Question

我遇到了这个问题。我写了这个PHP来输出指定文件夹中的子文件夹的内容。代码工作得很好，它显示了子文件夹中的所有图像。我遇到的问题是，在echo $underfolder上，它只打印出第一个子文件夹而没有关闭DIV class=image。

怎么了？我疯了

 <div class="folders">
<?php
global $current_user;
$current_user = wp_get_current_user();  
$current_user_nambre = $current_user->user_login;
$sottocartelle = glob("wp-content/uploads/$current_user_nambre/*", GLOB_ONLYDIR);

	 
	foreach ($sottocartelle as $value) {
    echo "<div class='products'>";
		$underfolder = $value;
		$contenuto = scandir($value);
		echo $underfolder;
		foreach ($contenuto as $value) {
			
		echo "<div class='image;' style=\"display:inline-block;\">";
			echo "<a href='/$underfolder/$value'>download</a>";
            echo "<img src='/$underfolder/$value' height=\"100\height=\"100\" width=\"100\>";
            echo '</div>';
			
		
			}
		echo '</div>';
}
	 
	 ?>
	 </div>

Answer 1

检查img标记中的引号，并$value重新分配$value。在第二个foreach()

中设置不同的名称

<div class="folders">
    <?php
    global $current_user;
    $current_user        = wp_get_current_user();
    $current_user_nambre = $current_user->user_login;
    $sottocartelle       = glob( "wp-content/uploads/$current_user_nambre/*", GLOB_ONLYDIR );
    foreach ( $sottocartelle as $value ) {
        echo "<div class='products'>";
        $contenuto = scandir( $value );
        echo $value;
        foreach ( $contenuto as $value1 ) {
            echo '<div class="image;" style="display:inline-block;">';
            echo "<a href='/$value/$value1'>download</a>";
            echo "<img src='/$value/$value1' height='100' height='100' width='100'>";
            echo '</div>';
        }
        echo '</div>';
    }
    ?>
</div>

此代码应该可以正常工作。

DIV没有关闭foreach

1 个答案: