This is a very popular interview question of the database, many times I faced this the question in SQL Server interview.
Problem:Suppose we have the following table (employee)
id | deprt_id | emp_name |
1 | 1 | Dilip |
2 | 1 | Anil |
3 | 1 | Mamta |
4 | 2 | Ashish |
5 | 2 | Naina |
And I want to result set as below
deprt_id | emp_name |
1 | Dilip,Anil,Mamta |
2 | Ashish,Naina,Kartikey |
To achieve this above result-set we will work on the PostgreSQL query. As new versions of the database come along, some new features and enhancements are also added. According to the different versions, whatever features are available here, I will give detailed information about it.
Create table employee
CREATETABLEemployee ( id intGENERATEDBYDEFAULTASIDENTITY, deprt_idtext, emp_nametext ); INSERTINTOemployee(deprt_id,emp_name)VALUES (1,'Dilip'); INSERTINTOemployee(deprt_id,emp_name)VALUES (1,'Anil'); INSERTINTOemployee(deprt_id,emp_name)VALUES (1,'Mamta'); INSERTINTOemployee(deprt_id,emp_name)VALUES (2,'Ashish'); INSERTINTOemployee(deprt_id,emp_name)VALUES (2,'Naina'); INSERTINTOemployee(deprt_id,emp_name)VALUES (2,'Kartikey'); |
PostgreSQL 9.0 or later
The function string_agg(expression, delimiter) is used to concatenate string value of a column, run the following query.
SELECTdeprt_id,string_agg(emp_name,',') FROMemployee GROUPBYdeprt_id; |
ORDER BY clause in any aggregate expression string_agg(column_name, ',', order by column_name).
SELECTdeprt_id,string_agg(emp_name,','ORDERBYemployee) FROMemployee GROUPBYdeprt_id; |
PostgreSQL 8.4 or later
The function array_to_string() can be used to give the desired result.
SELECTdeprt_id,array_to_string(array_agg(emp_name),',') FROMemployee GROUPBYdeprt_id; |
