Laravel5：复杂的 hasManyThrouh 自引用关系（带有别名和 SQL 计算）

Question

我正在 laravel 重建一个项目，我当前的问题是定义一个复杂的自引用 hasManyThrough 关系与 table 别名和 SQL 计算。

该关系应该根据匹配标签的向下总和来查找相关商家。模型拥有的相同标签越多，它们就越相关。

到目前为止，还不错。在我的旧项目中，我只是写下了以下 suitable SQL 查询：

SELECT source_merchant.id, target_merchant.id, COUNT(target_merchant.id) /
((
    (SELECT COUNT(*) FROM tagged WHERE model = 'Merchant' AND model_id = source_merchant.id) + 
    (SELECT COUNT(*) FROM tagged WHERE model = 'Merchant' AND model_id = target_merchant.id) 
) /2 ) as similarity 
FROM merchants source_merchant 
LEFT JOIN tagged source_merchant_tags ON (
    source_merchant.id = source_merchant_tags.model_id AND 
    source_merchant_tags.model = 'Merchant'
) 
INNER JOIN tagged target_merchant_tags ON (
    source_merchant_tags.tag_id = target_merchant_tags.tag_id 
    AND (source_merchant_tags.model = 'Merchant' AND target_merchant_tags.model = 'Merchant') 
    AND (source_merchant_tags.model_id != target_merchant_tags.model_id)
)  
LEFT JOIN merchants target_merchant ON (
    target_merchant_tags.model_id = target_merchant.id AND target_merchant_tags.model = 'Merchant'
) 
WHERE source_merchant.id = 2 
GROUP BY source_merchant.id, target_merchant.id 
ORDER BY similarity DESC
LIMIT 5

最好是捕捉类似

的东西

public function related_merchants() {
    return $this->hasManyThroug(relations_stuff_i_cannot_imagine...)
        ->selectRaw("SELECT source_merchant.id, target_merchant.id, COUNT(target_merchant.id) /
        ((
            (SELECT COUNT(*) FROM tagged WHERE model = 'Merchant' AND model_id = source_merchant.id) + 
            (SELECT COUNT(*) FROM tagged WHERE model = 'Merchant' AND model_id = target_merchant.id) 
        ) /2 ) as similarity")
        ->groupBy('source_merchant.id', 'target_merchant.id ')
        ->orderBy('similarity')
        ->limit(5);
}

就是这样 :-) 不幸的是，我找不到解决方案，因为我不知道如何在 hasManyThrough()...

中定义 suitable 关系参数

编辑 - 尝试将 Laravel 查询构建为 :

public function getRelatedMerchantsAttribute() {

    return $this->from('merchants AS source_merchant')
        ->selectRaw('source_merchant.id, target_merchant.id, COUNT(target_merchant.id) /
            ((
                (' . DB::table('tagged')->whereRaw("model = 'Merchant' AND model_id = source_merchant.id")->count() . ') +
                ' .  DB::table('tagged')->whereRaw("model = 'Merchant' AND model_id = target_merchant.id")->count() . ')
                /2 ) AS similarity')
        ->lefJoin('tagged AS source_merchant_tags', function ($join) {
            $join->on('source_merchant.id', '=', 'source_merchant_tags.model_id')
            ->on('source_merchant_tags.model', '=', 'Merchant');
        })
        ->join('tagged AS target_merchant_tags', function ($join) {
            $join->on('source_merchant_tags.tag_id', '=', 'target_merchant_tags.tag_id')
            ->on('source_merchant_tags.model', '=', 'Merchant')
            ->on('target_merchant_tags.model', '=', 'Merchant')
            ->on('source_merchant_tags.model_id', '!=', 'target_merchant_tags.model_id');
        })
        ->leftJoin('merchants AS target_merchant', function ($join) {
            $join->on('target_merchant_tags.model_id ', '=', 'target_merchant.id')
            ->on('target_merchant_tags.model', '=', 'Merchant');
        })
        ->whereRaw('source_merchant.id = ?', [ $this->id ])
        ->groupBy('source_merchant.id', 'target_merchant.id ')
        ->orderBy('similarity')
        ->limit(5)
        ->get();

}

由于以下原因，此解决方案尚未生效：

SQLSTATE[42S22]: Column not found: 1054 Unknown column 'merchants.id'
in 'where clause' (SQL: select count(*) as aggregate from `tagged`
where model = 'Merchant' AND model_id = merchants.id)

Edit - 使用会产生以下 Laravel 错误：

SQLSTATE[HY093]: 参数号无效。 SQL:

select source_merchant.id, target_merchant.id, 
COUNT(target_merchant.id) /
((
    (select COUNT(*) from `tagged` where `model` = Merchant and `model_id` = source_merchant.id) +
    (select COUNT(*) from `tagged` where `model` = Merchant and `model_id` = target_merchant.id)
) /2 ) AS similarity
from `merchants` as `source_merchant`
left join `tagged` as `source_merchant_tags` on 
    `source_merchant`.`id` = `source_merchant_tags`.`model_id`
    and `source_merchant_tags`.`model` = Merchant
inner join `tagged` as `target_merchant_tags` on 
    `source_merchant_tags`.`tag_id` = `target_merchant_tags`.`tag_id` 
    and `source_merchant_tags`.`model` = Merchant
    and `target_merchant_tags`.`model` = 2
    and `source_merchant_tags`.`model_id` != `target_merchant_tags`.`model_id` 
left join `merchants` as `target_merchant` on 
    `target_merchant_tags`.`model_id` = `target_merchant`.`id`
    and `target_merchant_tags`.`model` = ?
where `source_merchant`.`id` = ? 
group by `source_merchant`.`id`, `target_merchant`.`id`
order by `similarity` desc
limit 5

我什至想知道为什么这里使用当前模型 ID：and target_merchant_tags.model = 2...

看看这张截图。条件参数 'Merchant' 成为当前模型 ID，在本例中为 2（选定文本）。红色圆圈中的两个参数保持为空。这里有什么问题？

Answer 1

我认为恋爱关系不是适合您情况的正确选择。源商户和目标商户在 Laravel/Eloquent 意义上并不真正相关，因为公共标签仅用于确定订单。

我只是将您的原始 SQL 转换为 Laravel 查询并将您的方法重命名为 getRelatedMerchantsAttribute。然后您可以使用 $merchant->relatedMerchants.

访问它们

试试这个查询：

$sourceCount = DB::table('tagged')
    ->selectRaw('COUNT(*)')
    ->where('model', 'Merchant')
    ->where('model_id', DB::raw('source_merchant.id'));
$targetCount = DB::table('tagged')
    ->selectRaw('COUNT(*)')
    ->where('model', 'Merchant')
    ->where('model_id', DB::raw('target_merchant.id'));
$this->from('merchants AS source_merchant')
    ->addBinding($sourceCount->getBindings(), 'select')
    ->addBinding($targetCount->getBindings(), 'select')
    ->selectRaw('source_merchant.id, target_merchant.id, COUNT(target_merchant.id) /
    ((
        (' . $sourceCount->toSql() . ') +
        (' .  $targetCount->toSql() . '))
        /2 ) AS similarity')
    ->leftJoin('tagged AS source_merchant_tags', function ($join) {
        $join->on('source_merchant.id', '=', 'source_merchant_tags.model_id')
            ->where('source_merchant_tags.model', '=', 'Merchant');
    })
    ->join('tagged AS target_merchant_tags', function ($join) {
        $join->on('source_merchant_tags.tag_id', '=', 'target_merchant_tags.tag_id')
            ->where('source_merchant_tags.model', '=', 'Merchant')
            ->where('target_merchant_tags.model', '=', 'Merchant')
            ->on('source_merchant_tags.model_id', '!=', 'target_merchant_tags.model_id');
    })
    ->leftJoin('merchants AS target_merchant', function ($join) {
        $join->on('target_merchant_tags.model_id', '=', 'target_merchant.id')
            ->where('target_merchant_tags.model', '=', 'Merchant');
    })
    ->where('source_merchant.id', $this->id)
    ->groupBy('source_merchant.id', 'target_merchant.id')
    ->orderByDesc('similarity')
    ->limit(5)
    ->get();

Laravel5：复杂的 hasManyThrouh 自引用关系（带有别名和 SQL 计算）

Laravel5: Complex hasManyThrouh self referencing relation (with aliases and SQL calculations)

mysql

model

has-many-through

relation

laravel-5