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Overview

The FROM clause is used to specify which table or joins are required for the query/statement (e.g., SELECT statement) to return or obtain data.

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Syntax

There must be at least one table listed in the FROM clause. See the following syntax: 
query FROM table_name;
If two or more tables are listed in the FROM clause, these tables are joined using JOIN, RIGHT JOIN, LEFT JOIN, or OUTER JOIN, depending on the operations to be queried as seen in the syntax below: 
FROM table1_name
[ { JOIN
  | LEFT JOIN
  | RIGHT JOIN
  | OUTER JOIN } table2_name
ON table1_name.column1 = table2_name.column1 ]
The examples below are executed in the public schema, the default schema in Oxla. You can also create, insert, and display a table from other schemas - click here for more info.

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Example

We’ll start by looking at how to use the FROM clause with only a single table. There is a client table, and we want to know the client’s name and the city where the company is based. 
CREATE TABLE client (
  client_id int,
  client_name text,
  client_origin text
);
INSERT INTO client 
    (client_id, client_name, client_origin) 
VALUES 
    (181891,'Oxla','Poland'),
    (181892,'Google','USA'),
    (181893,'Samsung','South Korea');

SELECT * FROM client;
It will create a table as shown below: 
+------------+--------------+------------------+
| client_id  | client_name  | client_origin    |
+------------+--------------+------------------+
| 181891     | Oxla         | Poland           |
| 181892     | Google       | USA              | 
| 181893     | Samsung      | South Korea      |
+------------+--------------+------------------+
  1. Run the following query:
SELECT client_name, client_origin FROM client;
  1. You will get a list of the client’s data for a successful result:
+--------------+------------------+
| client_name  | client_origin    |
+--------------+------------------+
| Oxla         | Poland           |
| Google       | USA              | 
| Samsung      | South Korea      |
+--------------+------------------+
If two or more tables are listed in the FROM clause, please refer to these sections for more examples related to this: JOIN, RIGHT JOIN, LEFT JOIN, or OUTER JOIN.

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FROM - Sub Queries

FROM clause is also used to specify a sub-query expression. The relation created from the sub-query is then used as a new relation on the other query.
More than one table can be defined by separating it with a comma (,).

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Syntax

Here is an example of the sub-query syntax that uses a FROM clause: 
SELECT X.column1, X.column2, X.column3 
FROM table_2 as X, table_1 as Y
WHERE conditions (X.column, Y.column);
  1. The sub-query in the first FROM clause will select the columns from the specific table using a new temporary relation (SELECT X.column1, X.column2, X.column3 FROM ).
  2. Set the tables into a new temporary relation (table_2 as X, table_1 as Y).
  3. Next, the query is evaluated, selecting only those rows from the temporary relation that fulfill the conditions stated in the WHERE clause.

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Example

We want to find a product whose price exceeds all categories’ average budget.  product table 
CREATE TABLE product (
  id int,
  product text,
  category text,
  price int
);
INSERT INTO product 
    (id, product, category, price) 
VALUES 
    (445747,'Court vision women’s shoes nike','Shoes', 8000),
    (445641,'Disney kids h&m','Shirt', 6500),
    (477278,'Defacto adidas','Hat', 8500),
    (481427,'Sophie shopping bag','Bag', 6500),
    (411547,'Candy skirt zara','Skirt', 6500),
    (488198,'Slim cut skirt hush puppies','Skirt', 7600);

SELECT * FROM product;
It will create a table as shown below: 
+---------+----------------------------------+-----------+--------+
| id      | product                          | category  | price  |
+---------+----------------------------------+-----------+--------+
| 445747  | Court vision women’s shoes nike  | Shoes     | 8000   |
| 445641  | Disney kids h&m                  | Shirt     | 6500   |
| 477278  | Defacto adidas                   | Hat       | 8500   |
| 481427  | Sophie shopping bag              | Bag       | 6500   |
| 411547  | Candy skirt zara                 | Skirt     | 6500   |
| 488198  | Slim cut skirt hush puppies      | Skirt     | 7600   |
+---------+----------------------------------+-----------+--------+
category table 
CREATE TABLE category (
  categoryName text,
  budget int
);
INSERT INTO category 
    (categoryName, budget) 
VALUES 
    ('Shoes', 7000),
    ('Shirt', 9000),
    ('Bag', 8000),
    ('Skirt', 7500),
    ('Hat', 7000);

SELECT * FROM category;
It will create a table as shown below: 
+---------------+----------+
| categoryName  | budget   |
+---------------+----------+
| Shoes         | 7000     |
| Shirt         | 9000     |
| Bag           | 8000     |
| Skirt         | 7500     |
| Hat           | 7000     |
+---------------+----------+
  1. Run the following query to know and ensure the average value of all category’s budgets:
select avg(budget) as avgBudget from category;
  1. The average budget of all categories from the category table is 7700. 
+--------------------+
| avgbudget          |
+--------------------+
| 7700.000000000000  |
+--------------------+
  1. Now, run the following query:
  • We specify the product table as P and the budget’s average value from the category table as C.
  • We will display the product’s name, category, and price.
  • We set the conditions where the product’s price exceeds the budget’s average value.
select P.product, P.category, P.price from
(select avg(budget) as avgBudget from category) as C, product as P
where P.price > C.avgBudget;
➡️ The output will display “court vision women’s shoes nike” and “Defacto adidas” as the products with a price of more than 7700. 
+------------------------------------+-----------+----------+
| product                            | category  | price    |
+------------------------------------+-----------+----------+
| court vision women`s shoes nike    | shoes     | 8000     |
| Defacto adidas                     | hat       | 8500     |
+------------------------------------+-----------+----------+