执行完全外连接时出错
执行完全外连接时出错
提问于 2017-05-27 01:25:00
product表：
╔══════════════════════════════════╦═══════════╦═══════════════════╦════════════════════╦════╗
║ 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 ║
╚══════════════════════════════════╩═══════╩═════════╩═════════════════════╝..。我想做一个FULL OUTER JOIN (只有当键在两个表中都存在时，我才能正确地理解？)使用梅杜
$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我还在控制台中尝试了这些查询，但得到了一个语法错误：
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 ║
╚═══════════════════╩══════╩═════════╩═════════╩════╩═══════╩═════════╩═════════════════════╝回答 1
Stack Overflow用户
回答已采纳
发布于 2017-05-27 01:33:21
Mysql不支持FULL OUTER JOIN，您的逻辑似乎是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的ref都是相同的，您的预期结果只有一个记录，根据Medoo文档，代码应该如下所示：
$data = $database->select(
"product",
[
"[><]inventory" => "ref"
],
"*");对于所有的表，请尝试如下：
$data = $database->select(
"product",
[
"[><]inventory" => "ref",
"[><]detail" => "ref",
"[><]moredetails" => "ref",
"[><]info" => "ref",
"[><]images" => "ref",
"[><]features" => "ref",
"[><]categories" => "ref",
"[><]tags" => "ref"
],
["product.*", "inventory.*"]);页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/44212044
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