MySQL Orderby a number, Nulls last

Question

Currently I am doing a very basic OrderBy in my statement.

SELECT * FROM tablename WHERE visible=1 ORDER BY position ASC, id DESC

The probl

Answer 1

For a DATE column you can use:

---

NULLS last:

ORDER BY IFNULL(`myDate`, '9999-12-31') ASC

Blanks last:

ORDER BY IF(`myDate` = '', '9999-12-31', `myDate`) ASC

Answer 2

To achieve following result :

1, 2, 3, 4, NULL, NULL, NULL.

USE syntax, place -(minus sign) before field name and use inverse order_type(Like: If you want order by ASC order then use DESC or if you want DESC order then use ASC)

SELECT * FROM tablename WHERE visible=1 ORDER BY -position DESC

Answer 3

Try using this query:

SELECT * FROM tablename
WHERE visible=1 
ORDER BY 
CASE WHEN position IS NULL THEN 1 ELSE 0 END ASC,id DESC

Answer 4

Why don't you order by NULLS LAST?

SELECT * 
FROM tablename
WHERE visible = 1 
ORDER BY position ASC NULLS LAST, id DESC 

Answer 5

You can swap out instances of NULL with a different value to sort them first (like 0 or -1) or last (a large number or a letter)...

SELECT field1, IF(field2 IS NULL, 9999, field2) as ordered_field2
  FROM tablename
 WHERE visible = 1
 ORDER BY ordered_field2 ASC, id DESC

Answer 6

This is working fine:

SELECT * FROM tablename ORDER BY position = 0, position ASC;

---

position
1 
2
3
0
0