Question

我正在重建laravel中的项目，我当前的问题是定义一个复杂的自引用hasManyThrough与表别名和SQL计算的关系。

此关系应根据匹配标记的向下总和找到相关商家。模型具有的标签越多，它们就越相关。

到目前为止，这么好。在我的旧项目中，我刚刚写下了以下合适的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

最好是抓住像

这样的东西

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);
}

那就是:-)不幸的是，我找不到解决方案，因为我不知道，如何在hasManyThrough()中定义合适的关系参数...

修改 - 尝试将Laravel查询构建为suggested：

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，此解决方案尚未运行：

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)

修改 - 使用this query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

我甚至想知道为什么在这里使用当前的模型ID：and target_merchant_tags.model = 2 ...

看一下这个截图。条件参数＆＃39;商家＆＃39;成为当前的模型ID，在这种情况下它是2（所选文本）。红色圆圈中的两个参数保持为空。这里有什么问题？

Answer 1

我不认为关系是你情况的正确选择。在Laravel / Eloquent意义上，源和目标商家并不真正相关，因为常用标签仅用于确定订单。

我只是将原始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

尝试此查询：

$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：Complex hasManyThrouh自引用关系（带别名和SQL计算）

1 个答案: