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我有五个表需要接收正确的信息到 Wordpress 项目中的高级自定义菜单

这些是我需要的五个表和列

wp_term_taxonomy         - Need: term_id, taxonomy WHERE: taxonomy="nav_menu"
wp_terms                 - Need: term_id, name WHERE: term_id matches wp_term_taxonomy.term_id
wp_term_relationships    - Need: object_id, term_taxonomy_id WHERE: term_taxonomy_id matches wp_term_taxonomy.term_id
wp_postmeta              - Need: post_id, meta_key, meta_value WHERE: post_id matches wp_term_relationships.object_id AND meta_key="_menu_item_object_id"
wp_posts                 - Need: id, post_title, post_status, guid, post_parent, post_type WHERE: id matches wp_postmeta.meta_value

But that is not it I then need to:

wp_posts                 - Need: guid, post_parent, post_type WHERE: post_parent matches wp_posts.id AND post_type="attachment"
wp_postmeta              - Need: post_id, meta_key, meta_value WHERE: post_id matches wp_posts.id AND meta_key="description"

我希望这有点道理。

我要做的基本上是，创建一个下拉菜单，其中包含 WordPress 自定义菜单功能中的页面列表，获取每个页面的特色图像，以及它们的自定义字段描述，其中我有一些文本要显示。

带有样式的最终菜单如下所示：

到目前为止，我已经成功地使菜单工作，但不是非常好的代码类型：

<ul>
<?php 
    $getMenus = mysql_query('SELECT term_id, taxonomy FROM wp_term_taxonomy WHERE taxonomy="nav_menu"');
    while ($addMenus = mysql_fetch_assoc($getMenus)) { 
        $menus_id = $addMenus['term_id'];
?>
    <?php 
        $getTerms = mysql_query('SELECT term_id, name FROM wp_terms WHERE term_id='.$menus_id.'');
        while ($addTerms = mysql_fetch_assoc($getTerms)) { 
    ?>
        <li>
            <span class="menu-sub-headline"><?php echo $addTerms['name']; ?></span>
            <ul>
                <?php 
                    $getTermsRelationship = mysql_query('SELECT object_id, term_taxonomy_id FROM wp_term_relationships WHERE term_taxonomy_id='.$menus_id.'');
                    while ($addTermsRelationship = mysql_fetch_assoc($getTermsRelationship)) {

                    $termsRelationship = $addTermsRelationship['object_id'];

                    $getMetaRelationship = mysql_query('SELECT post_id, meta_key, meta_value FROM wp_postmeta WHERE post_id='.$termsRelationship.' and meta_key="_menu_item_object_id"');
                    while ($addMetaRelationship = mysql_fetch_assoc($getMetaRelationship)) { 

                    $metaKeyValue = $addMetaRelationship['meta_value'];
                ?>
                <?php
                    $result = mysql_query('SELECT id, post_title, post_status, guid, post_parent FROM wp_posts WHERE id='.$metaKeyValue.'');
                    while ($row = mysql_fetch_assoc($result)) {
                ?>
                <li>
                <span><a href="<?php echo $row['guid']; ?>"><?php echo $row['post_title']; ?></a></span>
                <?php $thumb = $row['id']; ?>
                <ul class="menu-sub-sub-item-ul">
                    <li>
                        <span class="menu-product-headline"><?php echo $row['post_title']; ?></span>
                        <?php $getThumb = mysql_query('SELECT guid, post_parent, post_type FROM wp_posts WHERE post_parent='.$thumb.' AND post_type="attachment"');
                            while ($addThumb = mysql_fetch_assoc($getThumb)) {
                        ?>
                            <img src="<?php echo $addThumb['guid']; ?>"/>
                        <? } ?>
                        <?php $getMeta = mysql_query('SELECT post_id, meta_key, meta_value FROM wp_postmeta WHERE post_id='.$thumb.' AND meta_key="description"'); 
                            while ($addMeta = mysql_fetch_assoc($getMeta)) {
                        ?>
                            <p><?php echo $addMeta['meta_value']; ?></p>
                            <a href="<?php echo $row['guid']; ?>"><img src="/wp-content/themes/mygind-co/images/design/read_more.png"/></a>
                        <?php } ?>
                    </li>
                </ul>
                <?php }}} ?>    
            </ul>
        </li>
    <? } ?>
<?php } ?>
</ul>

希望你们中的一些人可以帮助我获得相同的结果，但是通过更好的查询，甚至可以向我解释如何正确使用连接。我对 SQL 很陌生，这是我知识非常有限的原因。在此之前，我已经阅读了加入的说明，并自己进行了尝试，但似乎这个菜单对我来说有点太难试错了。

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1 回答 1

2

尽管我讨厌在 WordPress 中推荐直接 SQL 查询（如果可以的话，您应该始终尝试使用query_posts()），这可能是您的唯一选择。也就是说，您将需要执行两个复杂的查询。

首先，您需要运行查询以获取自定义菜单中的页面。使用您的问题中列出的要求...

SELECT wtt.term_id AS term_id, wtt.taxonomy AS taxonomy, wt.name AS name, wtr.object_id AS object_id, wtr.term_taxonomy_id AS term_taxonomy_id, meta.post_id as post_id, meta.meta_key as meta_key, meta.meta_value AS meta_value, posts.post_title AS post_title, posts.post_status AS post_status, posts.guid AS guid, posts.post_parent AS post_parent, posts.post_type AS post_type
    FROM wp_term_taxonomy AS wtt
    INNER JOIN wp_terms AS wt ON wt.term_id = wtt.term_id
    INNER JOIN wp_terms_relationships AS wtr ON wtr.term_taxonomy_id = wtt.term_id
    INNER JOIN wp_postmeta AS meta ON meta.post_id = wtr.object_id
    INNER JOIN wp_posts AS posts ON posts.id = meta.meta_value
    WHERE wtt.taxonomy = "nav_menu" AND meta.meta_key = "_menu_item_object_id"

这应该为您提供具有您说需要的适当值的帖子集合。然后，您需要在循环中运行该集合并执行附加查询以在第二组数据中获取您想要的信息。

///psuedocode
for( ... ) {

    SELECT posts.guid AS guid, posts.post_parent AS post_parent, posts.post_type AS post_type, meta.post_id AS post_id, meta.meta_key AS meta_key, meta.meta_value AS meta_value
        FROM wp_posts AS posts
        INNER JOIN wp_postmeta AS meta ON meta.post_id = posts.id
        WHERE posts.post_parent = XXX AND posts.post_type = "attachment" AND meta.meta_key = "description"

}

在这种情况下，XXX是第一个查询返回的特定帖子的 ID（在for(){ }循环中迭代）。

现实情况是，如果您可以更清楚地了解您实际需要哪些数据，您的查询可能会大大简化。由于第一个查询只是为了获取菜单中的帖子列表，因此您可能不需要那么大的SELECT语句。我之所以将其包括在内，是因为您在问题中将这些值中的每一个都表示为“需要”。

于 2011-03-18T14:27:30.200 回答