无法从数据库中获取图像和信息

I am trying to display image and details from database where i have stored. I have got code which connects to database but doesn't fetch. Edited: Thanks guys i was able to fetch but I have incurred anther problem. I have stored image as blob. so when i queried it displayed some mumbojumbo language. screenshot of the result screenshot of how i stored image ....... echo"Connected to db" ;

if (isset($_POST['info'])){
    $info= 'info';
    if ($info== 1){
        $sql= "SELECT name,img,price FROM product WHERE price<400";
        $result=mysqli_query($conn,$sql);
        while ($row = mysqli_fetch_assoc($result)){
            echo "<div class= 'row'>";
            echo    "<div class = 'col-sm-3'>";
            echo        "<p>".$row['img']."</p><br/>";
            echo        "<h3>".$row['name']."</h3><br/>";
            echo        "<h4>£".$row['price']."</h4><br/>";
            echo    "</div>";
            echo "</div>";
        }
    }

exit();

Here is my jquery code where it prints the the output of checkbox and it echo connected to database but can't fetch.

$(".id_price").each(function() {
if ($(this).is(":checked")) {
    var check = $(this).val();
    console.log(check);
    $.post('database.php',{info:check},function(response){
        $(".product").html(response).show(); 

    });

}

});

    echo"Connected to db" ;

if (isset($_POST['info'])){
    $info= 'info'; // <----------------
    if ($info== 1){
        $sql= "SELECT name,img,price FROM product WHERE price<400";
        $result=mysqli_query($conn,$sql);
        while ($row = mysqli_fetch_assoc($result)){
            echo "<div class= 'row'>";
            echo    "<div class = 'col-sm-3'>";
            echo        "<p>".$row['img']."</p><br/>";
            echo        "<h3>".$row['name']."</h3><br/>";
            echo        "<h4>£".$row['price']."</h4><br/>";
            echo    "</div>";
            echo "</div>";
        }
    }

exit();

You set $info == "info" and then check if $info == 1

Try this:

    .......
echo"Connected to db" ;

    if (isset($_POST['info'])){
                $sql= "SELECT name,img,price FROM product WHERE price<400";
                $result=mysqli_query($conn,$sql);
                while ($row = mysqli_fetch_assoc($result)){
                    echo "<div class= 'row'>";
                    echo    "<div class = 'col-sm-3'>";
                    echo        "<p>".$row['img']."</p><br/>";
                    echo        "<h3>".$row['name']."</h3><br/>";
                    echo        "<h4>£".$row['price']."</h4><br/>";
                    echo    "</div>";
                    echo "</div>";

            }
        }
        exit();

$info= 'info';
if ($info== 1){

$info will always equal 'info' so it will not attempt the query.