我试图改变div的背景图像使用PHP和jquery.php变量不获取值

Question

I wrote the code for changing the background image of a div. I am able to upload the image and image is visible in the folder. The image path is getting uploaded in mysql database. But the image is not displaying as background in the div. The php variable value(location) for the specified path of image is not passing to the div. Can any one please let me know my mistake?

<?php

session_start();

include_once ('dataconnect.php');

$db_conx = mysqli_connect("localhost", "root", "", "productiondata");

$uid = $_SESSION['user_id'];

$username = $_SESSION['user_name'];

if (isset($_FILES['file'])) {

    global $location;

    $imagename = $_FILES["file"]["name"];

    $imagepath = "images/";

    $location = $imagepath . $_FILES["file"]["name"];

    move_uploaded_file($_FILES["file"]["tmp_name"], "$imagepath" . $_FILES["file"]["name"]);


    $sql = "INSERT INTO image(user_id,imagepath,imagename) VALUES('$uid','$imagepath','$imagename')";


    $query = mysqli_query($db_conx, $sql);

    $sql1 = "SELECT * FROM image";


    $query1 = mysqli_query($db_conx, $sql1);

    while ($row = mysqli_fetch_array($query1)) {

        $imagename = $row['imagename'];

        $imagepath = $row['imagepath'];

        echo $location;

    }

}

jquery code

$('document').ready(function () {


            $('.js_p-uploader').click(function () {

                $('#file').click();

                $('#file').change(function (e) {

                    e.preventDefault();

                    var blob = this.files[0];

                    var formData = new FormData();

                    formData.append('file', blob);

                    $.ajax({

                       url: "profile.php",

                        type: "POST",

                        data: formData,

                        mimeType: "multipart/form-data",

                        processData: false,

                        contentType: false,

                        success: function (location) {



                        }


                    });

                });

            });


        });

div class="profile-cover-wrapper" style="background-image:url('?php echo 

$location ;?');"input type="file" id="file" name ="file"

Answer 1

Returned data from an Ajax call cannot be used outside, or cannot be used as a PHP variable. What you can do is to change the src attribute of your <img> inside the success: function() :

// AJAX CALL HERE
success: function(location) {
    $('.profile-cover-wrapper').css("background-image", "url("+location+")");
}