2

I have a tables in MySQL one is test table columns like this

  id refid name userdefined
  1  0       A
  2  0       B 
  3  0       C
  4  1       A1   abc
  5  1       A2   cde
  6  2       B1
  7  2       B2
  8  3       C3
  9  3       c4
  10 3       c5
  11 2       C7 
  12 2       C8  lmn
  13 11       c9
  14 11      c10

Using the above table I am creating the dynamic menu using a PHP function. I have one more table, it has the login fields and data like this:

id username password  field3
1  john      john     1,3,4,5,6,7,8,9

What i want is if John is logged in, how do I only show the menu items in field3. I am new to PHP. I am showing all menu items using a function in PHP please help me, thanks in advance.

<?php

//this is php function for creating menu
$sql2 = mysqli_query($conn, 'select * from login');
$row = mysqli_fetch_array($sql2);
$menu_items = explode(',', $row['field3']);

function submenu($parentid = 0)
{
    global $conn;
    $sql = mysqli_query($conn, "SELECT * FROM test WHERE refid=" . $parentid . " AND id IN " . ($menu_items));
    {
        $rowcount = mysqli_num_rows($sql);
        if ($rowcount > 0) {
            echo '<ul>';
        }

        while ($row = mysqli_fetch_array($sql, MYSQLI_ASSOC)) {
            if ($row['refid'] == 0) {
                echo '<li class="limain">' . $row['name'];
                submenu($row['id']);
                echo '</li>';

            } else {
                if ($row['userdefined']) {
                    echo '<li class="lichild"><a href="' . $row['userdefined'] . '">' . $row['name'] . '</a>';
                } else {
                    echo '<li class="lichild">' . $row['name'];
                }
                submenu($row['id']);
                echo '</li>';

            }

        }

        if ($rowcount > 0) {
            echo '</ul>';
        }
    }
}

?>
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3 Answers 3

Reset to default
1 
$sql2=mysqli_query($conn,'select * from login');
$row = mysqli_fetch_array($sql2);
//storing the field values in menuitems variables
$menu_items =  $row['field3'];


//pass the menuitem variables to query

function submenu($parentid=0){
global $conn;
$sql=mysqli_query($conn,"SELECT * FROM test WHERE refid=".$parentid ." AND 
id in ($menu_items)");
 {
   $rowcount=mysqli_num_rows($sql);
   if($rowcount>0){
    echo '<ul>';
   }
         while($row=mysqli_fetch_array($sql,MYSQLI_ASSOC))
        {
          if($row['refid']==0)
          {
            echo '<li class="limain">'.$row['name'];
             submenu($row['id']);
             echo '</li>';
        }
          else{
            if($row['userdefined']){
           echo '<li class="lichild"><a 
   href="'.$row['userdefined'].'">'.$row['name'].'</a>';
         }else{
           echo '<li class="lichild">'.$row['name'];
         }
           submenu($row['id']);
           echo '</li>';
        }

      }
  if($rowcount>0){
    echo '</ul>';
    }
   }
}
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Comments

0

Your function is mostly correct in principle, I had a bit of trouble following your function so I changed a few things to improve readability, but this should work.

The only issue I found that would prevent this from working was your if($refid), which is checking for any data not == to null.

0 == null //true 0 === null //false

If I understood your table correctly, entries with a refid of 0 are parent items. In that case they're the only menu items passing into that part of the function, so I changed it from if ($refid) to if ($refid === 0)

With regards to you wanting to you only wanting to show specific menu items, you'll need to run a separate query outside of the function to get the list you made. And then add a check in you submenu() function.

I didn't have a way to test this, but it should work.

//Query to get list of menu items
$menu_items = explode(',', $row['menus']);

function submenu($parentid=0, $menu_items=null) {
    global $conn;
    $sql=mysqli_query($conn,"SELECT * FROM test WHERE refid=".$parentid);

    $rowcount=mysqli_num_rows($sql);
    if($rowcount > 0) {
        echo '<ul>';
    }

    while($row=mysqli_fetch_array($sql,MYSQLI_ASSOC)) {

        if($menu_items !== null || !in_array($id, $menu_items)) {
            continue;
        }

        $id = $row['id'];
        $refid = $row['refid'];
        $userdefined = $row['userdefined'];
        $name = $row['name'];

        if($refid === 0) {
            echo "<li class='limain'>{$name}";
            submenu($id, $menu_items);
            echo '</li>';
        } else if ($refid > 0) {
            if($userdefined) {
               echo "<li class='lichild'><a href='{$userdefined}'>{$name}</a>";
            } else {
               echo "<li class='lichild'>{$name}";
            }
            submenu($id, $menu_items);
            echo '</li>';
        }
    }

    if($rowcount > 0) {
        echo '</ul>';
    }
}
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4 Comments

i think you didn't read the second part of question.anyway thanks for help.
@itsme, apologies for that, I've corrected my answer to better respond to your question.
no problem. Would you mind voting my answer, or marking it as correct please :). Could do with the reputation.
it is actually not working as expected thats y i din't accept it.thanks for the help.
0

You must protect the php files too, not only the menu items. If the user know the php filename, can access with a direct link to protected area.

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1 Comment

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