MySql conditional order by

Question

I have this table (simplified):

CREATE TABLE `my_table` (
  `id` INT NOT NULL AUTO_INCREMENT ,
  `item_name` VARCHAR(45) NULL ,
  `price` DECIMAL(10,0) NULL ,
  PRIMARY KEY (`id`) )

I need to select all items from the table, ordered this way:
1. items with price > 0.00 first, ordered by price ASC
2. items with price = 0.00 last, ordered by id

I tried this:

    SELECT * 
    FROM my_table 
    WHERE 1  
    ORDER BY 
      CASE price WHEN !0.00 THEN price 
       ELSE id
      END 
    ASC

And i get results like

item_name | price
----------|-------
foo       | 150,00
bar       |   0,00
baz       | 500,00
hum       |   0,00

How do I build the query to have

item_name | price
----------|-------
foo       | 150,00
baz       | 500,00
bar       |   0,00
hum       |   0,00

?

Thank you for your time

Answer 1

Both queries just give me first the items with price = 0.00.

I need to have first items with price > 0, then the items with price = 0

Answer 2

This will do the trick..

 SELECT * 
    FROM my_table 
    WHERE 1  
    ORDER BY 
      CASE price WHEN 0 THEN 1
       ELSE -1
      END ASC, price asc, id asc

Answer 3

You can also use the following:

SELECT * 
FROM my_table 
WHERE 1  
ORDER BY price=0, price, id;

The part 'price=0' will be 1 for items with zero price, 0 for items with non-zero price. As the default sort order is ASC, non-zero items are now placed first.

The next bit of the order-by clause means that non-zero items are then sorted by price (again ascending). If any items with non-zero price have the same price, they will be further sorted by id, but we don't care about that.

The last part is only there for items where price=0. As all these items do have the same price, the effect is to sort all the zero-priced items by id.