how to fetch menu from database with its subcategory

This is my menu

              <ul class="sub-menu">
                <li><a href="product-category.html">Product 1</a></li>
                <li><a href="product-category.html">Product 2</a>
                  <ul class="sub-menu">
                    <li><a href="product-category.html">P 21</a></li>
                    <li><a href="product-category.html">P 22</a></li>
                    <li><a href="product-category.html">P 23</a></li>
                    <li><a href="product-category.html">P 24</a></li>
                  </ul>
                </li>
                <li><a href="product-category.html">Product 3</a></li>
                <li><a href="product-category.html">Product 4</a></li>
              </ul>

Product1, Product2, Product3 are categories from category table. P 21, P 22, P 23, P 24 are subcategory of Product 2 in subcategory. All category dont have subcategory. category id saved in subcategory table.

CREATE TABLE IF NOT EXISTS category ( cid int(11) NOT NULL AUTO_INCREMENT, cname varchar(255) NOT NULL, PRIMARY KEY ( cid ) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;

CREATE TABLE IF NOT EXISTS subcategory ( sid int(11) NOT NULL AUTO_INCREMENT, sname varchar(255) NOT NULL, cid int(11) NOT NULL, PRIMARY KEY ( sid ) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;

How can i show it in menu by fetctching data from MySql with single query?

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please show me architecture of data base.In general it takes 2 cycles with a condition,first for category and second for subcategory. file.php

<ul class="sub-menu">
<?php
 foreach($array_category as $category):?>
   <li><a href="<?php echo $category['link'];?>"><?php echo $category['name'];?></a>
     <?php  if( $exist_subcategory):?>
           <ul>
           <?php foreach($array_subcategory as $subcategory):?>
                 <li><a href="<?php echo $subcategory['link'];?>"><?php echo $subcategory['name'];?></a> 
           <?php endforeach;?>
           </ul>
           <?php endif;?> 
      </li>
      <?php endforeach;?>
    </ul>

The only problem is the condition to enter the correct condition need to know to data base ahritecture.

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您可以通过替换您的表名和所有特定的详细信息来使用此查询

SELECT * FROM `parenttable` p WHERE p.foreignkey IN (SELECT c.foreignkey FROM `childtable` c WHERE c.foreignkey = p.foreignkey)

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