在PHP和MySQL中显示URL中的img

Question

I'm trying to grab an img URL from a mysql database and display the image itself, not the URL link.

Is there a way to concatenate in some HTML into the php script I'm running below?

Right now it displays the image URL just fine but would like it to display the actual image it's calling.

<?php
            $db = new mysqli('localhost', 'root', 'root', 'testdb');

            if($db->connect_errno > 0){
                die('Unable to connect to database [' . $db->connect_error . ']');
            }

            $sql = <<<SQL
                SELECT *
                FROM `site_img`
            SQL;

            if(!$result = $db->query($sql)){
                die('There was an error running the query [' . $db->error . ']');
            }
            while($row = $result->fetch_assoc()){
                echo $row['img_id'] . ' ' . $row['img_url'] . '<br />';
            }
            echo 'Total results: ' . $result->num_rows;

            // Free result set
            mysqli_free_result($result);

            mysqli_close($db);
            ?>

Answer 1

You could insert the html img tag inside your loop.

<?php
            $db = new mysqli('localhost', 'root', 'root', 'testdb');

            if($db->connect_errno > 0){
                die('Unable to connect to database [' . $db->connect_error . ']');
            }

            $sql = <<<SQL
                SELECT *
                FROM `site_img`
            SQL;

            if(!$result = $db->query($sql)){
                die('There was an error running the query [' . $db->error . ']');
            }
            while($row = $result->fetch_assoc()){
?>
            <img name="<?php echo $row['img_id'] ?>" src="<?php $row['img_url'] ?>" /> <br />
<?php
            }
            echo 'Total results: ' . $result->num_rows;

            // Free result set
            mysqli_free_result($result);

            mysqli_close($db);
?>