Laravel Eloquent 条件与影响输出的“with”关系数据

Question

这是我的代码：

$search = request()->get('search');

    if(Auth::user()->hasRole('admin') || true) {
        list($orderBy, $orderDirection) = explode('.', request()->get('sort_by'));
        $prestations = Prestation::with(
            [
            'service' => function($service) use($search){
                $service->select(['id','name']);
                ->where('name', 'regexp', "/$search/i"); <--- HERE THE PROBLEM, i get some "null" values in the output, it's not a really condition to display or not
            },
            'facility' => function($facility) {
                $facility->select(['id','name']);
            }
            ]
        )
            ->where('name', 'regexp', "/$search/i") <-- I want to do this with service name
            ->orderBy($orderBy, $orderDirection)
            ->simplePaginate(50);

        $res = [
            'results' => $prestations,
            'total' => Prestation::all()->count(),
        ];

        return $res;

我想对我的“服务：名称”做一个 where 条件，这会真正影响输出，并且不显示值为“null”的数据，例如 ->where('name', 'regexp', "/$search/i")，但我不知道如何访问“with”关系的“service:name”。

谢谢你，新年快乐！

Answer 1

您可以为此使用 whereHas

$prestations = Prestation::whereHas('service', function($subQuery) use($search) {
                $subQuery->where('name', 'like', '%' . $search . '%');
            })->with(
    [
        'service' => function($service) use($search){
            $service->select(['id','name'])->where('name', 'like', '%' . $search . '%');
        },
    ]
)->orderBy($orderBy, $orderDirection)->simplePaginate(50);

Answer 2

您可以使用 LIKE 运算符来做到这一点：

$prestations = Prestation::with(
        [
            'service' => function($service) use($searchTerm){
                $service->select(['id','name'])->where('name', 'LIKE', "%$searchTerm%");
            },
        ]
    )->orderBy($orderBy, $orderDirection)->simplePaginate(50);

1. 如果要搜索包含搜索词的记录，请使用LIKE '%$searchTerm%'
2. 如果要搜索以搜索词开头的记录，请使用LIKE '$searchTerm%'
3. 如果您想搜索以搜索词结尾的记录，请使用LIKE '%$searchTerm'

希望对你有所帮助