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());

如果你想在数据库中插入一条记录，而相同邮件的记录不存在，那么你可以这样做:

$user = User::updateOrCreate(
    ['email' => Input::get('email')],
    ['first_name' => 'Test', 'last_name' => 'Test']
);

updateOrCreate方法的第一个参数列出了唯一标识关联表中的记录的列，而第二个参数由要插入或更新的值组成。

你可以在这里查看文档:Laravel upserts doc

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!

$user = User::where('email', '=', Input::get('email'))->first();
if ($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。

最短工作选项:

// if you need to do something with the user 
if ($user = User::whereEmail(Input::get('email'))->first()) {

    // ...

}

// otherwise
$userExists = User::whereEmail(Input::get('email'))->exists();

要知道是否有记录是很简单的

$user = User::where('email', '=', Input::get('email'))->get();
if(count($user) > 0)
{
echo "There is data";
}
else
echo "No data";