我尝试了不同的东西来添加无害的子菜单,但我无法做到。这是笔:
SELECT P.Ext_item_id, P.Ext_shop_id, P.Ext_date_id, S.SALES
FROM PROMO P
INNER JOIN SHOP SH ON P.Ext_shop_id = SH.Ext_shop_id
INNER JOIN ITEM I ON P.Ext_item_id = I.Ext_item_id
INNER JOIN [DATE] D ON P.Ext_date_id = D.Ext_date_id
INNER JOIN SALES S
ON SH.shop_id=S.shop_id AND I.item_id = S.item_id AND D.date_id = S.date_id
答案 0 :(得分:0)
你很亲密。
HTML:
<li><a class="nav-link" href="">За матурата</a>
<ul class="submenu">
<li ><a class="nav-link" href="">Матурата в цифри</a></li>
<li ><a class="nav-link" href="">Конспект</a></li>
<li ><a class="nav-link" href="">ЧЗВ</a></li>
</ul>
</li>
CSS:
.submenu {display:none}
li:hover > .submenu {
display:block;
position:absolute;
}