Error doing a Full Outer Join

Question

product table:

╔══════════════════════════════════╦═══════════╦═══════════════════╦════════════════════╦════╗
║               ref                ║    mfr    ║       pnum        ║        ssku        ║ id ║
╠══════════════════════════════════╬═══════════╬═══════════════════╬════════════════════╬════╣
║ 6541_aten_2a-130g                ║ Aten      ║ 2A-130G           ║ 2A-130G            ║  6 ║
║ 7466_eaton_5sc1000i              ║ Eaton     ║ 5SC1000I          ║                    ║  8 ║
║ 8214_ivanti-uk_template-material ║ IVANTI UK ║ TEMPLATE MATERIAL ║ 000000000003616655 ║  4 ║
║ 8361_aywun_92sfan1               ║ Aywun     ║ 92SFAN1           ║ 92SFAN             ║  9 ║
║ 9824_autodesk_00100-000000-9880  ║ AUTODESK  ║ 00100-000000-9880 ║ 00100-000000-9880  ║  5 ║
╚══════════════════════════════════╩═══════════╩═══════════════════╩════════════════════╩════╝

inventory table:

╔══════════════════════════════════╦═══════╦═════════╦═════════════════════╗
║               ref                ║ scost ║ instock ║        date         ║
╠══════════════════════════════════╬═══════╬═════════╬═════════════════════╣
║ 6541_aten_2a-130g                ║    26 ║       0 ║ 2017-05-27 10:45:23 ║
║ 7466_eaton_5sc1000i              ║   489 ║       0 ║ 2017-05-27 10:45:23 ║
║ 8214_ivanti-uk_template-material ║     0 ║       0 ║ 2017-05-27 10:45:23 ║
║ 8361_aywun_92sfan1               ║     4 ║       0 ║ 2017-05-27 10:45:23 ║
║ 9824_autodesk_00100-000000-9880  ║   738 ║       0 ║ 2017-05-27 10:45:23 ║
╚══════════════════════════════════╩═══════╩═════════╩═════════════════════╝

... and I'm looking to do a FULL OUTER JOIN (get columns from both tables only if key exists in both if I understand correctly?) using Medoo:

$data = $database->select("product", [
    "[<>]inventory" => ["ref" => "ref"],
]);

Error:

Invalid argument supplied for foreach() in /var/www/html/vendor/catfan/medoo/src/Medoo.php on line

I also tried these queries in the console but getting a syntax error:

SELECT *
FROM product
FULL OUTER JOIN product ON product.ref = inventory.ref;

and

SELECT * FROM `product`, * FROM `inventory` 
WHERE product.`ref` = inventory.`ref`;

Expected result:

╔═══════════════════╦══════╦═════════╦═════════╦════╦═══════╦═════════╦═════════════════════╗
║        ref        ║ mfr  ║  pnum   ║  ssku   ║ id ║ scost ║ instock ║        date         ║
╠═══════════════════╬══════╬═════════╬═════════╬════╬═══════╬═════════╬═════════════════════╣
║ 6541_aten_2a-130g ║ Aten ║ 2A-130G ║ 2A-130G ║  6 ║    26 ║       0 ║ 2017-05-27 10:45:23 ║
╚═══════════════════╩══════╩═════════╩═════════╩════╩═══════╩═════════╩═════════════════════╝

Answer 1

Mysql does not support FULL OUTER JOIN, and your logic seems to be INNER JOIN:

SELECT *
FROM product
INNER JOIN inventory ON product.`ref` = inventory.`ref`;

But I don't know why in two tables the ref's are all the same, your expected result only have one record, and according to Medoo document, the code should be like following:

$data = $database->select(
   "product",
    [
       "[><]inventory" => "ref"
    ],
    "*");

For all of your tables, try this:

$data = $database->select(
   "product",
    [
       "[><]inventory" => "ref",
       "[><]detail" => "ref",
       "[><]moredetails" => "ref",
       "[><]info" => "ref",
       "[><]images" => "ref",
       "[><]features" => "ref",
       "[><]categories" => "ref",
       "[><]tags" => "ref"
    ],
    ["product.*", "inventory.*"]);