錯誤做一個完全外部聯接

product表：錯誤做一個完全外部聯接

╔══════════════════════════════════╦═══════════╦═══════════════════╦════════════════════╦════╗ 
║    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表：

╔══════════════════════════════════╦═══════╦═════════╦═════════════════════╗ 
║    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 ║ 
╚══════════════════════════════════╩═══════╩═════════╩═════════════════════╝

...，我希望從兩個表做一個FULL OUTER JOIN（得到列只有當關鍵使用Medoo存在於兩個，如果我理解正確）：

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

錯誤：

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

我也嘗試過這些查詢的控制檯，但得到一個語法錯誤：

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

和

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

預期結果：

╔═══════════════════╦══════╦═════════╦═════════╦════╦═══════╦═════════╦═════════════════════╗ 
║  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 ║ 
╚═══════════════════╩══════╩═════════╩═════════╩════╩═══════╩═════════╩═════════════════════╝

來源

2017-05-27 3zzy

每個'ref'小號eems在兩個表中是相同的。爲什麼你的預期結果只有一個記錄。而你的sql'加入'同一個表'產品'。 – Blank

@Forward啊，是這個問題嗎？我是MySQL的新手，所以沒有關於關係數據庫的線索。現在我明白了，列名應該是不同的我猜，這就是爲什麼它不工作。 – 3zzy

@Forward這是我的數據庫結構：https://pastebin.com/6A0E1xUM – 3zzy

MySQL不支持FULL OUTER JOIN，和你的邏輯似乎是INNER JOIN：

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

但我不知道爲什麼兩個表中ref的都是一樣的，您預期的結果只有一個記錄，根據Medoo文檔，代碼應該像如下：

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

對於所有的表的，試試這個：

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

來源

2017-05-27 01:33:21 Blank

太棒了，按預期工作！謝謝！ – 3zzy

只注意到它顯示第一行兩次的唯一問題。 – 3zzy

錯誤做一個完全外部聯接

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