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laravel - 如何在 laravel 的查询中编写 if 而不破坏链和序列

问题描述

我有以下查询，我需要查询中的 if 语句，我该怎么做。以下查询获取基于制造商的产品列表。如果在此查询之后尝试 if 然后使用 get(); 但这对我不起作用

询问：

  $productsFeatured = Product::select('products.low_emitting_material_prerequisite_file',
        'manufacturers.phone','manufacturers.address','manufacturers.linkedin_url','manufacturers.city', 
        'products.specs_file','products.specs_file_url','products.low_emitting_material_credit_file', 
        'products.id', 'products.name', 'products.logo', 'products.manufacturer_id',
        'products.division_id', 'products.section_id', 'products.website', 'divisions.name as division_name', 'divisions.code as division_code',
        'sections.name as section_name', 'sections.code'   
         )
                ->whereIn('products.status', ['active'])
                ->where(function ($query) {
                    $query->where('products.low_emitting_material_prerequisite_file', '!=', '');
                })
                ->where('users.user_type', 'manufacturer_paid')
                ->leftJoin('products_selected_leeds', 'products.id', '=', 'products_selected_leeds.product_id')
                ->leftJoin('sections', 'products.section_id', '=', 'sections.id')
                ->leftJoin('divisions', 'products.division_id', '=', 'divisions.id')
                ->join('manufacturers', 'products.manufacturer_id', '=', 'manufacturers.id')
                ->join('users', 'manufacturers.user_id', '=', 'users.id')
                ->orderBy('divisions.code', 'asc')
                ->orderBy('sections.code', 'asc')
                ->groupBy('products.id');
                ->get()

我想要的是

        $productsFeatured = Product::select('products.low_emitting_material_prerequisite_file',
        'manufacturers.phone','manufacturers.address','manufacturers.linkedin_url','manufacturers.city', 
        'products.specs_file','products.specs_file_url','products.low_emitting_material_credit_file', 
        'products.id', 'products.name', 'products.logo', 'products.manufacturer_id',
        'products.division_id', 'products.section_id', 'products.website', 'divisions.name as division_name', 'divisions.code as division_code',
        'sections.name as section_name', 'sections.code'   
         )
         $productsFeatured ->whereIn('products.status', ['active'])
                if($request->chps_approved ==  '63'){
                    $productsFeatured->where(function ($query) {
                        $query->where('products.low_emitting_material_prerequisite_file', '!=', '');
                    })
                }

                ->where('users.user_type', 'manufacturer_paid')
                ->leftJoin('products_selected_leeds', 'products.id', '=', 'products_selected_leeds.product_id')
                ->leftJoin('sections', 'products.section_id', '=', 'sections.id')
                ->leftJoin('divisions', 'products.division_id', '=', 'divisions.id')
                ->join('manufacturers', 'products.manufacturer_id', '=', 'manufacturers.id')
                ->join('users', 'manufacturers.user_id', '=', 'users.id')
                ->orderBy('divisions.code', 'asc')
                ->orderBy('sections.code', 'asc')
                ->groupBy('products.id');
                 ->get()

标签： laravellaravel-5eloquent

解决方案

你可以：

$productsFeatured ->whereIn('products.status', ['active']);

if($request->chps_approved ==  '63'){
    $productsFeatured->where(function ($query) {
        $query->where('products.low_emitting_material_prerequisite_file', '!=', '');
    })
}

$productsFeatured->where('users.user_type', 'manufacturer_paid')
    ->leftJoin('products_selected_leeds', 'products.id', '=', 'products_selected_leeds.product_id')
    ->....
    ->get();

可链接方法只是返回原始类实例（在本例中为 $productsFeatured），然后用于调用下一个方法。

IE

$class->methodOne()->methodTwo();

等同于

$class->methodOne();
$class->methodTwo();

提供已使方法可链接的方法： PHP 方法链接？

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