如何在单个HTTP响应中使用PHP从Mysql服务4个图像（Blob）？

I am trying to display 4 images using PHP and MySQL database. I have to display the 4 images as rows.

I use the table cars with fields (id_car, car_image1, car_image2, car_image3, car_image4), with all the images being blob datatype.

$id = $_GET['id'];

$link = mysql_connect("localhost", "root", "");
mysql_select_db("cars_database");
$sql = "SELECT car_image1, car_image2, car_image3, car_image4 FROM cars WHERE id_car='$id'";
$result = mysql_query("$sql");

mysql_close($link);
while($row=mysql_fetch_array($result))
{
    header('Content-type: image/jpeg');
    echo $row['car_image1'];
    echo $row['car_image2'];
    echo $row['car_image3'];
    echo $row['car_image4'];    
}

I can only display 1 image and not the other images. Since I am a newbie to this technology I need help.

That's how HTTP works (at least, current version): you cannot have a URL that points to more than one resource simultaneously. Your script needs to send one image.

Simply, call it four times:

<img src="/get-image.php?id=1">
<img src="/get-image.php?id=2">
<img src="/get-image.php?id=3">
<img src="/get-image.php?id=4">

You need to define path for images.

<?php
    while($row=mysql_fetch_array($result))
    {
     header('Content-type: image/jpeg');    
?>  
    <img src="path<?php echo $row['car_image1']; ?>" alt="" />
    <img src="path<?php echo $row['car_image2']; ?>" alt="" />
    <img src="path<?php echo $row['car_image3']; ?>" alt="" />
    <img src="path<?php echo $row['car_image4']; ?>" alt="" />
<?php
    }
?>