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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T16:07:01+00:00 2026-05-23T16:07:01+00:00

I’m trying to produce a page that would function a bit like a digital

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I’m trying to produce a page that would function a bit like a digital sticker album. My SQL knowledge isn’t very good so I need a bit of help finishing off what I have so far.

I want to display a list of all available stickers but then also show whether or not a user has a sticker. So the idea is to display a list of empty boxes(items) and then display an image inside the box if the user owns the sticker.

The two relevant tables I have are called “items” and “inventory”. Items contains all the available stickers and inventory contains the stickers owned by the users.

Here are the available columns:

CREATE TABLE items (
    id INT UNSIGNED PRIMARY KEY AUTO_INCREMENT,
    name VARCHAR(255)
) Engine=InnoDB;

CREATE TABLE inventory (
    userid INT UNSIGNED,
    itemid INT UNSIGNED,
    -- FOREIGN KEY (userid) REFERENCES users (id),
    FOREIGN KEY (itemid) REFERENCES items (id),
    UNIQUE (itemid, userid)
) Engine=InnoDB;

I’m open to suggestions on the best way to go about doing this using PHP and MySQL, but I think what I need is a query to return a list of all the item names and then another column to flag whether the user has the item. I can then loop through the items in php and then use a conditional based on the second column to show if the sticker is there or not.

So far i’ve got as far as the below query but it needs to only show items for the current user. Sticking in a ‘where’ clause doesn’t work either as it then only shows the inventory items and not the NULL items (that is, it doesn’t include all items).

SELECT items.name, inventory.userid
  FROM items
    LEFT JOIN inventory ON items.id = inventory.itemid

SELECT items.name, inventory.userid
  FROM items
    LEFT JOIN inventory ON items.id = inventory.itemid
  WHERE inventory.userid = '$userid'
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T16:07:02+00:00Added an answer on May 23, 2026 at 4:07 pm

    Try moving the userid test to the JOIN condition:

    SELECT items.name, inventory.userid
      FROM items
        LEFT JOIN inventory 
          ON items.id = inventory.itemid AND inventory.userid=?
    

    That way, when the join condition fails (such as when the item is in another user’s inventory), the item itself is still included in the result, but with a null userid. It can also make use of an appropriately defined index on table inventory.

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