get hierarchical data from mysql

Question

I want to display menu from mysql database. This is what I tried so far and don't know what is the error for. I appreciate any help . I posted the table that I want to display and the error that I got

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\\wamp\\www\\second _try\\eCRA.php on line 18

<?php
    function display_menu($parent, $level) {
        $query=mysql_query("SELECT n.id, n.menu_name, n.link, d.count FROM 'menu' n 
        LEFT OUTER  JOIN(SELECT parent, COUNT (*) AS count FROM 'menu' GROUP BY
        parent) d ON  n.id=d.parent  W HERE n.parent=".$parent);
        echo"<ul>";

        while($row = mysql_fetch_assoc($query)) {
            if($row['count']>0) {
                echo"<li><a href='" .$row['link']. "'>". $row['menu_name'] ."</a>";
                display_menu($row['id'], $level + 1);
                echo"</li>";
            }//f
            elseif($row['count']==0) { 
                echo"<li><a href='". $row['link'] ."'>". $row['menu_name'] ."</a></li>";
            }else; 
        }//w
        echo"</ul>";
    }

    display_menu(0,2);
?>

![error][1] ![Table][2]

Answer 1

On line 11 you have $row['count'] which means you are retrieving the column count which is not in in your database columns (As per your attached image), I think you want this.

if(mysql_num_rows($query)>0){
//do stuff here 
}
elseif(mysql_num_rows($query)>==0){
//do stuff here 
}

Answer 2

Here's what's wrong:

$query=mysql_query("SELECT n.id, n.menu_name, n.link, d.count FROM 'menu' n 
    LEFT OUTER  JOIN(SELECT parent, COUNT (*) AS count FROM 'menu' GROUP BY
    parent) d ON  n.id=d.parent  W HERE n.parent=".$parent);

You're using single quotes for referencing to a table name. You cannot do that. You'd need to use backticks for that!

Like so:

$query=mysql_query("SELECT n.id, n.menu_name, n.link, d.count FROM `menu` n 
    LEFT OUTER JOIN (SELECT parent, COUNT (*) AS count FROM `menu` GROUP BY
    parent) d ON  n.id=d.parent  W HERE n.parent=".$parent);

I would also like to point out that your code is possibly vulnerable to SQL injection. Look into that, if you don't mind.

Answer 3

You're using the wrong syntax. FROM 'menu' should be with backticks instead of single quotes:

  FROM `menu` 

Answer 4

mysql_query fails to get the result use mysql_error() to find the problem

 function display_menu($parent, $level) {
    $query=mysql_query("SELECT n.id, n.menu_name, n.link, d.count FROM 'menu' n 
    LEFT OUTER  JOIN(SELECT parent, COUNT (*) AS count FROM 'menu' GROUP BY
    parent) d ON  n.id=d.parent  W HERE n.parent=".$parent) or die(mysql_error());
    echo"<ul>";

    while($row = mysql_fetch_assoc($query)) {
        if($row['count']>0) {
            echo"<li><a href='" .$row['link']. "'>". $row['menu_name'] ."</a>";
            display_menu($row['id'], $level + 1);
            echo"</li>";
        }//f
        elseif($row['count']==0) { 
            echo"<li><a href='". $row['link'] ."'>". $row['menu_name'] ."</a></li>";
        }else; 
    }//w
    echo"</ul>";
}

display_menu(0,2);

Answer 5

Assuming parent is numeric, try this...

SELECT n.id
     , n.menu_name
     , n.link
     , d.count 
  FROM menu n 
  LEFT 
  JOIN 
     ( SELECT parent 
            , COUNT (*) count 
         FROM menu
        GROUP 
           BY parent
     ) d 
    ON n.id = d.parent 
 WHERE n.parent = $parent;