PostgreSQL - FULL JOIN Keyword
The PostgreSQL FULL JOIN keyword (or sometimes called FULL OUTER JOIN) is used to combine column values of two tables based on the match between the columns and returns all rows of the both tables. If there is no match in any table, the result table will contain NULL value.
Syntax
The syntax for using FULL JOIN keyword in PostgreSQL is given below:
SELECT table1.column1, table1.column2, table2.column1, table2.column2, ... FROM table1 FULL JOIN table2 ON table1.matching_column = table2.matching_column;
Example:
Consider database tables called Employee and Contact_Info with the following records:
Table 1: Employee table
EmpID | Name | City | Age | Salary |
---|---|---|---|---|
1 | John | London | 25 | 3000 |
2 | Marry | New York | 24 | 2750 |
3 | Jo | Paris | 27 | 2800 |
4 | Kim | Amsterdam | 30 | 3100 |
5 | Ramesh | New Delhi | 28 | 3000 |
6 | Huang | Beijing | 28 | 2800 |
Table 2: Contact_Info table
Phone_Number | EmpID | Address | Gender |
---|---|---|---|
+1-8054098000 | 2 | Brooklyn, New York, USA | F |
+33-147996101 | 3 | Grenelle, Paris, France | M |
+31-201150319 | 4 | Geuzenveld, Amsterdam, Netherlands | F |
+86-1099732458 | 6 | Yizhuangzhen, Beijing, China | M |
+65-67234824 | 7 | Yishun, Singapore | M |
+81-357799072 | 8 | Koto City, Tokyo, Japan | M |
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To full join Employee and Contact_Info tables based on matching column EmpID, the query is given below. This will fetch Name and Age columns from Employee table and Address column from Contact_Info table.
SELECT Employee.Name, Employee.Age, Contact_Info.Address FROM Employee FULL JOIN Contact_Info ON Employee.EmpID = Contact_Info.EmpID;
This will produce the result as shown below:
Name Age Address John 25 Marry 24 Brooklyn, New York, USA Jo 27 Grenelle, Paris, France Kim 30 Geuzenveld, Amsterdam, Netherlands Ramesh 28 Huang 28 Yizhuangzhen, Beijing, China Yishun, Singapore Koto City, Tokyo, Japan -
To fetch all fields of a table, table.* keyword is used, for example - to fetch all fields of the Employee table, Employee.* is used in the below query:
SELECT Employee.*, Contact_Info.Address FROM Employee FULL JOIN Contact_Info ON Employee.EmpID = Contact_Info.EmpID;
This result of the following code will be:
EmpID Name City Age Salary Address 1 John London 25 3000 2 Marry New York 24 2750 Brooklyn, New York, USA 3 Jo Paris 27 2800 Grenelle, Paris, France 4 Kim Amsterdam 30 3100 Geuzenveld, Amsterdam, Netherlands 5 Ramesh New Delhi 28 3000 6 Huang China 28 2800 Yizhuangzhen, Beijing, China Yishun, Singapore Koto City, Tokyo, Japan
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