Using Arrays to Create Menu

[Ben2]

Hello,

 

I hope someone can help me with this as I really cant seem to get it to work!

 

I am wanting to create a navigation menu from data on a SQL database table, I know how I need to present the data to create the menu but I can not get it to work.

 

Here is an example of what I need:

//get the top menu
$sql = "SELECT name, link FROM menu WHERE category = 'none'";
$result = mysql_query($sql,$conn);
while($row=mysql_fetch_assoc($result))
{
  $topmenu[] = $row['name'];
}

//go through the top menu names
foreach($topmenu as $top)
{
       //echo the current top meny name
echo $top."<br />\n";

//get menu name and link if it matched the current top menu name
$sql = "SELECT name, link FROM menu WHERE category = ".$top."";
$result2 = mysql_query($sql,$conn);
while($row2=mysql_fetch_assoc($result2))
{
                //show the menu name and link
	echo $row2['name']."<br />\n";
	echo $row2['link']."<br />\n";
}
     
        //create space between top menu names
        echo "<br /><br />";
}

 

This was my first attempt which did not work at all but since then I have tried multiple things such as putting all the data into multiple arrays, in to one array etc but I couldnt get anything to produce the result that I need.

 

what I am hoping to produce is something like:

 

Home

Information

    Info1

    Info2

    Info3

About us

    About us1

    About us1

etc

etc

 

Once I can produce the data in this way I can then add the html, css etc to create the menu.

 

I will be very greateful for any help and advice that you can provide.

 

Kind Regards

Ben

 

[litebearer]

cant seem to get it to work!

Means what, exactly?  Nothing displays? You get error message?

[Ben2]

cant seem to get it to work!

Means what, exactly?  Nothing displays? You get error message?

 

Thank you for the quick reply,

 

Sorry I forgot to include the error message I get from that code.

Request Topic
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /blah/blah/blah/blah.com/blah/blah/blah.php on line 27

Home
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /blah/blah/blah/blah.com/blah/blah/blah.php on line 27

Information
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /blah/blah/blah/blah.com/blah/blah/blah.php on line 27

Tips
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /blah/blah/blah/blah.com/blah/blah/blah.php on line 27

Opinions
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /blah/blah/blah/blah.com/blah/blah/blah.php on line 27

 

Line 27 is:

while($row2=mysql_fetch_assoc($result2))

 

Kind Regards

Ben

[MarPlo]

Try the second $sql like this:

$sql = "SELECT `name`, `link` FROM `menu` WHERE `category`='$top'";

[Ben2]

Try the second $sql like this:

$sql = "SELECT `name`, `link` FROM `menu` WHERE `category`='$top'";

 

Thank you!!

 

I keep making silly mistakes like this lol

 

Once again thank you for your help.

 

Kind Regards

Ben

[Pikachu2000]

Here's a useful tip: When using double quotes to enclose a string, there isn't any need to use string concatenation to include variables. The two strings below are functionally identical. There's a big difference in readability and potential for error though.

<?php
$field1 = 'name';
$field2 = 'address';
$table = 'users';
$array['index'] = 'Frank';
$num = 10;

$string1 = "SELECT `$field1`, `$field2` FROM `$table` WHERE `$field1` = '{$array['index']}' LIMIT $num";
// AND //
$string2 = "SELECT `" . $field1 . "`, `" . $field2 . "` FROM `" . $table . "` WHERE `" . $field1 . "` = '" . $array['index'] . "' LIMIT " . $num;

echo "<br>String1: $string1<br>String2: $string2";
?>
Returns:
String1: SELECT `name`, `address` FROM `users` WHERE `name` = 'Frank' LIMIT 10
String2: SELECT `name`, `address` FROM `users` WHERE `name` = 'Frank' LIMIT 10

 

EDIT: Added <?php tags for syntax highlighting . . .