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  1. Practice
  2. SQL
  3. Basic Join
  4. Ollivander's Inventory
  5. Discussions

Ollivander's Inventory

  • chenjesu + 10 comments

    Hermione's reasoning is flawed. The wand chooses the wizard, not the other way around.

    • davi418 + 0 comments

      LOL

    • Hari_MBIH + 0 comments
      [deleted]
    • torab_shaikh + 0 comments

      hahaha ;)

    • purplegal1976 + 0 comments

      You win the internet.

    • xy2286 + 0 comments

      the ultimate truth!

    • dannywin19901102 + 0 comments

      LOL It's definately a flaw.

    • te9dinne + 0 comments

      Lol, that's so true!

    • hariom7793 + 1 comment

      rofl...

      • jagatsemwal + 1 comment

        Complicated yet simple :) Below is my code in MySQL

        SELECT temp2.I, temp2.A, temp2.WNN, temp2.P FROM (SELECT MIN(W1.COINS_NEEDED) AS WN, WP1.AGE as AG, W1.POWER AS PW FROM WANDS W1 INNER JOIN WANDS_PROPERTY WP1 ON W1.CODE=WP1.CODE 
 GROUP BY W1.POWER, WP1.AGE ORDER BY W1.POWER DESC, WP1.AGE DESC) temp1
INNER JOIN
(SELECT W.ID AS I, MIN(W.COINS_NEEDED) AS WNN, WP.AGE as A, W.POWER AS P  FROM WANDS W INNER JOIN WANDS_PROPERTY WP ON W.CODE=WP.CODE 
WHERE WP.IS_EVIL=0
GROUP BY W.POWER, WP.AGE, W.ID ORDER BY W.POWER DESC, WP.AGE DESC) temp2
ON temp1.WN=temp2.WNN AND temp1.PW=temp2.P AND temp1.AG=temp2.A;
        • akshaypanchal23 + 2 comments

          Here is my simple solution

          select a.id, b.age, a.coins_needed, a.power from Wands a inner join Wands_Property b on a.code=b.code where b.is_evil!=1 and a.coins_needed=(select min(Wands.coins_needed) from Wands inner join Wands_Property on Wands.code=Wands_Property.code where Wands_Property.age=b.age and Wands.power=a.power) order by a.power desc,b.age desc

          • zdravko2 + 0 comments

            Could you explain why is it that if I have 

            Wands_Property.id = a.id

            in the final where clause I get duplicates (power and age) in the result.

          • anarchtheist + 0 comments

            This is a little less nested than what I came up with, good job! I tried to figure out a clever way to OVER(PARTITION BY but came up short.

    • jayakrishnanadi1 + 0 comments

      haha..crct..

    • sheatsej + 0 comments

      so true!