如何在单个查询中显示外键值？

I have two tables one is image and another one is tags.

image_id=primary key(IMAGE TABLE),fk_image_id=foreign key(TAGS TABLE).

Expected output: if example tagname two, image one to display single image and tagname (more names).

But now I get suppose two tagname, get two image separate ..how to solve this?

I using implode function but its coming " invalid arguments passed".

include_once("config.php");
        $result=mysqli_query($mysqli,"SELECT * FROM image,tags WHERE image_id=fk_image_id  ORDER BY creation_dt DESC LIMIT 5 ");
        while($res = mysqli_fetch_array($result)) {
            $tagname=$res['tag_txt'];
            echo $tagname;
            echo "<tr>"."<img  src='http://localhost:8080/memes/".$res['path_txt']."' width='380' height='280' style='padding: 10px;'  />"."</tr>";
        }