Display the specific part of the string in PostgreSQL 9.3 - string

Question

I have a string to modify as per the requirements.
For example:
The given string is:
str1 varchar = '123,456,789';
I want to show the string as:
'456,789'
Note: The first part (delimited) with comma, I want to remove from string and show the rest of string.
In SQL Server I used STUFF() function.
SELECT STUFF('123,456,789',1,4,'');
Result:
456,789
Question: Is there any string function in PostgreSQL 9.3 version to do the same job?

Answer 1

you can use regular expressions:
select substring('123,456,789' from ',(.*)$');
The comma matches the first comma found in the string. The part inside the brackets (.*) is returned from the function. The symbol $ means the end of the string.
A alternative solution without regular expressions:
select str, substring(str from position(',' in str)+1 for length(str)) from
(select '123,456,789'::text as str) as foo;

Answer 2

You could first turn the string to array and return second and third cell:
select array_to_string((regexp_split_to_array('123,456,789', ','))[2:3], ',')
Or you could use substring-function with regular expressions (pattern matching):
SELECT substring('123,456,789' from '[0-9]+,([0-9]+,[0-9]+)')
[0-9]+ means one or more digits
parentheses tell to return that part from the string
Both solutions work on your specific string.

Answer 3

Your The SQL Server example indicates you just want to remove the first 4 characters, which makes the rest of your question seem misleading because it completely ignores what's in the string. Only the positions matters.
Be that as it may, the simple and cheap way to cut off leading characters is with right():
SELECT right('123,456,789', -4);
SQL Fiddle.