PHP - While Statement In Drop Down Menu?

Hi i currenlty have adrop box filled with companies so the user can select which company they woudl like services from but the default is currently 0 and is to selecvt all firms but im unsure how to do this.

Current code: Code: [Select]
<td>Taxi Firm</td><td> <select name="taxifirm">
<option value="0" selected>All Taxi Firms</option>
<?php
 $sql = mysql_query("SELECT * FROM taxi_Firms"); 
while($row = mysql_fetch_array($sql)){
$uid = $row["Firm_ID"];
$username = $row["Firm_Name"]; 
echo '<option value="'.$uid.'">'.$username.'</option>' ;

}
?>

amny help is welcomed

thanx

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 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

im fairly new to php so tend to do trial and error..... more error than trial tbh.

im wondering if it is possible to get a drop down menu to fill from a mysql database and to arrange it alphabetically.   i have created the menu just dont know how to arrange it better.

also  how can i use the items id in drop menu to load other info from that row on the database.

hope  you can help me.

Code: [Select]
<FORM>
<?php $result = mysql_query( "SELECT * FROM movie_info" )
;
echo "<select name= film onChange='submit()' >film name</option>";
while
($nt=mysql_fetch_array($result)){
?>

<?php

echo "<option value='$nt[id]'>$nt[title] </option>";

}
?>
</select>
</FORM>

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  

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
 

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

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 am trying to get the drop downs to work in this order on the second <li> 903,896,898,1513 but they won't. I think it's picking up ascending order maybe? Is there another php/include I should use? It's a WordPress site.

Code: [Select]
<div id="navigation">
<div id="navbar_menu">
<ul id="nav_menu">
<li><?php wp_list_pages('include=3&title_li=&depth=2');?></li>
<li><?php wp_list_pages('include=2,903,896,898,1513&title_li=&depth=2');?></li> 
<li><?php wp_list_pages('include=23,911,913&title_li=&depth=1');?></li>
<li><?php wp_list_pages('include=25,921,925,927&title_li=&depth=1');?></li>             
<li><?php wp_list_pages('include=27&title_li=&depth=2');?></li>
<li><?php wp_list_categories('include=3&title_li=&depth=2');?></li>
<li><?php wp_list_pages('include=407&title_li=&depth=2');?></li> 
</ul>

</div>
</div>
Thanks in advance!

[attachment deleted by admin]

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 am using jquery .change function to perform an operation when a month is selected from a drop down menu. The change works but I am unable to update the value of the drop down menu with the updated month. My drop down shows the starting value as default even on change. Can anyone help. Following is the code snippet that does change and then the drop down menu form.

Code: [Select]
$("#monthName").change(function()
{
alert($("#monthName").val());
if ($("#post").val() == 1)
{
$("#monthselect").submit();
}
});
Code: [Select]
                   <form id="monthselect" action="<?=$_SERVER['PHP_SELF']?>" method="get">
                        <input id="post" type="hidden" name="post" value="1">
                        <label>SELECT MONTH</label>
                        <select id="monthName" name="monthName">
                                                     
                                <option value="January">January</option>
                                <option value="February">February</option>
                                <option value="March">March</option>
                                <option value="April">April</option>
                                <option value="May">May</option>
                                <option value="June">June</option>
                                <option value="July">July</option>
                                <option value="August">August</option>
                                <option value="September">September</option>
                                <option value="October">October</option>
                                <option value="November">November</option>
                                <option value="December">December</option>

                        </select>
                    </form>
Even after the change, January shows up by default even if I select say June or July. I tried something like following but did not work.

Code: [Select]
$("#monthName option[value=" + $("#monthName").val() +"]").attr("selected","selected") ;

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?

Hi guys, I had a problem before which is similar to the problem im about to describe, however Pikachu helped me solve the last problem so a big thanks to him! However this problem is slightly different and im struggling to find a solution. I will first describe the problem I had which pikachu helped me solve as it would be easier for me to describe this past problem.

I have an update form, so a user can edit existing records which are pulled from a database. When the user changes the values in a text field then submits the form, the values in the text fields updates fine. However if they change the values in the text box and then tries to submit the form but there was an error, the values in the text fields go back to the values from the database and the user has to change the text fields again.

Well this problem was solved with the help of pikachu and the this is what I did...

<?php if(isset($_POST['title'])){echo htmlspecialchars($_POST['title']);} else if (isset($title)) {echo htmlspecialchars($title);}?>

Again, the above works fine however the problem I have now is the same problem as described above however it relates to drop down menus and not text fields. I can't seem to figure out how to retain the values in a drop down menu the same way I have for text fields. Below is the code that im using that I thought would work however it isnt working, the values of the drop down menu (if there was an error) are going back to the values from the database. Heres the code...

<option value="Psychopathic" <?php if ((isset($_POST['category']))&&($_POST['category'] == 'Psychopathic')) { echo ' selected=selected'; }  else if ((isset($category))&&($category == 'Psychopathic')) { echo ' selected=selected'; } ?>>Psychopathic</option>

Does anybody know what i am doing wrong and how i can fix this problem? I understand what i need to do but i cant seem to get the code to do what i want to do and thats to retain values of a drop down menu when the page is returned with an error.

Any help would be much appreciated.

AJay

Hi I'm new here, also I'm new in php. What I need to do is to write registration form with drop down menu. Actually, I need to have drop down menu which would write data into database when someone fill registration form. Also I need when someone choose one of the city name in drop down menu, to display certain image in specific php page. Can anyone help me.
Here is my code for registration form :
register.php
<?PHP

//Database Information

$dbhost = "";
$dbname = "";
$dbuser = "";
$dbpass = "";

//Connect to database

mysql_connect ( $dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error());
mysql_select_db($dbname) or die(mysql_error());

   
$name = $_POST['name'];
$email = $_POST['email'];   
$username = $_POST['username'];
$password = md5($_POST['password']);
$grad = $_POST['grad'];

// lets check to see if the username already exists

$checkuser = mysql_query("SELECT username FROM users WHERE username='$username'");

$username_exist = mysql_num_rows($checkuser);
if($username_exist > 0){
    echo "I'm sorry but the username you specified has already been taken.  Please pick another one.";
    unset($username);
    include 'register.html';
    exit();
}
$checkemail = mysql_query("SELECT email FROM users WHERE email='$email'");
$email_exist = mysql_num_rows($checkemail);
if($email_exist > 0){
    echo "Email koji ste izabrali vec postoji.";
    unset($email);
    include 'register.html';
    exit();
}
$checkname = mysql_query("SELECT name FROM users WHERE name='$name'");
$name_exist = mysql_num_rows($checkname);
if($name_exist > 0){
    echo "Ime koje ste izabrali vec postoji.";
    unset($name);
    include 'register.html';
    exit();
}
// lf no errors present with the username
// use a query to insert the data into the database.

$query = "INSERT INTO users (name, email, username, password, grad) VALUES('$name', '$email', '$username', '$password','$grad')";

mysql_query($query) or die(mysql_error());
mysql_close(); 
if(empty($name) || empty($email) || empty($username))
{
   echo "Nisu uneseni svi podaci.";
}
/*else if(empty(($password))) //ovaj uslov ne prolazi !!!
{
   echo "Niste uneli lozinku.";
}*/
else
{
   echo "Uspesno ste se registrovali";
}
?>

and for form :
register.html
<form name="login" method="post" action="register.php">
<table border="0" width="225" align="center">
    <tr>
        <td width="219" bgcolor="#ff6600">
            <p align="center"><font color="white"><span style="font-size:14pt;"><b>Registration</b></span></font></p>
        </td>
    </tr>
    <tr>
        <td width="555">
            <table border="0" width="555" align="center">
                    <tr>
                        <td width="116"><span style="font-size:12pt;">Name:</span></td>
                        <td width="156"><input type="text" name="name" maxlength="100"></td>   
                  <td width="200"><p align="left"><span style="font-size:11pt;"><i>*Ukucajte vase puno ime i prezime*</i></span></p></td>
                    </tr>
               
               <tr>
                        <td width="116"><span style="font-size:12pt;">Email:</span></td>
                        <td width="156"><input type="text" name="email" maxlength="100"></td><br/>
                  
                    </tr>
                <tr>
                    <td width="116"><span style="font-size:12pt;">Username:</span></td>
                    <td width="156"><input type="text" name="username"></td><br/>
               
                </tr>
                <tr>
                    <td width="116"><span style="font-size:12pt;">Password:</span></td>
                    <td width="156"><input type="password" name="password"></td><br/>
               
                </tr>
                <tr>
                    <td width="116">&nbsp;</td>
                        <td width="156">
                            <p align="right"><input type="submit" name="submit" value="Submit"></p>
                        </td>
                </tr>
            <tr>
               <td>
               <select name = "dropdown" value="options">
               <option value "register.php">Izaberi grad</option>
               <option value = "Beograd">Beograd</option>
               <option value = "Nis">Nis</option>
               <option value = "Novi Sad">Novi Sad</option>
               </select>
               </td>
            </tr>
            </table>
        </td>
    </tr>
    <tr>
        <td width="219" bgcolor="#ff6600">&nbsp;</td>
    </tr>
   
</table>
</form>

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!!!