Question

我试图创建一个表单，当您提交表单时，您将保持在同一页面上并将用户输入发送到Process.php和我的数据库。我面临的问题是页面刷新或打开页面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;

}


?>

的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

也许这会解决问题（从this other SO question复制回答，并带有@HarveyARamer的积分）：

我最好的猜测是您要添加form提交侦听器在实际呈现表单之前。尝试包装你的jQuery $(document).ready(function () {});

Answer 2

可能你对表单采取了行动。如果您对表单提供操作，它将打开您在操作中指定的页面。尝试删除表单的动作属性如果你给了

Answer 3

尝试在ajax成功之后阻止默认，而不是在启动ajax

之前

$(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

<强> HTML

使用type as按钮代替提交

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

<强>的Javascript

On Click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');
            }
          });

        });

在提交表单不起作用时使用Ajax保持在同一页面上

4 个答案: