Question

我的代码是这样的：

<?php
public function getFavoriteStore($param = null)
{
    $num = 20;
    $q = $param['q'];
    $location = $param['location'];

    $result = $this->store_repository->whereHas('favorites', function ($query) {
        $query = $query->where('stores.status', '=', 1)
              ->where('favorites.favoritable_type', 'like', 'App\\\Models\\\Store');
        if(isset($location))
           $query = $query->where('stores.address', 'like', "%$location%");

        if(isset($q)) {
            $query = $query->where(function ($query) use ($q) {
                $query->where('stores.name', 'like', "%$q%")
                      ->where('stores.address', 'like', "%$q%", 'or');
            });
        }

        return $query;
    })->paginate($num);

    return $result;
}

它有效

但是，条件是{if(isset($location))＆amp; if(isset($q))）不起作用

似乎还有错误

有没有人可以帮助我？

我遵循本教程：https://laravel.com/docs/5.3/eloquent-relationships#querying-relationship-existence

Answer 1

您需要在第一个闭包中添加use()：

public function getFavoriteStore($param = null)
{
    $num = 20;
    $q = $param['q'];
    $location = $param['location'];

    $result = $this->store_repository->whereHas('favorites', function ($query) use($q, $location) {
        $query->where('stores.status', '=', 1)
              ->where('favorites.favoritable_type', 'like', 'App\\\Models\\\Store');
        if(isset($location))
           $query->where('stores.address', 'like', "%$location%");

        if(isset($q)) {
            $query->where(function ($query) use ($q) {
                $query->where('stores.name', 'like', "%$q%")
                      ->where('stores.address', 'like', "%$q%", 'or');
            });
        }
    })->paginate($num);

    return $result;
}

并且无需在闭包函数中分配和返回$query变量。

如何添加查询关系存在的条件？ Laravel 5.3

1 个答案: