如何在php，mysql中創建多級菜單？

Question

我想在php中創建一個多級菜單，但它不起作用。有誰知道爲什麼下面的代碼不能在if close中生成第一個html代碼？

<?php 
include "db.php"; 
if(isset($_POST["category"])){ 
$category_query="SELECT * FROM categories WHERE part='Product' "; 
$run_query=mysqli_query($con,$category_query); 

if(mysqli_num_rows($run_query)>0){ 
    while($row=mysqli_fetch_array($run_query)){ 
     $cat_id=$row["cat_id"]; 
     $cat_name=$row["cat_name"]; 
     for ($i=0; $i<count($cat_id); $i++) { 
      $category_query2="SELECT * FROM categories WHERE parent_id='$cat_id[$i]' "; 
      $run_query2=mysqli_query($con,$category_query2); 
      $j=0; 
      if(mysqli_num_rows($run_query2)>0){ 
       $j++; 
       echo " 
        <li><a class='dropdown-button d' href='#' data-activates='dropdown2' data-hover=\"hover\" data-alignment=\"left\">$cat_name</a></li> 
        "; 

      } else { 
       echo " 
        <li><a herf='2#'>$cat_name</a></li> 
       "; 
      } 

     } 

    } 

} 
} 
?> 

Answer 1

我認爲代碼應該像這樣改變。

if(mysqli_num_rows($run_query2)>0){` 
    $j++; 
    echo "HTML code 'attribute' ".$cat_name."HTML code";` 
} else {` 
    echo "HTML code".$cat_name."HTML code";` 
}