Question

我在Laravel中有3个模型：data is data is {"product":{"description":"","productPrice":"","isTaxable":""},"quantity":"","discount":"","finalPrice":"","lineItemTotal":""} 8:352 data is {"delete":" ","id":21,"quantity":12,"discount":12,"product":{"id":501,"upc":null,"description":"Eggs","restockQty":26,"productPrice":40000,"isTaxable":false,"productCost":11000,"image":"","productCode":"01 B 04","packing":20,"productWholesalePrice":20000},"productId":501,"salesOrder":null,"salesOrderId":8,"lineTaxes":9,"isTaxable":true,"lineItemTotal":999,"finalPrice":9999}，Contracter和Contractor_Areas

Contractor_Trade_Type可以具有贸易类型（水暖，电力等）以及按邮政编码工作的区域。

所以我需要选择所有Contractor和某个contractors和trade。

因此，代码需要对postcode Contractor_Trade_Type（在这种情况下为id）和此处的邮政区域16进行以下操作，并给我匹配的承包商输入ID和邮政编码。

但是它说：

找不到
trade_type_id列

SS

模型如下：

$con = App\Contractor::with('Contractor_Trade_Types')->where('trade_type_id' , 16 )->with( 'contractor_areas' )->where('contractor_areas.postcode','ss')->first();
print_r( $con );

Answer 1

据我所知，您正在尝试链接with()和where()子句以便在关系中进行搜索。

您可以在with()查询方法中使用closure添加子查询：

$con = App\Contractor::with(['contractor_trade_types' => function ($query) {
        $query->where('id', 16);
    }])
    ->with(['contractor_areas' => function($query) {
        $query->where('id', $area_id);
    }])
    ->first();

print_r($con);

此外，这些检查可以添加到query scope中以在您的应用程序中使用：

$con = App\Contractor::withTradeType(16)->withArea('ss')->first();

承包商模型（例如App \ Contractor.php）

public function scopeWithTradeType($query, $trade_type_id)
{
    return $query->with(['contractor_trade_types' => function ($q) {
        $q->where('id', $trade_type_id);
    }]);
}

public function scopeWithArea($query, $area_id)
{
    return $query->with(['contractor_areas' => function($q) {
        $q->where('id', 'ss');
    }]);
}

laravel跨多个表搜索

1 个答案:

承包商模型（例如App \ Contractor.php）