错误做一个完全外部联接

product表：错误做一个完全外部联接

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

...，我希望从两个表做一个FULL OUTER JOIN（得到列只有当关键使用Medoo存在于两个，如果我理解正确）：

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

错误：

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

我也尝试过这些查询的控制台，但得到一个语法错误：

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

和

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

预期结果：

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

来源

2017-05-27 3zzy

每个'ref'小号eems在两个表中是相同的。为什么你的预期结果只有一个记录。而你的sql'加入'同一个表'产品'。 – Blank

@Forward啊，是这个问题吗？我是MySQL的新手，所以没有关于关系数据库的线索。现在我明白了，列名应该是不同的我猜，这就是为什么它不工作。 – 3zzy

@Forward这是我的数据库结构：https://pastebin.com/6A0E1xUM – 3zzy

MySQL不支持FULL OUTER JOIN，和你的逻辑似乎是INNER JOIN：

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

但我不知道为什么两个表中ref的都是一样的，您预期的结果只有一个记录，根据Medoo文档，代码应该像如下：

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

对于所有的表的，试试这个：

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

来源

2017-05-27 01:33:21 Blank

太棒了，按预期工作！谢谢！ – 3zzy

只注意到它显示第一行两次的唯一问题。 – 3zzy

错误做一个完全外部联接

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