PHP - Mysql Menu Issues

Hi everyone,

I hope I explain my problem well enough. I have created a cms with the help of a tutorial for my website, it allows me to click on the page from a menu and shows me the results. The menu list is taken from my mysql database, so for example i have homepage and recent news in my mysql table and these are what are shown in the menu.

What I want is for the menu to to be images that can be clicked to take you to the correct page.

I have attached a print screen to show you what it looks like at the minute and what I want the menu to look like.

I have no idea if this is even possible can someone please help me out?

Hopefully I will hear a reply, I will send my code if needed when I know whether it is possible or not.

Thanks in hope

i am storing my menu in the database, i want to be able to output it by priority, heres so far what i have. I have no idea were to start.
Database dump:
--
-- Table structure for table `menu`
--
CREATE TABLE IF NOT EXISTS `menu` (
  `menu_access_lvl` int(2) NOT NULL,
  `priority` int(11) NOT NULL,
  `name` varchar(200) NOT NULL,
  `comment` text NOT NULL,
  `location` text NOT NULL,
  `creator_id` varchar(255) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
--
-- Dumping data for table `menu`
--
INSERT INTO `menu` (`menu_access_lvl`, `priority`, `name`, `comment`, `location`, `creator_id`) VALUES
(0, 1, 'Home Page', 'Home page', 'index.php', 'admin'),
(0, 3, 'Contact', 'Contact', 'index.php?PG=contact', 'admin'),
(0, 2, 'Events & Meetings', 'Events & Meetings', 'index.php?PG=events', 'admin'),
(0, 4, 'About', 'About', 'index.php?PG=about', 'admin'),
(2, 5, 'Admin', 'Admin', 'index.php?PG=admin', 'admin');

And here is the php code displaying it

       //gets the role of the user if set, otherwise role = 0
   if(isset($_SESSION['SESS_MEMBER_ID']))$lvl = $_SESSION['SESS_ROLE']; else $lvl = 0;

   // this loads the menu buttons that correspond to the users role
   $menuqry="SELECT * FROM menu WHERE menu_access_lvl<='$lvl'";
   $menuresult=mysql_query($menuqry);
   
   while($row = mysql_fetch_array($menuresult)){
      echo "<li class=\"menuitem\"><a href=\"".$row['location']."\">".$row['name']."</a></li>";
   }


What this currently displays:
Home
Contact
Events & meetings
About

I want it to be according to priority in the menu like:
Home
Events & meetings
Contact
About

I have a table in my db with all my menus and submenues of my site, I'd like to show them in some kind of menu containing all of them
each entry in db has the columns ID,NAME,TEXT,LEVEL,DEPENDENT
level has a value of 1 or 2 depending if it's an item or a subitem
dependent has a reference to the id of the section it belongs to

Using dreamweaver I created two recordsets based on the level, they are called row_sections (only where level = 1) and row_subsections (only where level = 2)

here is the code I tried to use:

    <?php do { ?>
      <p><a href="test.php?sec=<?php echo $row_sections['id'];?>"><?php echo $row_sections['name']; ?></a></p>  //prints the section text
                    <?php do { ?>
                    <?php
                    if($row_subsections['dependent'] == $row_sections['id']) {
                    echo $row_subsections['name']."<br>"; //tries to print all subitems corresponding to the actual item
                    }
                    ?>
                    <?php } while ($row_subsections = mysql_fetch_assoc($subsections)); ?>
            <?php } while ($row_sections = mysql_fetch_assoc($sections)); ?>

Here's the output:

Quote

section 1 title

subsection 1
subsection 2

section 2 title


both section 1 and 2 should have two different subitems.. where is the problem?

if I add the line <?php echo $row_sections['id']; ?> right before the second <?php do{ ?> it shows the corresponding ID for BOTH sections so I can't understand the problem...

Alright so I created a MySQL database that has 5 tables each named a brand of a dirt bike.

In each table their are 2 fields, one INDEX_ID and MODELS.

Under that for rows I have every model bike named.

A quick question before I get onto what I want to do:

Can data be added underneath the rows?

For example, users will be able to submit information about each bike model. If each bike model is a row, can there be a category past that row or do I need to make each field a model name and just have a ton of fields and have the rows be the information users submit.

To make it easier to understand I'll post the SQL code for the brand Honda:

CREATE TABLE `Honda` (
  `INDEX_ID` int(3) NOT NULL auto_increment,
  `MODELS` varchar(20) collate latin1_general_ci NOT NULL,
  PRIMARY KEY  (`INDEX_ID`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci AUTO_INCREMENT=23 ;

--
-- Dumping data for table `Honda`
--

INSERT INTO `Honda` VALUES(1, 'CR85');
INSERT INTO `Honda` VALUES(2, 'CR125');
INSERT INTO `Honda` VALUES(3, 'CR250');
INSERT INTO `Honda` VALUES(4, 'CRF100');
INSERT INTO `Honda` VALUES(5, 'CRF150');
INSERT INTO `Honda` VALUES(6, 'CRF230');
INSERT INTO `Honda` VALUES(7, 'CRF250X');
INSERT INTO `Honda` VALUES(8, 'CRF250R');
INSERT INTO `Honda` VALUES(9, 'CRF450X');
INSERT INTO `Honda` VALUES(10, 'CRF450R');
INSERT INTO `Honda` VALUES(11, 'CRF50');
INSERT INTO `Honda` VALUES(12, 'CRF70');
INSERT INTO `Honda` VALUES(13, 'CRF80');
INSERT INTO `Honda` VALUES(14, 'XR650');
INSERT INTO `Honda` VALUES(15, 'CR500');
INSERT INTO `Honda` VALUES(16, 'XR100');
INSERT INTO `Honda` VALUES(17, 'XR200');
INSERT INTO `Honda` VALUES(18, 'XR250');
INSERT INTO `Honda` VALUES(19, 'XR400');
INSERT INTO `Honda` VALUES(20, 'XR50');
INSERT INTO `Honda` VALUES(21, 'XR70');
INSERT INTO `Honda` VALUES(22, 'XR80');

Anyway I still need to figure out how to have this under a form that a user can use to select the bike they want to submit information about. A code like this perhaps?:


<?
$connection = mysql_connect("localhost","user","pass");
$fields = mysql_list_fields("dbname", "table", $connection);
$columns = mysql_num_fields($fields);
echo "<form action=page_to_post_to.php method=POST><select name=Field>";
for ($i = 0; $i < $columns; $i++) {
echo "<option value=$i>";
echo mysql_field_name($fields, $i);
}
echo "</select></form>";
?>


Thanks.

your MySQL server version:  5.1.36

Code: [Select]
SELECT * FROM game_weapons

Table: Picture Attached

the EXPLAIN output for your query, if applicable:
I wish to connect a drop down menu's selection of weapons with the correlated table information.

What do I want to happen:
Click on a drop down Menu and have a list of weapons, these weapons are associated with a set number in the database and passed onto the next screen.   (The larger number wins, this part I have figured out).

               <p>
               <select name="weapon2" style="font-size:20px;font-family:Arial;width:275px">
               <option value="power">Power</option>
               <option value="intelligence">Intelligence</option>
               <option value="speed">Speed</option>
               <option value="reserve">Reserve</option>
               </select>
               </p>


I do have the battle code figured out!  (This should be the last step).

I wonder whether someone can help me please.

I've found http://www.plus2net.com/php_tutorial/ajax-listbox.php tutorial to create a drop down menu using mySQL table data, which, in turn returns a list of results on the page.

Following this tutorial I've put together the tables in my database and the required scripts as shown in the tutorial with the one exception, the "z_db.php" file, which I've assumed to be:

Code: [Select]
<?php mysql_connect("host", "user", "password")or die(mysql_error()); mysql_select_db("database"); ?>

The problem I have, is that when I try and run this, I receive the following error: Quote

Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /homepages/2/d333603417/htdocs/development/catsearch.php on line 91 which is this line in the search form: echo "</head><body onload="ajaxFunction()";>";.


I must admit I've guessed as to the structure of the 'z_db.php' file should look like because this is not shown so perhaps this is the problem.

I just wondered wether someone could perhaps take a look at this please and let me know where I've gone wrong.


Many thanks and kind regards

I have a mysql table with the structure of
Code: [Select]
ID  Menu_Name   Parent_ID
1   Finance     NULL
2   Business    NULL
3   Investment  1
4   Trading     2
How can I create a html <ul><li> list based on the parent?

Hello everyone,

So what I'm trying to do is have a dropdown menu displaying a number of <options> for people to select and to update that selection to the database, easy enough right? But I want that option to be displayed as the "selected" option when the page is revisited or refreshed and I just can't figure it out!!! (Permission to bang head on desk?) It would seem like it sould be a really basic thing to do but it's got me completely and a lot of menus around the site are going to rely on this so I came to you guys for help.

A simple example would be like the facebook edit profile page, the user selects whether they are Male or Female, the database gets updated and when you return the option you selected before is the one that appears as if selected="selected" had been done. I've tried everything I can think of (all be it from a learners perspective) with no joy, ive managed to get the database connection sorted, the tables done, the login with unique id $_SESSION, logout etc... so then when I got to this I thought... easy LOL yeah right.

Some of this probably doesnt even make sense but I'll show you the kind of things I've tried...


            <select name="gender" size="1" id="gender">
              <option value="male" <?php if ($gender == "male") {echo 'selected="selected"';} ;?>>Male</option>
              <option value="female" <?php if ($gender == "female") {echo 'selected="selected"';} ;?>>Female</option>
            </select>



OR


     <select name="gender" id="gender">
              <option value="" selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option>
              <option value="male" selected="<?php if ($gender == "male") {echo "selected";} else {echo "";} ;?>">Male</option>
              <option value="female" selected="<?php if ($gender == "female") {echo "selected";} else {echo "";} ;?>">Female</option>
            </select>


OR



            <select name="gender" size="1" id="gender">
              <option selected="<?php if (!isset($gender)) {echo "selected";} ;?>">Select</option>
              <option value="<?php if ($gender == "Male") {echo "selected";} else {echo "male";} ;?>">Male</option>
              <option value="<?php if ($gender == "Female") {echo "selected";} else {echo "female";} ;?>">Female</option>
            </select>


OR

            <select name="gender" id="gender">
              <option value="male"><?php if ($gender == "male") {echo "Male";} ;?></option>
              <option value="female"><?php if ($gender == "female") {echo "Female";} ;?></option>
            </select>


Honestly man, I've got no idea.


The other thing is, I have more than 1 dropdown menu in the same form (5 in total) and if I use 2 or more selecting different options as I go I get a blank screen.  And one more, if I have selected Male and it updates the users row and I resubmit Male again it's blank screen time again, lol.

Any help would be tremendous and greatly appreciated.

Thanks very much, Learner

P.S  Man!

Hi all,

I am currently learning PHP and have the homework to produce a function that can delete a row in a MySQL database table by clicking on an item in a drop-down menu in a web page.

The code I have produced up until now is this:

<!DOCTYPE HTML>
<html lang="de">
<head>
	<meta charset="utf-8" />
	<title>E3_Artikel_Löschen</title>

</head>
<body>
<form method = "GET">

<?php
	$anr="";
	
	try {
	$pdo = new PDO ('mysql:dbname=bestelldatenbank;host=localhost;charset=utf8', 'root', '');
	} catch (PDOException $error){
		die ($error->getMessage());
		}		
?>

	<div>
		<p>
			<label for="artikel">Artikel: </label>
				<select id="artikel" name="artikel">
					<?php
						$sqlSelect = "SELECT anr, name FROM artikel ORDER BY anr ASC";
						foreach ($pdo->query($sqlSelect) as $row) {
							echo "<option value=$row[0]>$row[0] | $row[1]</option>\n";
							$anr = $row[0];

						}
					?>
				</select>
			
			<input type = "submit" value = "Delete row" />
			
		</p>
	</div>

<?php
	function artLoeschen($anr) {
		echo "Function called $anr";
		if(isset($_GET[$anr])) {
			$anr = $_GET[$anr];
			$sqlDelete = $pdo->query("DELETE FROM artikel WHERE anr = :anr");
			if ($stmt = $pdo->prepare($sqlDelete)) {
				$stmt->bindParam(':anr', $anr);
				$stmt->execute();
			}
			echo "<h2><b>Artikel gelöscht!</b></h2>";
		}
	}
?>

</form>
</body>
</html>

So, I have observed the following when I run the script in a browser:

1.  The HTML works as expected and I get a drop-down list with the article number and description of each item in the affected table.

2.  I can click on an item in the list and it populates the top item in the drop-down list.

3.  When I click delete row, the selected item is not deleted.

4.  There are no error messages returned but the function is not executed (at least not as I would like to expect).

 

I have obviously missed something or made a mistake in my code. 

I would be very grateful for any help...this is driving me mad! :)

Regards,

Kevin

Hi all

I need to combine these two scripts:

Firstly, the following decides which out of the following list is selected based on its value in the mySQL table:

<select name="pack_choice">
<option value="Meters / Pack"<?php echo (($result['pack_choice']=="Meters / Pack") ? ' selected="selected"':'') ?>>Meters / Pack (m2)</option>
<option value="m3"<?php echo (($result['pack_choice']=="m3") ? ' selected="selected"':'') ?>>Meters / Pack (m3)</option>
<option value="Quantity"<?php echo (($result['pack_choice']=="Quantity") ? ' selected="selected"':'') ?>>Quantity</option>
</select>

Although this works OK, I need it also to show dynamic values like this:

select name="category">
<?php
$listCategories=mysql_query("SELECT * FROM `product_categories` ORDER BY id ASC");
while($categoryReturned=mysql_fetch_array($listCategories)) {
echo "<option value=\"".$categoryReturned['name']."\">".$categoryReturned['name']."</option>";
}
?>
</select>

I'm not sure if this is possible?

Many thanks for your help.

Pete

I am working on a project where I want a select form to display information from a MySQL table. The select values will be different sports (basketball,baseball,hockey,football) and the display will be various players from those sports. I have set up so far two tables in MySQL. One is called 'sports' and contains two columns. Once called 'category_id' and that is the primary key and auto increments. The other column is 'sports' and contains the various sports I mentioned. For my select menu I created the following code.


<?php
#connect to MySQL
$conn = @mysql_connect( "localhost","uname","pw")
or die( "You did not successfully connect to the DB!" );

#select the specified database
$rs = @mysql_SELECT_DB ("test", $conn )
or die ( "Error connecting to the database test!");
?>

<html>
<head>Display MySQL</head>
<body>
<form name="form2" id="form2"action="" >
<select name="categoryID">
<?php
$sql = "SELECT category_id, sport FROM sports ".
"ORDER BY sport";

$rs = mysql_query($sql);

while($row = mysql_fetch_array($rs))
{
  echo "<option value=\"".$row['category_id']."\">".$row['sport']."</option>\n  ";
}
?>
</select>

</form>
</body>
</html>




this works great. I also created another table called 'players' which contains the fields 'player_id' which is the primary key and auto increments, category_id' which is the foreign key for the sports table, sport, first_name, last_name.

The code I am using the query and display the desired result is as follows


<html>
<head>
<title>Get MySQL Data</title>
</head>
<body>
<?php
#connect to MySQL
$conn = @mysql_connect( "localhost","uname","pw")
or die( "Err:Db" );

#select the specified database
$rs = @mysql_SELECT_DB ("test", $conn )
or die ( "Err:Db");

#create the query
$sql ="SELECT *
FROM sports
INNER JOIN players ON sports.category_id = players.category_id
WHERE players.sport = 'Basketball'";

#execute the query
$rs = mysql_query($sql,$conn);

#write the data
while( $row = mysql_fetch_array( $rs) )
{
echo ("<table border='1'><tr><td>");
echo ("Caetegory ID:  " . $row["category_id"] );
echo ("</td>");
echo ("<td>");
echo ( "Sport: " .$row["sport"]);
echo ("</td>");
echo ("<td>");
echo ( "first_name: " .$row["first_name"]);
echo ("</td>");
echo ("<td>");
echo ( "last_name: " .$row["last_name"]);
echo ("</td>");
echo ("</tr></table>");
}
?>
</body>
</html>





this also works fine. All I need to do is tie the two together so that when a particular sport is selected, the query will display below in a table. I know I need to change my WHERE clause to a variable. This is what I need help with.

thanks

create table mimi (mimiId int(11) not null, mimiBody varchar(255) );
<?php
//connecting to database
include_once ('conn.php');
$sql ="SELECT mimiId, mimiBody FROM mimi";

$result = mysqli_query($conn, $sql );

$mimi = mysqli_fetch_assoc($result);

$mimiId ='<span>No: '.$mimi['mimiId'].'</span>';

$mimiBody ='<p class="leading text-justify">'.$mimi['mimiBody'].'</p>';
?>
//what is next?

i want to download pdf or text document after clicking button or link how to do that

Hi All, Am sure this is simple but can't work it out. I have the following which creates 3 menus:

<p>Drag items from one menu to another</p>
<table>
	<tr>
		<td valign="top">Menu 1</td>
		<td valign="top">Menu 2</td>
		<td valign="top">Menu 3</td>
	</tr>
	<tr>
		<td valign="top">
		<ul class="sortable" id="menu1">
			<li id="id_1">Item 1</li>
			<li id="id_2">Item 2</li>
		</ul>		
		</td>
		<td valign="top">
		<ul class="sortable" id="menu2">
			<li id="id_3">Item 3</li>
			<li id="id_4">Item 4</li>
		</ul>		
		</td>
		<td valign="top">
		<ul class="sortable" id="menu3">
			<li id="id_5">Item 5</li>
			<li id="id_6">Item 6</li>
		</ul>		
		</td>
	</tr>
</table>

This uses the following to show an array of each menu:

$(function() {
 $("ul.sortable").sortable({
    connectWith: '.sortable',
    update: function(event, ui) {
    $('#menu_choice').empty().html( $('.sortable').serial() );
    }
  });
});

(function($) {
$.fn.serial = function() {
 var array = [];
 var $elem = $(this); 
    $elem.each(function(i) {
      var menu = this.id;
        $('li', this).each(function(e) {
                array.push( menu + '['+e+']=' + this.id  );
        });
    });
    return array.join('&');      
    }
 
})(jQuery);

What I'd like is to simply grab the values of menu 1 only, as the array shows all menus. What am I doing wrong as can't seem to just return menu1. Going forward, I then want to store the returned array into a PHP cookie if this is possible?   Many thanks

So i have a <drop down> menu, and a <nav> menu on my left side of the page. I have a problem when i click at the first column of my drop down menu and it explores submenu, the submenu mixes with the nav menu that is under the drop down menu on the left. i could solve it with margin-top of the nav menu but i don't like the empty space beetwen them. I tried with putting overflow:visible; in CSS of my dropdown menu but it is still the same. drop down menu is seen but there is stil seen nav menu under and they are mixed. So basicly i want my dropdown menu to be priority,  so when i clicks on my first column (only first column is problem because there under is nav menu the other are open nice) it will explore submenu and the part of nav menu that covers the submenu will be hidden.   Here is the screen shot:     Here is the code of head <dropdown> menu if someone finds the problem.

#menu, #menu ul {
	 
    margin: 0;
    padding: 0;
    list-style: none;
}

#menu {
	 
    width: 900px;
	margin-top:20px;
    margin-left:auto;
	margin-right:auto;
    border: 1px solid #222;
    background-color: #111;
    background-image: linear-gradient(#444, #111);
    border-radius: 6px;
    box-shadow: 0 1px 1px #777;
}

#menu:before,
#menu:after {
    content: "";
    display: table;
}

#menu:after {
    clear: both;
}

#menu {
    zoom:1;
}

#menu li {
	 
    float: left;
    border-right: 1px solid #222;
    box-shadow: 1px 0 0 #444;
    position: relative;
}

#menu a {
	 
    float: left;
    padding: 12px 30px;
    color: #999;
    text-transform: uppercase;
    font: bold 12px Arial, Helvetica;
    text-decoration: none;
    text-shadow: 0 1px 0 #000;
}

#menu li:hover > a {
    color: #fafafa;
}

*html #menu li a:hover { /* IE6 only */
    color: #fafafa;
}

#menu ul {
	
    margin: 20px 0 0 0;
    _margin: 0; /*IE6 only*/
    opacity: 0;
    visibility: hidden;
    position: absolute;
    top: 38px;
    left: 0;
    z-index: 1;    
    background: #444;   
    background: linear-gradient(#444, #111);
    box-shadow: 0 -1px 0 rgba(255,255,255,.3);  
    border-radius: 3px;
    transition: all .2s ease-in-out;
}

#menu li:hover > ul {
	
    opacity: 1;
    visibility: visible;
    margin: 0;
}

#menu ul ul {
	 
    top: 0;
    left: 150px;
    margin: 0 0 0 20px;
    _margin: 0; /*IE6 only*/
    box-shadow: -1px 0 0 rgba(255,255,255,.3);      
}

#menu ul li {
	 
    float: none;
    display: block;
    border: 0;
    _line-height: 0; /*IE6 only*/
    box-shadow: 0 1px 0 #111, 0 2px 0 #666;
}

#menu ul li:last-child {  
  
    box-shadow: none;    
}

#menu ul a { 
    
    padding: 10px;
    width: 130px;
    _height: 10px; /*IE6 only*/
    display: block;
    white-space: nowrap;
    float: none;
    text-transform: none;
}

#menu ul a:hover {
    background-color: #0186ba;
    background-image: linear-gradient(#04acec, #0186ba);
}

#menu ul li:first-child > a {
    border-radius: 3px 3px 0 0;
}

#menu ul li:first-child > a:after {
	 
    content: '';
    position: absolute;
    left: 40px;
    top: -6px;
    border-left: 6px solid transparent;
    border-right: 6px solid transparent;
    border-bottom: 6px solid #444;
}

#menu ul ul li:first-child a:after {
	 
    left: -6px;
    top: 50%;
    margin-top: -6px;
    border-left: 0; 
    border-bottom: 6px solid transparent;
    border-top: 6px solid transparent;
    border-right: 6px solid #3b3b3b;
}

#menu ul li:first-child a:hover:after {
    border-bottom-color: #04acec; 
}

#menu ul ul li:first-child a:hover:after {
    border-right-color: #0299d3; 
    border-bottom-color: transparent;   
}

#menu ul li:last-child > a {
    border-radius: 0 0 3px 3px;
}

Here is the code of <nav> menu if someone finds the problem.

#cssmenu {
	
	
	width:15%;
  padding: 0;
  margin-top: 50px;
  margin-left:auto;
  margin right:auto;
  float:left;
  border: 0;
  line-height: 1;
}
#cssmenu ul,
#cssmenu ul li,
#cssmenu ul ul {
  list-style: none;
  margin: 0;
  padding: 0;
}
#cssmenu ul {
  position: relative;
  z-index: 597;
  float: left;
}
#cssmenu ul li {
  float: left;
  min-height: 1px;
  line-height: 1em;
  vertical-align: middle;
  position: relative;
}
#cssmenu ul li.hover,
#cssmenu ul li:hover {
  position: relative;
  z-index: 599;
  cursor: default;
}
#cssmenu ul ul {
  visibility: hidden;
  position: absolute;
  top: 100%;
  left: 0px;
  z-index: 598;
  width: 100%;
}
#cssmenu ul ul li {
  float: none;
}
#cssmenu ul ul ul {
  top: -2px;
  right: 0;
}
#cssmenu ul li:hover > ul {
  visibility: visible;
}
#cssmenu ul ul {
  top: 1px;
  left: 99%;
}
#cssmenu ul li {
  float: none;
}
#cssmenu ul ul {
  margin-top: 1px;
}
#cssmenu ul ul li {
  font-weight: normal;
}
/* Custom CSS Styles */
#cssmenu {
  width: 200px;
  background: #333333;
  font-family: 'Oxygen Mono', Tahoma, Arial, sans-serif;
  zoom: 1;
  font-size: 12px;
}
#cssmenu:before {
  content: '';
  display: block;
}
#cssmenu:after {
  content: '';
  display: table;
  clear: both;
}
#cssmenu a {
  display: block;
  padding: 15px 20px;
  color: #ffffff;
  text-decoration: none;
  text-transform: uppercase;
}
#cssmenu > ul {
  width: 200px;
}
#cssmenu ul ul {
  width: 200px;
}
#cssmenu > ul > li > a {
  border-right: 4px solid #1b9bff;
  color: #ffffff;
}
#cssmenu > ul > li > a:hover {
  color: #ffffff;
}
#cssmenu > ul > li.active a {
  background: #1b9bff;
}
#cssmenu > ul > li a:hover,
#cssmenu > ul > li:hover a {
  background: #1b9bff;
}
#cssmenu li {
  position: relative;
}
#cssmenu ul li.has-sub > a:after {
  content: '+';
  position: absolute;
  top: 50%;
  right: 15px;
  margin-top: -6px;
}
#cssmenu ul ul li.first {
  -webkit-border-radius: 0 3px 0 0;
  -moz-border-radius: 0 3px 0 0;
  border-radius: 0 3px 0 0;
}
#cssmenu ul ul li.last {
  -webkit-border-radius: 0 0 3px 0;
  -moz-border-radius: 0 0 3px 0;
  border-radius: 0 0 3px 0;
  border-bottom: 0;
}
#cssmenu ul ul {
  -webkit-border-radius: 0 3px 3px 0;
  -moz-border-radius: 0 3px 3px 0;
  border-radius: 0 3px 3px 0;
}
#cssmenu ul ul {
  border: 1px solid #0082e7;
}
#cssmenu ul ul a {
  font-size: 12px;
  color: #ffffff;
}
#cssmenu ul ul a:hover {
  color: #ffffff;
}
#cssmenu ul ul li {
  border-bottom: 1px solid #0082e7;
}
#cssmenu ul ul li:hover > a {
  background: #4eb1ff;
  color: #ffffff;
}
#cssmenu.align-right > ul > li > a {
  border-left: 4px solid #1b9bff;
  border-right: none;
}
#cssmenu.align-right {
  float: right;
}
#cssmenu.align-right li {
  text-align: right;
}
#cssmenu.align-right ul li.has-sub > a:before {
  content: '+';
  position: absolute;
  top: 50%;
  left: 15px;
  margin-top: -6px;
}
#cssmenu.align-right ul li.has-sub > a:after {
  content: none;
}
#cssmenu.align-right ul ul {
  visibility: hidden;
  position: absolute;
  top: 0;
  left: -100%;
  z-index: 598;
  width: 100%;
}
#cssmenu.align-right ul ul li.first {
  -webkit-border-radius: 3px 0 0 0;
  -moz-border-radius: 3px 0 0 0;
  border-radius: 3px 0 0 0;
}
#cssmenu.align-right ul ul li.last {
  -webkit-border-radius: 0 0 0 3px;
  -moz-border-radius: 0 0 0 3px;
  border-radius: 0 0 0 3px;
}
#cssmenu.align-right ul ul {
  -webkit-border-radius: 3px 0 0 3px;
  -moz-border-radius: 3px 0 0 3px;
  border-radius: 3px 0 0 3px;
}

Edited by Dorkmind, 26 November 2014 - 05:00 PM.