$user = User::where('email', '=', Input::get('email'))->first();
if ($user === null) {
   // user doesn't exist
}

可以写成

if (User::where('email', '=', Input::get('email'))->first() === null) {
   // user doesn't exist
}

这将返回true或false，而不分配临时变量，如果这是你在原始语句中使用$user的全部目的。

创建下面的方法(为自己)，以检查给定的记录id是否存在于Db表中。

private function isModelRecordExist($model, $recordId)
{
    if (!$recordId) return false;

    $count = $model->where(['id' => $recordId])->count();

    return $count ? true : false;
}

// To Test
$recordId = 5;
$status = $this->isModelRecordExist( (new MyTestModel()), $recordId);

有帮助!

我解决了这个问题，使用empty()函数:

$user = User::where('email', Input::get('email'))->get()->first();
//for example:
if (!empty($user))
    User::destroy($user->id);

$user = User::where('email', request('email'))->first();
return (count($user) > 0 ? 'Email Exist' : 'Email Not Exist');

if($user->isEmpty()){
    // has no records
}

Eloquent使用集合。 参见以下链接:https://laravel.com/docs/5.4/eloquent-collections

如果你想插入一个唯一的记录，你可以使用laravel验证:

$validated = $request->validate([
    'title' => 'required|unique:usersTable,emailAddress|max:255',
]);

但是你也可以用这些方法:

1:

if (User::where('email',  $request->email)->exists())
{
  // object exists
} else {
  // object not found
}

2:

$user = User::where('email',  $request->email)->first();

if ($user)
{
  // object exists
} else {
  // object not found
}

3:

$user = User::where('email',  $request->email)->first();

if ($user->isNotEmpty())
{
  // object exists
} else {
  // object not found
}

4:

$user = User::where('email',  $request->email)->firstOrCreate([
      'email' => 'email'
],$request->all());