如何在 laravel 的“whereIn”中使用“if”条件答案

【问题标题】：How can I used "if" condition in "whereIn" in laravel如何在 laravel 的“whereIn”中使用“if”条件
【发布时间】：2017-04-28 09:27:57
【问题描述】：

以下是我使用的查询，并且对我来说工作正常。

    if($t_requested['category_id'])
    {
        $t_query_condition['p.category_id'] = $t_requested['category_id'];
    }
    if($t_requested['brand_id'])
    {
        $t_query_condition['p.brand_id'] = $t_requested['brand_id'];
    }

    $browseData = DB::table('products as p')
        ->leftjoin('categories as cp', 'p.category_id', '=', 'cp.id')
        ->leftjoin('brands as bp', 'p.brand_id', '=', 'bp.id')
        ->select('p.id as product_id','p.name as product_name','cp.title as category_name','bp.name as brand_name','p.product_weight',
               'p.short_description','p.product_price','p.special_price','p.stock_quantity')
        ->orderBy('p.created_by', "desc")
        ->offset(0)
        ->limit(10)
        ->where($t_query_condition)
        ->get();

但现在我在“category_id”和“brand_id”中有多个 id，我想使用 whereIn 但它使用了 of 条件。如果我得到 category_id 或 brand_id 为空，则跳过。

提前致谢。

【问题讨论】：

Manish Patoliya: laravel 5.2 ??
不，它在 5.4 @KetanAkbari

标签： php jquery model-view-controller laravel-5.2

【解决方案1】：

试试这个查询希望你能得到结果

        $browseData = DB::table('products as p')
       ->select('p.id as product_id','p.name as product_name','cp.title as category_name','bp.name as brand_name','p.product_weight',
               'p.short_description','p.product_price','p.special_price','p.stock_quantity')

        ->leftjoin('categories as cp', 'p.category_id', '=', 'cp.id')
        ->leftjoin('brands as bp', 'p.brand_id', '=', 'bp.id')
        ->orderBy('p.created_by', "desc");
       if($t_requested['category_id'])
         {
          $browseData->whereIn('p.category_id',$t_requested['category_id']);
         }
      if($t_requested['brand_id'])
         {
         $browseData->whereIn('p.brand_id',$t_requested['brand_id']);
         }

      $result=browseData->offset(0)->limit(10)->get();

【讨论】：

【解决方案2】：

我认为这很简单，你必须传递一组值而不是一个值

我已经尝试编写选项来实现这一点，你必须遵循适合你的选项

if($t_requested['category_id'])
{
   $t_query_condition['p.category_id'] = $t_requested['category_id']; // it should be array OR

   $t_query_condition['p.category_id'] = explode(",",$t_requested['category_id']); // if you get value comma saperated

$t_query_condition['p.category_id'] = explode(",",$t_requested['category_id']); // if you get value comma saperated

$t_query_condition['p.category_id'] = [$t_requested['category_id']]; // if you want to convert one value into array
        }

【讨论】：