Question

我有点新的PHP和我有一个问题，应该是小但我无法弄明白。

我有一个wordpress页面，每个帖子都有2个链接。我想只在href =＆＃34;＆＃34;中有地址时显示链接。否则我希望他们保持隐形/消失。

<span class="play"><a rel="lightbox" href="<?php swf_file_url(); ?>">Play</a></span>
<span class="download"><a href="<?php swf_file_url(); ?>" download>Download</a></span>

任何人都知道一种简单的方法吗？

＆＃13;

add_action("admin_init", "swf_init");
add_action('save_post', 'save_swf_link');
function swf_init(){
        add_meta_box("my-swf", "swf", "swf_link", "game", "normal", "low");
        }
function swf_link(){
        global $post;
        $custom  = get_post_custom($post->ID);
        $link    = $custom["link"][0];
        $count   = 0;
        echo '<div class="link_header">';
        $query_swf_args = array(
                'post_type' => 'attachment',
                'post_mime_type' =>'application/x-shockwave-flash',
                'post_status' => 'inherit',
                'posts_per_page' => -1,
                );
        $query_swf = new WP_Query( $query_swf_args );
        $swf = array();
        echo '<select name="link">';
        echo '<option class="swf_select">SELECT SWF FILE</option>';
        foreach ( $query_swf->posts as $file) {
           if($link == $swf[]= $file->guid){
              echo '<option value="'.$swf[]= $file->guid.'" selected="true">'.$swf[]= $file->guid.'</option>';
                 }else{
              echo '<option value="'.$swf[]= $file->guid.'">'.$swf[]= $file->guid.'</option>';
                 }
                $count++;
        }
        echo '</select><br /></div>';
        echo '<p>Kies een swf document om te laten zien op de website.</p>';
        echo '<div class="swf_count"><span>Files:</span> <b>'.$count.'</b></div>';
}
function save_swf_link(){
        global $post;
        if (defined('DOING_AUTOSAVE') && DOING_AUTOSAVE){ return $post->ID; }
        update_post_meta($post->ID, "link", $_POST["link"]);
}
add_action( 'admin_head', 'swf_css' );
function swf_css() {
        echo '<style type="text/css">
        .swf_select{
                font-weight:bold;
                background:#e5e5e5;
                }
        .swf_count{
                font-size:9px;
                color:#0066ff;
                text-transform:uppercase;
                background:#f3f3f3;
                border-top:solid 1px #e5e5e5;
                padding:6px 6px 6px 12px;
                margin:0px -6px -8px -6px;
                -moz-border-radius:0px 0px 6px 6px;
                -webkit-border-radius:0px 0px 6px 6px;
                border-radius:0px 0px 6px 6px;
                }
        .swf_count span{color:#666;}
                </style>';
}
function swf_file_url(){
        global $wp_query;
        $custom = get_post_custom($wp_query->post->ID);
        echo $custom['link'][0];
}

＆＃13;

Answer 1

http://php.net/manual/en/control-structures.alternative-syntax.php

变化

function swf_file_url(){
        global $wp_query;
        $custom = get_post_custom($wp_query->post->ID);
        echo $custom['link'][0];
}

到

function swf_file_url($return = false){
        global $wp_query;
        $custom = get_post_custom($wp_query->post->ID);
        if($return) return $custom['link'][0];
        echo $custom['link'][0];
}

然后使用：

<?if(swf_file_url(1)): ?>
<span class="play"><a rel="lightbox" href="<?php swf_file_url(); ?>">Play</a></span>
<span class="download"><a href="<?php swf_file_url(); ?>" download>Download</a></span>
<?php endif; ?>

Answer 2

您的问题并不清楚，但要检查swf_file_url();是否包含值，您可以使用empty()，即：

if(!empty(swf_file_url())){
  $swfUrl = swf_file_url();
  echo <<< EOF
  <span class="play"><a rel="lightbox" href="$swfUrl">Play</a></span>
  <span class="download"><a href="$swfUrl" download>Download</a></span>
EOF;
}

php if语句检测[href =＆＃34;＆＃34;]

2 个答案: