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?
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.');
}
}