无法获得 concat 值和 php 无法正常工作

Question

所以我的问题是为什么jquery concat 不起作用，而且它没有发布并且无法插入database

我尝试更改代码并阅读参考仍然无法获得足够的信息

这是我的 jquery

    var uid = $('#lname').val() + $('fname').val() + $('#datecreated').val(moment().format('YYYY')); 
    var datecreated = $('#datecreated').val(moment().format('YYYY'));
    var fname = $('#fname').val();
    var lname = $('#lname').val();
    var email = $('#email').val();
    var password = $('#pass').val();
    var passcheck = false;

这是我的 ajax

    if (uid && fname && lname && email && password && datecreated)
                    {
                        var form = $(this);
                        var formData = new FormData(this);
                        $(".formcontent").hide();
                        $.ajax({
                          url : form.attr('action'),
                          type: form.attr('method'),
                          data: form.serialize(),
                          data: formData,
                          dataType: 'json',
                          cache: false,
                          contentType: false,
                          processData: false,
                          success:function(response)
                          {

这是我的完整代码php，我不知道问题是否与xampp有关。我现在连续 3 天解决这个问题，我不知道问题出在哪里

    valid['success'] = array('success' => true, 'messages' => array());
    $uid = $_POST ['uid'];
    $pass  =  $_POST['pass'];
    $fname = $_POST['fname'];
    $lname = $_POST['lname'];
    $email = $_POST['email'];
    $datecreated = $_POST['datecreated'];
        if ($_POST)
    {
        if(true)
        {
            $sqlmail = "SELECT * FROM acc WHERE (email = '$email') AND acc_stat < 3";
            $resmail = $connect->query($sqlmail);
            if($resmail->num_rows > 0) 
            {
                while($row = $resmail->fetch_array())
                {
                    if($email === $row['email'])
                    {
                        $valid['messages'] = "Email address is already taken";
                    }

                }
                $valid['success'] = false;
                $connect->close();
                echo json_encode($valid);
            }
            else
            {
                $sql = "INSERT INTO 'acc' ('uid', 'password', 'lname', 'fname', 'email', 'acc_type', 'acc_stat','date_create')  VALUES ('$uid', '$pass', '$fname', '$lname', '$email', '3', '1','$datecreated')";

                if($connect->query($sql) === TRUE) 
                {
                    $valid['success'] = true;
                    $valid['messages'] = "Account registration successful.";                
                    $connect->close();
                    echo json_encode($valid);
                }
                else 
                {
                    $valid['success'] = false;
                    $valid['messages'] = "Network connection not stable. Please try again later.";
                    $connect->close();
                    echo json_encode($valid);       
                }

            }
        }
        else
        {
            $valid['success'] = false;
            $valid['messages'] = "No internet connection.";
            $connect->close();
            echo json_encode($valid);   
        }
    }

Answer 1

您的代码中有一个小错误，请修复它，它应该可以正常工作。错误在

行

var uid = $('#lname').val() + $('fname').val() + $('#datecreated').val(moment().format('YYYY'));

将其更改为var uid = $('#lname').val() + $('#fname').val() + $('#datecreated').val(moment().format('YYYY'));。

错过# 的愚蠢错误。还有一件事，你不会在 php 端收到 uid 参数，只是因为你没有用表单发送它。将其附加到FormData 作为给定，

var uid = $('#lname').val() + $('#fname').val() + $('#datecreated').val(moment().format('YYYY'));.
var formData = new FormData(this);
formData.append('uid' , uid);

现在您将能够收到 uid 参数。

Answer 2

尝试使用这个：

 var uid = $('#lname').val() + $('#fname').val() + $('#datecreated').val(moment().format('YYYY')); 
    var datecreated = $('#datecreated').val(moment().format('YYYY'));
    var fname = $('#fname').val();
    var lname = $('#lname').val();
    var email = $('#email').val();
    var password = $('#pass').val();
    var passcheck = false;