How to replace a string value with single character - sql

Question

I want to mask the data except the last four characters.
For example:
If there is a varchar column (Name) with the value Rio De janerio, I want to mask it as xxx xx xxxerio.
If it is a numeric column (acc number) with value 123 453 6987,then I want it to be displayed as 000 000 6987.
I tried using right and replace function. But I could not find the solution.
Mask the data except last four characters/numbers

Answer 1

On Oracle, You can try below query to replace numbers with 0s -
SELECT REGEXP_REPLACE (SUBSTR('123 453 6987', 1, LENGTH('123 453 6987')-4), '\d', '0') || SUBSTR('123 453 6987', -4, 4) PROCESSED_STRING
FROM DUAL;
PROCESSED_STRING
000 000 6987
For Replacing strings with #s -
SELECT REGEXP_REPLACE (SUBSTR('Rio De janerio', 1, LENGTH('Rio De janerio')-4), '\S', '#') || SUBSTR('Rio De janerio', -4, 4) PROCESSED_STRING
FROM DUAL;
PROCESSED_STRING
### ## ###erio
Here is the fiddle.

Answer 2

This is not a complete solution but it is just a idea and a head up to move you in the right direction . You can modify the query as per your requirement .
In SQL SERVER :
create table place
(
place_name varchar(100),
place_pincode varchar(100)
)
insert into place values ('Rio De janerio',1234536987)
select substring(place_pincode,len(place_pincode)-3,len(place_pincode)),
case when isnumeric(place_pincode)=1 then
concat('000000',substring(place_pincode,len(place_pincode)-3,len(place_pincode))) end as place_pincode,
case when isnumeric(place_name)<>1 then
concat('XXXXXX',substring(place_name,len(place_name)-3,len(place_name))) end as place_name
from place

Answer 3

If you are using SQL Server 2016 and above you can use Translate function:
declare #name varchar(50) = 'Rio De janerio'
select translate(substring(#name,1,len(#name)-4),'abcdefghijklmnopqrstuvwxyz','**************************') + right(#name,4)
declare #i int = 1234536987
select translate(substring(convert(varchar(50),#i),1,len(#i)-4),'0123456789','0000000000') + right(#i,4)

Answer 4

Using SQLSERVER STUFF() and PATINDEX() will solve this question
FOR MASKING THE NAME
declare #str varchar(20)
declare #rcInt int
set #str ='Rio De janerio'
set #rcInt = 4;
with cte as
(select left(#str, len(#str)-#rcInt) as name
),
rn as
(select stuff(name, patindex('%[a-z]%',name), 1, 'X') as ch, patindex('%[a-z]%',name)as rn, name from cte
union all
select stuff(ch, rn+patindex('%[a-z]%',substring(name, rn+1, len(name))), 1, 'X') as ch, rn+patindex('%[a-z]%',substring(name, rn+1, len(name)))as rn, name from rn
where patindex('%[a-z]%',substring(name, rn+1, len(name)))>0
)
select top 1 ch + right(#str, #rcInt) from rn order by rn desc
FOR TELEPHONE NUMBER (as function)
create function ufn_MaskPhoneNum(
#str varchar(20),
#rcInt int
)
returns #tblReturn table(maskedStr varchar(100))
as
begin
with cte as
(select left(#str, len(#str)-#rcInt) as name
),
rn as
(select stuff(name, patindex('%[0-9]%',name), 1, 'X') as ch, patindex('%[0-9]%',name)as rn, name from cte
union all
select stuff(ch, rn+patindex('%[0-9]%',substring(name, rn+1, len(name))), 1, 'X') as ch,
rn+patindex('%[0-9]%',substring(name, rn+1, len(name)))as rn, name from rn
where patindex('%[0-9]%',substring(name, rn+1, len(name)))>0
)
insert into #tblReturn
select top 1 ch + right(#str, #rcInt) from rn order by rn desc
return;
end
sample:
select * from ufn_MaskPhoneNum('123 453 6987', 4)