提交表單后無法使用Ajax停留在同一頁面上

Question

我試圖創建一個表單，當您提交表單時，您將停留在同一頁面上，並將用戶輸入發送到Process.php和我的數據庫。 我面臨的問題是頁面刷新或打開頁面Process.php

我的表格

<form action="process.php" method="post" class="copy" id="formid" enctype="multipart/form-data">

    Project name: <input type="text" name="name"> <br>


        Video: 
        <input type="text" rows="1" cols="40" name="video">
    <br> 
    Svar 1<input type="text" name="answer1"/> 
    <select name="point1">
      <option value="1">1</option>
      <option value="2">2</option>
      <option value="3">3</option>
      <option value="4">4</option>
      <option value="5">5</option>
      <option value="6">6</option>
      <option value="7">7</option>
      <option value="8">8</option>
      <option value="9">9</option>
      <option value="10">10</option>
    </select>
    <br>
    Svar 2<input type="text" name="answer2"/> 
    <select name="point2">
      <option value="1">1</option>
      <option value="2">2</option>
      <option value="3">3</option>
      <option value="4">4</option>
      <option value="5">5</option>
      <option value="6">6</option>
      <option value="7">7</option>
      <option value="8">8</option>
      <option value="9">9</option>
      <option value="10">10</option>
    </select>
    <br>
    Svar 3<input type="text" name="answer3"/> 
    <select name="point3">
      <option value="1">1</option>
      <option value="2">2</option>
      <option value="3">3</option>
      <option value="4">4</option>
      <option value="5">5</option>
      <option value="6">6</option>
      <option value="7">7</option>
      <option value="8">8</option>
      <option value="9">9</option>
      <option value="10">10</option>
    </select>
    <br>
    Svar 4<input type="text" name="answer4"/> 
    <select name="point4">
      <option value="1">1</option>
      <option value="2">2</option>
      <option value="3">3</option>
      <option value="4">4</option>
      <option value="5">5</option>
      <option value="6">6</option>
      <option value="7">7</option>
      <option value="8">8</option>
      <option value="9">9</option>
      <option value="10">10</option>
    </select>
    <br>



       <br>
    <input type="submit" name="submit" value="create question" id="submit">
    </form>

Process.php

<?php
// Exempel 1: Lägga till 

if (isset($_POST['submit'])){

$localhost = "localhost";
$username = "root";
$password = "";

$connect = mysqli_connect($localhost, $username, $password)or 
die("Kunde inte koppla");

mysqli_select_db($connect, 'wildfire');


$name=$_POST['name'];
$video=$_POST['video'];
$answer1=$_POST['answer1'];
$answer2=$_POST['answer2'];
$answer3=$_POST['answer3'];
$answer4=$_POST['answer4'];

$point1=$_POST['point1'];
$point2=$_POST['point2'];
$point3=$_POST['point3'];
$point4=$_POST['point4'];




$sql1= "INSERT INTO question (answer, point) VALUES ('$answer1', '$point1')";

$result=$connect->query($sql1);

$sql2= "INSERT INTO question (answer, point) VALUES ('$answer2', '$point2')";

$result=$connect->query($sql2);

$sql3= "INSERT INTO question (answer, point) VALUES ('$answer3', '$point3')";

$result=$connect->query($sql3);

$sql4= "INSERT INTO question (answer, point) VALUES ('$answer4', '$point4')";

$result=$connect->query($sql4);

print $sql1;
print $sql2;
print $sql3;
print $sql4;

}


?>

使用Javascript

$(function () {

        $('form').on('submit', function (e) {

          e.preventDefault();

          $.ajax({
            type: 'post',
            url: 'process.php',
            data: $('form').serialize(),
            success: function () {
              alert('form was submitted');
            }
          });

        });

      });   

Answer 1

也許這可以解決問題（ 此其他SO問題的復制答案，以@HarveyARamer表示感謝）：

我最好的猜測是您要在實際呈現表單之前添加form提交偵聽器。 嘗試將jQuery包裝在$(document).ready(function () {}) ；

Answer 2

可能是您對表格采取了行動。 如果您向表單提供操作，它將打開您在操作中指定的頁面。 嘗試刪除表單的動作屬性（如果您給了

Answer 3

在ajax成功之后而不是在啟動ajax之前嘗試防止默認

$(function () {

    $('form').on('submit', function (e) {

      $.ajax({
        type: 'post',
        url: 'process.php',
        data: $('form').serialize(),
        success: function () {
          alert('form was submitted');
          e.preventDefault();
        }
      });

    });

  });   

Answer 4

HTML

使用類型作為按鈕，而不是提交

<input type="button" name="submit" value="create question" id="submit">

使用Javascript

單擊功能使用以下功能

$('#submit').click(function(e){

     e.preventDefault();

          $.ajax({
            type: 'post',
            url: 'process.php',
            data: $('form').serialize(),
            success: function () {
              alert('form was submitted');
            }
          });

        });