简单的PHP，回显<li> ... </ li>目录中的所有图像

问题描述：

I'm attempting to make a simple responsive gallery using bootstrap 3 and a simple php script to display all images in a specific directory I in response to another question on here.

The issue I'm having is echoing the the required <li></li> tags rather then <br></br> as demonstrated below.

here is how I currently display the images, within the <li></li>tag with a class to display the images layout/positioning, called in the <ul></ul> tag.

<div class="container">
          <ul class="row">
            <li class="col-lg-2 col-md-2 col-sm-3 col-xs-4">
                <img class="img-responsive" src="images/kitchens.jpg">
            </li>
            <li class="col-lg-2 col-md-2 col-sm-3 col-xs-4">
                <img class="img-responsive" src="images/garden.jpg">
            </li>
            <li class="col-lg-2 col-md-2 col-sm-3 col-xs-4">
                <img class="img-responsive" src="images/front.jpg">
            </li>
          </ul>
</div> <!-- /container -->

this is the PHP im using in response to question i found here https://*.com/a/19255786/3599850

<?php
$files = glob("images/*.*");

for ($i=0; $i<count($files); $i++) {
    $image = $files[$i];
    print $image ."<br />";
    echo             

    '<img src="'.$image .'" alt="Random image" />'."<br /><br />";
}

?>

This is my attempt :(

<?php
$files = glob("images/*.*");

for ($i=0; $i<count($files); $i++) {
    $image = $files[$i];
    print $image ."<br />";
    echo

    '<li class="col-lg-2 col-md-2 col-sm-3 col-xs-4">
    <img src="'.$image .'" alt="Random image" /></li>'."<li /><li />";
}

?>

答

You have too many closing </li> tags :

<?php
$files = glob("images/*.*");

for ($i=0; $i<count($files); $i++) {
    $image = $files[$i];
    echo '<li class="col-lg-2 col-md-2 col-sm-3 col-xs-4"><img  class="img-responsive" src="'.$image .'" /></li>';
}

?>

And you don't need the print $image ."<br />";

简单的PHP，回显 ... 目录中的所有图像

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