PHP - Drop Down Menu Order

helo
i a newbie in php. i need some help regarding my system. i want my data from database to display in drop down menu form.anyone have idea how to do that? i really need some help. i have search all over and couldn't find any solution. any suggestions would be helpful 

Hi, i know this is probably very basic but i have been banging my head and looking for tuts.

i have built a mysql php dropdown menu.

all displays fine. now, how do i get the menu to actualy take me to a new url?

the new url should be       www.mysite.com/"menu selection"



Code: [Select]
<?

include_once 'includes/db.php';

$result = mysql_query("select * from crimerate WHERE DISTRICT = 'Limpopo'", $con);

if (!$result) {
    die('Invalid query: ' . mysql_error());

}


$options="";

while ($row=mysql_fetch_array($result)) {

    $id=$row["id"];
    $crime=$row["CRIME"];
    $options.="<OPTION VALUE=\"$id\">$crime</option>";
}
?>


<SELECT NAME=crime>
<OPTION VALUE=0>Choose
<?=$options?>
</SELECT>

I'm trying to code a drop-down menu that has four options; one for pie,exp, root 3, and the golden ratio.(all math values)  Beside the drop-down menu, there is an option for user-inputted data, they must input only positive numbers. I must take there selected drop-down menu option and times it by the user inputted number. I'm not sure how to check which option the user chose.

    <form id="s" method="post">
      <select id="math" name="options">
        <option value="nothing">Please select a constant</option>
        <option value="pie">Pi</option>
    <option value="econs">e</option>
    <option value="ratio">Golden Ratio</option>
    <option value="root">Root 3</option>
  

      </select>

    <input type="submit" name="Submit" value="Send">

Everything That ive tried has failed

Hi, i wonder whether someone may be able to help me please.

I am using a combination of PHP and AJAX to create two drop down menus on a HTML form. The data is being pulled from a mySQL database with the options available in the second drop down dependent on the value selected in the first.

The initial drop down menu called 'detectors' and the behaviours for the second drop down menu, 'searchheads' are created with the following AJAX code:

Code: [Select]
Function AjaxFunction(detectorid)
            {
            var httpxml;
            try
            {
            // Firefox, Opera 8.0+, Safari
            httpxml=new XMLHttpRequest();
            }
            catch (e)
            {
            // Internet Explorer
            try
            {
            httpxml=new ActiveXObject("Msxml2.XMLHTTP");
            }
            catch (e)
            {
            try
            {
            httpxml=new ActiveXObject("Microsoft.XMLHTTP");
            }
            catch (e)
            {
            alert("Your browser does not support AJAX!");
            return false;
            }
            }
            }
            function stateck()
            {
            if(httpxml.readyState==4)
            {
           
            var myarray=eval(httpxml.responseText);
            // Before adding new we must remove previously loaded elements
            for(j=document.addfindstolocation.searchheads.options.length-1;j>=0;j--)
            {
            document.addfindstolocation.searchheads.remove(j);
            }
           
           
            for (i=0;i<myarray.length;i++)
            {
            var optn = document.createElement("OPTION");
            optn.text = myarray[i];
            optn.value = myarray[i];
            document.addfindstolocation.searchheads.options.add(optn);
           
            }
            }
            }
            var url="searchheaddetails.php";
            url=url+"?detectorid="+detectorid;
            url=url+"&sid="+Math.random();
            httpxml.onreadystatechange=stateck;
            httpxml.open("GET",url,true);
            httpxml.send(null);
            }
The following code is the file 'searchheaddetails.php' (as highlighted above) which populates the second drop down menu.

Code: [Select]
<?
$detectorid=$_GET['detectorid'];
require "config.php";
$q=mysql_query("SELECT * FROM searchheads WHERE detectorid='$detectorid' ORDER BY 'searchheadname' ASC");
echo mysql_error();
$myarray=array();
$str="";
while($nt=mysql_fetch_array($q)){
$str=$str . "\"$nt[searchheadname]\",";
}
$str=substr($str,0,(strLen($str)-1)); // Removing the last char , from the string
echo "new Array($str)";

?> 
And this is the section of my form that pulls together the two drop down menus.

Code: [Select]
<form name="addfindstolocation" method="post" id="addfindstolocation">
<div align="left">
                                        <select name=detectors id="detectorid" onchange="AjaxFunction(this.value);">
                                            <option value=''>Select One</option>
                                            <?
                                            require "phpfile.php";// connection to database
                                            $q=mysql_query("SELECT * from detectors WHERE userid='1'ORDER BY 'detectorname' ASC");
                                            while($n=mysql_fetch_array($q)){
                                            echo "<option value=$n[detectorid]>$n[detectorname]</option>";
                                            }
                                           
                                            ?>
                                      </select>
                                    </div>
                                </div>
                                <p align="left">
                                    <label></label>
                                    <label>Search Head Used</label></p>
                                <div>
                                    <div align="left">
                                        <select name=searchheads id="searchheadid">
                                        </select>
                                    </div>
The drop down menus work fine, but I'm having a little difficulty with the data that is being saved. For the 'detectors' drop down menu the data being saved upon a selection being made is the 'id' pertient to the relevant detector e.g. 'Detector1' is selected and the 'id' value of '1' is saved which is exactly what I want. However when it comes to the second drop down menu, the value saved is the text value that the user selects, rather than the 'id'. e.g. 'Deep Search Head ' rather than an 'id' of '1'. Could someone perhaps tell me please what I need to change so that the 'id' value is saved rather than the text value.

If it helps, the coding is taken from the following http://www.plus2net.com/php_tutorial/ajax_drop_down_list.php.

Many thanks and kind regards.

Chris

I did search this up but all of them were lists. I want to make a menu drop down like so.... Non-clicked... Clicked...   The grey boxes would be images (unless it is easier to code them). I am a complete noob so please don't use technical terms   Thanks

Hi I have a temporary web page with a drop down menu. Problem is to get rid of the gap on the drop down menus. Any help please       www.des-otoole.co.uk/top_menu  

Hello!

This is my first post here.

I'm usually do programming in Ruby (Rails) but for a special occasion I need PHP.

So hi PHP people! I come in peace   

My question is quite simple. How do you make a drop down menu? Most preferably with database populated fields in it.

In the Rails community, this really helped me make these menus in Rails: http://www.kahfei.com/?p=23

Could someone briefly write something like that too? (The place holders)

Orrr you guys can just explain how to do it. The ones I found on the internet were quite confusing and not explained enough.

Help anyone? Thank you

I have lots of tables for my products could i use the table name for a drop down menu? or is it not safe

Thanks for any help

Hi, I'm no pro at PHP but I am trying to get a drop down menu to a authenticate before moving to the next part of the form.

What I want is once a selection has been made, ONLY THEN can the user move on, OTHERWISE a message echo appears.

This is the html menu box

<select size="1" name="title">
                    <option>Please Select</option>
                    <option value="Mr">Mr</option>
                    <option value="Mrs">Mrs</option>
                    <option value="Miss">Miss</option>
                    <option value="Ms">Ms</option>
                    <option value="Dr">Dr</option>
                  </select>   

Then this is what I have in the form PHP:

$visitortitle = $_POST['visitortitle'];

if ( HOW DO I GET THIS PART TO AUTHENTICATE AN OPTION HAS BEEN SELECTED? ) {
echo "<p>Please enter a title correctly<br />before you try submitting the form again.</p>\n";
die ( '<a href="pef.html">click here go back and try again</a>' );
echo $id;}

If anyone can help me sort out this part of the form I can move on as the rest is working fine?

Thanks
Gary
 

Hey guys im looking to do a drop down menu which is pretty basic however, in this drop down menu it uses an array of 4 values eg Orange, Apples, Bananas, Strawberry's. For this menu if a user has previously chosen an Orange which is stored on the db then then when the drop down menu is loaded it should move Orange to the top of the list, the same applies for the other values. How could I do this? I know how to do the select statement and place the values in an array. from there I dont know quite how to approach it.


()
Thank you.

I'm having a problem with populating a second drop down menu that needs to be identical to the first from a mysql database.
I have about 20 items in the DB & I only want to show the first 9.
My code is: Code: [Select]
      <select name="a_type" style="width: 175px;"> <option value="" selected="selected"> None selected</option>
      <?php
$i=1;
while (($_produce = mysql_fetch_array($result_produce)) && ($i <=9 ))
{
$produce_type=$_produce["type"];
$produce_id=$_produce["id"];
$i++;
?>

<option value="<?php echo $produce_id; ?>"><?php echo $produce_id; ?> - <?php echo $produce_type; ?> </option>
<?php
}
?>
        </select>

That works fine in my first drop down menu but the second drop down menu with the same exact code will show items starting at 11 & higher
instead of 1 to 9. I haven't tried closing the DB & the reopening it which I assume would work but there are a lot of other variables & things I'd
have to check to make sure they don't get effected so I'd rather not.   Is there a simple way to solve this?   

I've followed a few tutorials, as I've never touched drop down before (more of a conventional navbar guy ). But none are dropping the nested list  :-\. I even took the code from one, no changes made, and it still didn't drop the menu        This is as far as I got on this tutorial (http://line25.com/tu...s-dropdown-menu) before coming here

#topbar-navigation
{
	width: 101%;
	margin: 0 auto;
	;
	margin-top: -8px;
	height: 56px;
	background: #ffffff;
	display: inline-block;
	text-align: center;
	min-width: 1100px;
	-webkit-box-shadow: 0 1px 8px 0px #b5b5b5;
	   -moz-box-shadow: 0 1px 8px 0px #b5b5b5;
	        box-shadow: 0 1px 8px 0px #b5b5b5;
}



#topbar-navigation ul
{
	display: inline-table;
	list-style: none;
	position: absolute;
	margin-left: auto;
	margin-right: auto;
	margin-top: 20px;
	top: 0;
	left: 765px;
	right: 0;
}


#topbar-navigation ul:after
{
	content: "";
	clear: both;
	display: block;
}


#topbar-navigation ul ul
{
	display: none;
}


#topbar-navigation ul li:hover > ul
{
	display: block;
}

	

	
#topbar-navigation ul li a
{
	display: block;
	padding: 25px 40px;
	color: #99a0a8;
	text-decoration: none;
	font-family: 'sinkin_sans400_regular';
	font-size: 12px;
}

<div id="topbar-navigation">
		<a href="index.php"><img src="images/moduniverselogo.png" class="site-logo" draggable="false" /></a>
			<ul>
			<li><a href="#">Welcome, <b style="color:#ff924e" ><?php echo $user->data['username'] ?></b></a></li>
				<ul>
					<li><a href="#">My Profile</a></li>
					<li><a href="#">Edit Account</a></li>
					<li><a href="#">Log Out</a></li>
				</ul>
			</ul>
	</div>

I want it nested to the right like on the site currently: http://i.imgur.com/llZPUbZ.png
Edited by Azercii, 20 June 2014 - 10:01 PM.

I'm not sure how this would be done with php or something else, what I want is to have a button and when the button is pressed it creates a drop down menu and if it's pressed again then it creates a second drop down menu.  I've been googling but haven't found anything yet if someone has an idea or a tutorial that they could direct me to that would be very helpful!!!!

Any help is greatly appreciated, thanks in advance!!!

Hi all

I am trying to create a menu that shows the initially selected value depending on which page the user is on:

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

There are three product categories; Stone, Wood and Accessories. in the SQL.

When on the stone.php page I need the option value 'stone' to stay selected.

Many thanks for your help

Pete

I am brand new to php and I don't know where I am going wrong with this.
I would like to have drop down menu which is populated from mySQL and which would allow user to select one option.
The selection would be used on the following page once the form is submitted.
Right now I have the following script and it populates the dropdown menu but it does not return value when selection is made.

<form action="databaseinputcheck2.php" method="post">
<?php
mysql_connect("localhost", "xxxxxxxxxxxx", "xxxxxxxxxxxxxx") or die(mysql_error());
echo "Connected to MySQL<br />";
?>
<p>Lotteries:
<?php
mysql_select_db("wpawlowski") or die(mysql_error());
// Get all the data from the "LotteryNames" table
$lotterynameresult = mysql_query("SELECT id, Name FROM LotteryNames") 
or die(mysql_error());  
?>
<select name="LotteryName" size="1">
<option selected value="">Select One</option>
<option value="">----------</option>
<?php
// keeps getting the next row until there are no more to get
while($currentlottery = mysql_fetch_array( $lotterynameresult )) {
$lid=$currentlottery['id'];
$lname=$currentlottery['Name'];
// Print out the contents of each row into drop down menu
echo "<option value='$lid'>$lname</option>\n";
//$options.="<OPTION VALUE=\"$lid\">".$lname; //Selection input line

} 
?>
<input type="submit" value="GO" />
</form>

I looked on the internet and it is my understanding that if I comment out the line starting withc "echo" and uncomment the one below the drop down menu should still be populated and the selection would be stored in the $options variable which in turn  could be transfered to databaseinputcheck2.php with
$_POST to be used there, but when I do uncomment that line and comment the echo line out my drop down menu is not populated, what am I doing wrong?