我刚来拉拉维尔。如何查找是否存在记录?

$user = User::where('email', '=', Input::get('email'));

我能做什么来查看$user是否有记录?


当前回答

$email = User::find($request->email);
If($email->count()>0)
<h1>Email exist, please make new email address</h1>
endif

其他回答

简单,舒适和易于理解的Validator

class CustomerController extends Controller
{
    public function register(Request $request)
    {

        $validator = Validator::make($request->all(), [
            'name' => 'required|string|max:255',
            'email' => 'required|string|email|max:255|unique:customers',
            'phone' => 'required|string|max:255|unique:customers',
            'password' => 'required|string|min:6|confirmed',
        ]);

        if ($validator->fails()) {
            return response(['errors' => $validator->errors()->all()], 422);
        }

这将检查用户表中是否存在所请求的电子邮件:

if (User::where('email', $request->email)->exists()) {
   //email exists in user table
}
$user = User::where('email', request('email'))->first();
return (count($user) > 0 ? 'Email Exist' : 'Email Not Exist');

在if语句中检查null可以防止Laravel在查询结束后立即返回404。

if ( User::find( $userId ) === null ) {

    return "user does not exist";
}
else {
    $user = User::find( $userId );

    return $user;
}

如果找到用户,它似乎会运行双重查询,但我似乎找不到任何其他可靠的解决方案。

Laravel 6或顶部:写表名,然后给出where子句条件,例如where('id', $request->id)

 public function store(Request $request)
    {

        $target = DB:: table('categories')
                ->where('title', $request->name)
                ->get()->first();
        if ($target === null) { // do what ever you need to do
            $cat = new Category();
            $cat->title = $request->input('name');
            $cat->parent_id = $request->input('parent_id');
            $cat->user_id=auth()->user()->id;
            $cat->save();
            return redirect(route('cats.app'))->with('success', 'App created successfully.');

        }else{ // match found 
            return redirect(route('cats.app'))->with('error', 'App already exists.');
        }

    }