我想要做的是每当用户选择图片并单击按钮时，它将把图像移动到特定的文件夹，并将链接保存到数据库的user_image列。

我的问题是，单击“提交”按钮后，图片的实际名称未保存在数据库列中。示例Oppa/upload/多数民众赞成在数据库中保存的值没有图片文件名。

我认为photo.php没有收到文件的值，任何人都可以帮助我解决它。

<input type='file' id="imageInput" name="imageInput" accept="image/*" />
<button  id="changePicture" name="changePicture">Submit</button>

脚本：

var data = {};
        data.imageInput = $('#imageInput').val();
        data.email = $('#email').val();

        $.ajax({
            type: "POST",
            url: "Oppa/view/photo.php",
            data: data,
            cache: false,
            success: function (response) {

            if (Number(response) == 1)
                {
                   $("#dialog-confirm-changedImage").dialog("open");
                }
            }
        });
            return false;

photo.php

<?php
include_once('../dbc/database.php');

$db = new Connection();
$db = $db->dbConnect();
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);


$email = isset($_POST['email']) ? $_POST['email'] : "";

$image = addslashes(file_get_contents($_FILES['imageInput']['tmp_name']));
$image_name = addslashes($_FILES['imageInput']['name']);
$image_size = getimagesize($_FILES['imageInput']['tmp_name']);

move_uploaded_file($_FILES["imageInput"]["tmp_name"], "Oppa/upload/" . $_FILES["imageInput"]["name"]);
$location = "Oppa/upload/" . $_FILES["imageInput"]["name"];


if(!empty($_POST['email'])) {

        $q = "UPDATE tbl_user SET user_image = '$location' WHERE user_email= :email ";
        $query = $db->prepare($q);
        $query->bindParam(':email', $email);
        $results = $query->execute();
        echo "1";
}

?>