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如何使用 Laravel 8 查询生成器,如 eloquent 进行搜索

[英]how to use Laravel 8 query builder like eloquent for searching

I'm developing a simple CRM with Laravel 8 that needs to search in query string.我正在使用 Laravel 8 开发一个需要在查询字符串中搜索的简单 CRM。

My query string would be look like:我的查询字符串如下所示:

http://127.0.0.1:8000/admin/users/supervisor?q=john&gender=m

Now the problem is how can I search in controller?现在的问题是如何在 controller 中搜索?

I want to do like this:我想这样做:

public function index (Request $request)
{
    $role = getRoleCode($request->role);
    $roles = Role::where('role', '=', $role);

    if ($request->q) {
        $roles->where('name', 'like', "%$request->$q%");
    }

    if ($request->gender) {
        $roles->where('gender', '=', $request->gender);
    }

    $role->orderBy('id', 'desc')->paginate(20);

    return view('admin.users.index', [
        'roles' => $roles,
        'role_name' => config('settings.roles')[$role],
        'role_en_name' => $request->role,
        'q' => $request->q,
        'gender' => $request->gender
    ]);
}

I wonder why its not working and what is the standard way to do this.我想知道为什么它不起作用以及执行此操作的标准方法是什么。

I've tried:我试过了:

Role::query();

but that didn't work either.但这也没有用。

I've also tried:我也试过:

$roles = Role::where('role', '=', $role)
    ->where('name', 'like', "%$request->$q%")
    ->where('gender', '=', $request->gender)
    ->orderBy('id', 'desc')
    ->paginate(20);

This codes works perfectly but we may not be sending the "q" or "gender" URL params.此代码完美运行,但我们可能不会发送“q”或“gender”URL 参数。

PS: Sorry for my bad English:) PS:对不起我的英语不好:)

If you want to conditionally add where statements to your query you can use the if statements like you are attempting to do or use the when method of the Eloquent/Query Builder:如果您想有条件地将where语句添加到查询中,您可以使用if语句,就像您尝试做的那样,或者使用 Eloquent/Query Builder 的when方法:

$roles = Role::where('role', $role)
    ->when($request->input('q'), fn ($query, $search) => $query->where('name', 'like', '%'. $search .'%'))
    ->when($request->input('gender'), fn ($query, $gender) => $query->where('gender', $gender))
    ->orderBy('id', 'DESC')
    ->paginate(20);

Laravel 8.x Docs - Queries - Conditional Clauses when Laravel 8.x Docs - Queries - Conditional Clauses when

work with when() it's better与 when() 一起工作会更好

->when(true, function($query){$query->where('x', x);})

Your route should be like this:你的路线应该是这样的:

Route::get('/admin/users/{role}', [YourController::class, 'index']);

Your index function will be like this:您的索引 function 将是这样的:

public function index (Request $request,$role)
{
   //old line $role = getRoleCode($request->role);
    $role = getRoleCode($role);
    $roles = Role::where('role', '=', $role);

    if ($request->q) {
        $roles->where('name', 'like', "%$request->$q%");
    }
    if ($request->gender) {
        $roles->where('gender', '=', $request->gender);
    }

    $role->orderBy('id', 'desc')->paginate(20);
    return view('admin.users.index', [
        'roles' => $roles,
        'role_name' => config('settings.roles')[$role],
        'role_en_name' => $request->role,
        'q' => $request->q,
        'gender' => $request->gender
    ]);
}

There is a mistake Mr Mohamed $roles instead of $role Mohamed 先生有一个错误是 $roles 而不是 $role

//$roles->orderBy('id', 'desc')->paginate(20)... //$roles->orderBy('id', 'desc')->paginate(20)...

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