如何回显多行HTML代码？

Question

<?php

$connection = mysql_connect('localhost', 'root', '1234');
mysql_select_db('database');

$query = "SELECT user_name FROM users";

while($row = mysql_fetch_array($result)){
echo '<div class="mosaic-block bar2" onmouseout="decolorit2()" onmouseover="colorit2()">'
        '<a target="_blank" class="mosaic-overlay">'
            '<div class="details">'
                '<p class="font_us">' . $row['user_name'] . '</p>   <br/>'
                '<p class="font_us1">technische Umsetzung</p>'
            '</div>'
        '</a>'
        '<div class="mosaic-backdrop"><img id="pic" src="../views/pictures/pic"/></div>'
    '</div>';

}
mysql_close();

?>

So I wanted to output the usernames of my database with some javascript effect, but it doesn't show. How do I have to echo this out?

Answer 1

echo 'foo
      bar
      baz';

No need to close the quote on each newline. Better though, simply go out of PHP mode:

<?php while ($row = mysql_fetch_array($result)) { ?>

<div>
    ...
       <?php echo $row['user_name']; ?>
    ...
</div>

<?php } ?>

Even better yet for readability (IMO):

<?php while ($row = mysql_fetch_array($result)) : ?>

    ...

<?php endwhile; ?>

Answer 2

You should use pdo or mysqli .
You did not concat your string.

 <?php

    $connection = mysql_connect('localhost', 'root', '1234');
    mysql_select_db('database');

    $query = "SELECT user_name FROM users";

    while($row = mysql_fetch_array($result)){
    echo '<div class="mosaic-block bar2" onmouseout="decolorit2()" onmouseover="colorit2()">'.
            '<a target="_blank" class="mosaic-overlay">'.
                '<div class="details">'.
                    '<p class="font_us">' . $row['user_name'] . '</p>   <br/>'.
                    '<p class="font_us1">technische Umsetzung</p>'.
                '</div>'.
            '</a>'.
            '<div class="mosaic-backdrop"><img id="pic" src="../views/pictures/pic"/></div>'.
        '</div>';

    }
    mysql_close();

    ?>