Delete duplicate records without creating a temporary table

Question

I have a table with many duplicate records:

shop
ID     tax_id
1      10
1      10
1      11
2      10
2      12
2      10
2      10

I want

Answer 1

Working solution.

//Sql query to find duplicates
SELECT id, tax_id, count(*) - 1 AS cnt 
  FROM shop 
  GROUP BY id
  HAVING cnt > 1

--- res

+------+--------+-----+
| id   | tax_id | cnt |
+------+--------+-----+
|    1 |     10 |   2 |
|    2 |     10 |   3 |
+------+--------+-----+


//Iterate through results with your language of choice
DELETE 
  FROM shop 
  WHERE id= 
    AND tax_id= 
  LIMIT 

---res (iterated)

+------+--------+
| id   | tax_id |
+------+--------+
|    1 |     10 |
|    1 |     11 |
|    2 |     12 |
|    2 |     10 |
+------+--------+

The two queries will require a small piece of php in order to carry out the deletes

$res = mysql_query("SELECT id, tax_id, count(*) - 1 AS cnt 
                      FROM shop 
                      GROUP BY id
                      HAVING cnt > 1")
while($row = mysql_fetch_assoc($res)){
    mysql_query("DELETE 
                   FROM shop 
                   WHERE id=".$row['id']."
                       AND tax_id=". $row['tax_id']."
                   LIMIT ".$row['cnt'] -1 . ");
}

Edit: Revisited this recently, for what it's worth, here's an alternative solution using a temporary column, removing the need for a scripting language.

ALTER TABLE shop ADD COLUMN place INT;

SET @i = 1

UPDATE shop SET place = @i:= @i + 1;

DELETE FROM shop WHERE place NOT IN (SELECT place FROM items GROUP BY id, tax_id);

ALTER TABLE shop DROP COLUMN place;