_,'y',1,1)), '[^y]+',1 , 1),2,0);
end if;
if instrb(myDate_,'m',1,1)>0 then
month_ := lpad(regexp_substr(substr(myDate_,instrb(myDate_,'y',1,1)+1,instrb(myDate_,'m',1,1)), '[^m]+',1 , 1),2,0);
end if;
if instrb(myDate_,'d',1,1)>0 then
day_ := lpad(regexp_substr(substr(myDate_,instrb(myDate_,'m',1,1)+1,instrb(myDate_,'d',1,1)), '[^d]+',1 , 1),2,0);
end if;
dbms_output.put_line(year_||month_||day_);
end;

display the content of a file split by a delimiter character

I am trying to display the content of a file, split by a delimiter character.
More exactly, starting from this topic, I am trying to display the result as:
bbb
aaa
qqq
ccc
but the data source to be taken from a file.
Until now, I tried:
DECLARE
l_bfile bfile;
BEGIN
l_bfile := bfilename(my_dir, my_file);
dbms_lob.fileopen(l_bfile);
FOR i IN
(SELECT TRIM(regexp_substr(TO_CHAR(l_bfile),'[^;]+',1,level) ) AS q
FROM dual
CONNECT BY regexp_substr(TO_CHAR(l_bfile),'[^;]+',1,level) IS NOT NULL
ORDER BY level
)
LOOP
dbms_output.put_line(i.q);
END LOOP;
EXCEPTION
WHEN No_Data_Found THEN
NULL;
END;
As result, I got
PL/SQL: ORA-00932: inconsistent datatypes: expected NUMBER got FILE
Can anyone give me a hint, please?
Have to write this as a new answer since this is too big for a comment to #SmartDumb:
Be advised the regex of the form '[^;]+' (commonly used for parsing delimited lists) fails when NULL elements are found in the list. Please see this post for more information: https://stackoverflow.com/a/31464699/2543416
Instead please use this form of the call to regexp_substr (note I removed the second element):
SELECT TRIM(regexp_substr('bbb;;qqq;ccc','(.*?)(;|$)',1,level, null, 1) ) AS q
FROM dual
CONNECT BY regexp_substr('bbb;;qqq;ccc','(.*?)(;|$)',1,level) IS NOT NULL
ORDER BY level
It may or may not be important in this example, it depends on if the order of the element in the string has importance to you or if you need to preserve the NULL. i.e. if you need to know the second element is NULL then this will work.
P.S. Do a search for external tables and see if that is a solution you could use. That would let you query a file as if it were a table.
You could possible try this if your file contains single line (hence the question about file structure):
DECLARE
utlFileHandle UTL_FILE.FILE_TYPE;
vLine varchar2(100);
BEGIN
utlFileHande := UTL_FILE.FOPEN(my_dir, my_file, 'r');
utl_file.get_line(utlFileHande, vLine);
FOR i IN
(SELECT TRIM(regexp_substr(vLine,'[^;]+',1,level) ) AS q
FROM dual
CONNECT BY regexp_substr(vLine,'[^;]+',1,level) IS NOT NULL
ORDER BY level
)
LOOP
dbms_output.put_line(i.q);
END LOOP;
utl_file.fclose(utlFileHande);
EXCEPTION
WHEN No_Data_Found THEN
utl_file.fclose(utlFileHande);
null;
END;

Equivalent regex in T-SQL with start/end input markers

I'm trying to get the following RegEx to work:
^[a-zA-Z][a-zA-Z ''.-]+[a-zA-Z]$
It should allow any alphas, space, apostrophe, full stop and hyphen as long as the beginning and last chars as alphas.
John - ok
John Smith - ok
John-Smith - ok
John.Smith - ok
.John Smith - not ok
John Smith. - not ok
When I use this in T-SQL it doesn't seem to work and I'm not sure if it the input start/end markers that are not compatible in T-SQL. How do I translate this to valid T-SQL?:
CREATE Function [dbo].[IsValidName](#value VarChar(MAX))
RETURNS INT
AS
Begin
DECLARE #temp INT
SET #temp = (
SELECT
CASE WHEN #value LIKE '%^[a-zA-Z][a-zA-Z ''.-]+[a-zA-Z]$%' THEN 1
ELSE 0
END
)
RETURN #Temp
End
T-SQL doesn't support regular expressions "out of the box". Depending on what environment you are using, there are different solutions, but none will probably be "pure T-SQL". In a Microsoft environment you can use CLR procedures to achieve this.
See SQL Server Regular expressions in T-SQL for some options.
I made SOMETHING like this to scrub data, to remove non-alpha characters, I've slightly modified it to fit your needs
CREATE Function [dbo].Func (#Temp VarChar(1000))
Returns VarChar(1000)
AS
BEGIN
DECLARE #Len INT = LEN(#Temp)
DECLARE #RETURN INT
Declare #KeepValues as varchar(50)
Set #KeepValues = '%[^a-z^ ]%'
IF PatIndex(#KeepValues, #Temp) = 1
BEGIN
Set #RETURN = 0
END
IF PATINDEX(#KeepValues, #Temp) = #Len
BEGIN
SET #RETURN = 0
END
IF PATINDEX(#KeepValues, #Temp) = 0
SET #RETURN = 1
IF #RETURN IS NULL
BEGIN
SET #Return = 1
END
RETURN #RETURN
END
This is assuming you would not need to do any sort of data scrubbing for restricted characters. If you need to scrub for restricted characters let me know we can add a little more in there but based on your dataset this will return the correct answers

PL/SQL optimize searching a date in varchar

I have a table, that contains date field (let it be date s_date) and description field (varchar2(n) desc). What I need is to write a script (or a single query, if possible), that will parse the desc field and if it contains a valid oracle date, then it will cut this date and update the s_date, if it is null.
But there are one more condition - there are must be exactly one occurence of a date in the desc. If there are 0 or >1 - nothing should be updated.
By the time I came up with this pretty ugly solution using regular expressions:
----------------------------------------------
create or replace function to_date_single( p_date_str in varchar2 )
return date
is
l_date date;
pRegEx varchar(150);
pResStr varchar(150);
begin
pRegEx := '((0[1-9]|[12][0-9]|3[01])[.](0[1-9]|1[012])[.](19|20)\d\d)((.|\n|\t|\s)*((0[1-9]|[12][0-9]|3[01])[.](0[1-9]|1[012])[.](19|20)\d\d))?';
pResStr := regexp_substr(p_date_str, pRegEx);
if not (length(pResStr) = 10)
then return null;
end if;
l_date := to_date(pResStr, 'dd.mm.yyyy');
return l_date;
exception
when others then return null;
end to_date_single;
----------------------------------------------
update myTable t
set t.s_date = to_date_single(t.desc)
where t.s_date is null;
----------------------------------------------
But it's working extremely slow (more than a second for each record and i need to update about 30000 records). Is it possible to optimize the function somehow? Maybe it is the way to do the thing without regexp? Any other ideas?
Any advice is appreciated :)
EDIT:
OK, maybe it'll be useful for someone. The following regular expression performs check for valid date (DD.MM.YYYY) taking into account the number of days in a month, including the check for leap year:
(((0[1-9]|[12]\d|3[01])\.(0[13578]|1[02])\.((19|[2-9]\d)\d{2}))|((0[1-9]|[12]\d|30)\.(0[13456789]|1[012])\.((19|[2-9]\d)\d{2}))|((0[1-9]|1\d|2[0-8])\.02\.((19|[2-9]\d)\d{2}))|(29\.02\.((1[6-9]|[2-9]\d)(0[48]|[2468][048]|[13579][26])|((16|[2468][048]|[3579][26])00))))
I used it with the query, suggested by #David (see accepted answer), but I've tried select instead of update (so it's 1 regexp less per row, because we don't do regexp_substr) just for "benchmarking" purpose.
Numbers probably won't tell much here, cause it all depends on hardware, software and specific DB design, but it took about 2 minutes to select 36K records for me. Update will be slower, but I think It'll still be a reasonable time.
I would refactor it along the lines of a single update query.
Use two regexp_instr() calls in the where clause to find rows for which a first occurrence of the match occurs and a second occurrence does not, and regexp_substr() to pull the matching characters for the update.
update my_table
set my_date = to_date(regexp_subtr(desc,...),...)
where regexp_instr(desc,pattern,1,1) > 0 and
regexp_instr(desc,pattern,1,2) = 0
You might get even better performance with:
update my_table
set my_date = to_date(regexp_subtr(desc,...),...)
where case regexp_instr(desc,pattern,1,1)
when 0 then 'N'
else case regexp_instr(desc,pattern,1,2)
when 0 then 'Y'
else 'N'
end
end = 'Y'
... as it only evaluates the second regexp if the first is non-zero. The first query might also do that but the optimiser might choose to evaluate the second predicate first because it is an equality condition, under the assumption that it's more selective.
Or reordering the Case expression might be better -- it's a trade-off that's difficult to judge and probably very dependent on the data.
I think there's no way to improve this task. Actually, in order to achieve what you want it should get even slower.
Your regular expression matches text like 31.02.2013, 31.04.2013 outside the range of the month. If you put year in the game,
it gets even worse. 29.02.2012 is valid, but 29.02.2013 is not.
That's why you have to test if the result is a valid date.
Since there isn't a full regular expression for that, you would have to do it by PLSQL really.
In your to_date_single function you return null when a invalid date is found.
But that doesn't mean there won't be other valid dates forward on the text.
So you have to keep trying until you either find two valid dates or hit the end of the text:
create or replace function fn_to_date(p_date_str in varchar2) return date is
l_date date;
pRegEx varchar(150);
pResStr varchar(150);
vn_findings number;
vn_loop number;
begin
vn_findings := 0;
vn_loop := 1;
pRegEx := '((0[1-9]|[12][0-9]|3[01])[.](0[1-9]|1[012])[.](19|20)\d\d)';
loop
pResStr := regexp_substr(p_date_str, pRegEx, 1, vn_loop);
if pResStr is null then exit; end if;
begin
l_date := to_date(pResStr, 'dd.mm.yyyy');
vn_findings := vn_findings + 1;
-- your crazy requirement :)
if vn_findings = 2 then
return null;
end if;
exception when others then
null;
end;
-- you have to keep trying :)
vn_loop := vn_loop + 1;
end loop;
return l_date;
end;
Some tests:
select fn_to_date('xxxx29.02.2012xxxxx') c1 --ok
, fn_to_date('xxxx29.02.2012xxx29.02.2013xxx') c2 --ok, 2nd is invalid
, fn_to_date('xxxx29.02.2012xxx29.02.2016xxx') c2 --null, both are valid
from dual
As you are going to have to do try and error anyway one idea would be to use a simpler regular expression.
Something like \d\d[.]\d\d[.]\d\d\d\d would suffice. That would depend on your data, of course.
Using #David's idea you could filter the ammount of rows to apply your to_date_single function (because it's slow),
but regular expressions alone won't do what you want:
update my_table
set my_date = fn_to_date( )
where regexp_instr(desc,patern,1,1) > 0