使用ajax将表单数据和文件传递给php

Question

这可能是之前被问到的,但是我在这里和谷歌搜索,我读过的每个答案都不起作用.

我必须解决的问题是制作一个包含名字,姓氏,电子邮件和图像的表格.然后将数据传递到数据库并将文件也上载到数据库.我按下提交后,目前我的代码没有做任何事情.在我添加文件框之前,它会将数据插入我的数据库.

HTML

<form id="myForm" method ="post" enctype="multipart/form-data">
    First Name: <input type="text" name="fname" id="fname"> <br>
    Last Name: <input type="text" name="lname" id="lname"> <br>
    Email:  <input type="text" name="email" id="email"> <br>
    Image: <input type="file" name="image" id="image"> <br>
    <button type="button" name="btnSubmit" id="btnSubmit"> Submit </button>
</form>

AJAX/JS

$("#btnSubmit").click(function(){
     var formData = new FormData($(this)[0]);
     $.ajax({
        type: 'POST',
        url: 'form2.php',
        data: formData,
         success: function (data) {
           alert(data)
         },
      });
  });

PHP

$upload = basename($_FILES['image']['name']);
$type = substr($upload, strrpos($upload, '.') + 1);
$size = $_FILES['image']['size']/1024; 

if ($_FILES["image"]["error"] > 0)
{
    echo "Error: " . $_FILES["image"]["error"] . "<br>";
}
else
{
    echo "Upload: " . $upload . "<br>";
    echo "Type: " . $type . "<br>";
    echo "Size: " . $size . " kB<br>";
}

$fname = $_POST['fname'];
$lname = $_POST['lname'];
$email = $_POST['email'];
echo "You Entered <br />";
echo "<b>First Name:</b> ". $fname . "<br />";
echo "<b>Last Name:</b> ". $lname . "<br />";
echo "<b>Email:</b> ". $email . "<br />";

Answer 1

默认情况下，表单会提交到指定的位置。为了阻止这种情况，您需要阻止它。你的 js 应该看起来像这样：

$("form#data").submit(function(event){
    event.preventDefault();
    ...
});