Question

我想将表单数据插入数据库。现在，我在做什么时，当我单击注册按钮时，它首先检查email是否已经存在，如果是，则显示一条现有消息，否则将表单数据插入数据库，并向用户发送确认邮件。那么，我该怎么做？请帮助我。

控制器：FirstController.php

public function register(Request $request)
{
    $user_id = date('YmdHis');

    $fname = $request->input('fname_sign');
    $lname = $request->input('lname_sign');
    $email = $request->input('email_sign');
    $pass = $request->input('pass_sign');
    $candidate_type = $request->input('candidate_type');
    $s_date = date('Y-m-d');
    $confirm_id = md5($user_id);

    $sql = DB::select('select * from user where email="'.$email.'"');
    if(count($sql) > 0)
    {
        echo "<p>Email id already exist. Please sign up with different email id</p>";
    }
    else
    {
        $data = array('user_id'=>$user_id,'fname'=>$fname, 'lname'=>$lname, 'email'=>$email, 'password'=>$pass, 'candidate_type'=>$candidate_type, 's_date'=>$s_date,'confirm_id'=>$confirm_id);
        DB::table('candidate')->insert($data);

        echo "<p>save successfully</p>";
    }
}

视图：

<script>
    $(document).ready(function(){
        $("#signup").click(function(e){
            e.preventDefault();
            fname_sign = $("#fname_sign").val();
            lname_sign = $("#lname_sign").val();
            email_sign = $("#email_sign").val();
            pass_sign = $("#pass_sign").val();
            candidate_type = $("#candidate_type").val();
            $.ajax({
                type:"POST",
                data:{"fname_sign":fname_sign, "lname_sign":lname_sign, "email_sign":email_sign, "pass_sign":pass_sign, "candidate_type":candidate_type, "_token":"{{csrf_token()}}"},
                url:"{{URL::to('register')}}",
                success:function(data){
                    $("#success_in").html(data);
                }
            });
        }); 
    });
</script>

使用laravel 5插入表单数据时发布500（内部服务器错误）

0 个答案: