PHP - Moved: Dynamic Drop Down Box

Hey Guys,

I'm building a site and I need a drop down menu created dynamicly....

This is the flat html code
Code: [Select]
<div class="art-nav">
            <div class="l"></div>
            <div class="r"></div>
            <ul class="art-menu">
<li><a href="URL 1"><span class="l"></span><span class="r"></span><span class="t">TOP LEVEL</span></a>
<ul>
<li><a href="URL 1">SUB CAT 1</a></li>
<li><a href="URL 2">SUB CAT 2</a></li>
<li><a href="URL 3">SUB CAT 3</a></li>
</ul>
</li>

<li class="art-menu-li-separator"><span class="art-menu-separator"></span></li>
<li><a href="URL 5"><span class="l"></span><span class="r"></span><span class="t">Contact Us</span></a></li>
            </ul>
            </div>

This is the code that shows a dyamic code, but not in the menu...


<?php wpsc_start_category_query(array('category_group'=> get_option('wpsc_default_category'), 'show_thumbnails'=> get_option('show_category_thumbnails'))); ?>

<a href="<?php wpsc_print_category_url();?>" class="wpsc_category_link"><?php wpsc_print_category_name();?></a> |

<?php wpsc_print_subcategory(); ?>
<?php wpsc_end_category_query(); ?>

hii,everyone...
 i am new to the forum ..so forgive me if i make some errors in describing my problem

i want to create a drop down menu...the values of which i want to retrieve from a table i have created in my database...
i have been able to acheive this...but i want a second drop down menu...in which i want the values to get filtered according to the selection of the first drop down menu...
i hav a table with the name test...which i am connecting to the first drop down menu
i hav created another table with the name testtype...for second menu..
the id of first table is the testid of second...
plzz help me with creating the second menu...

my code is
Code: [Select]
   <?php
$sql_result = "select test_series FROM test";
$result = mysql_query($sql_result); 
?>
<td>Select Test Series </td>
         <td>
        <select name="testseries" id="testseries" )">
        <option>Select</option>
        <?php  if(mysql_num_rows($result)) 
    { 
    while($row = mysql_fetch_assoc($result)) 
    { 
$test = $row['test_series'];?>
    <option><?php echo $test ?></option>
    <?php } 
 
    } 
    else {?>
    <option>No Names Present</option>"; 
   <?php } 
    ?>
    </select>
    </td>
    </tr>
   
     <tr>
<?php
$testseries = $_POST['testseries'];
 $query = "select id from test where test_series = '$testseries'";
 $resultt = mysql_query($query) or die(mysql_error());
 $row = mysql_fetch_array($resultt);
 $id = $row['id'];
   
$sql_result = "select test_name FROM testtype where testid = $id";
$result = mysql_query($sql_result); 
?>
<td>Select Test Name </td>
         <td>
        <select name="testtype">
        <option>Select</option>
        <?php  if(mysql_num_rows($result)) 
    { 
    while($row = mysql_fetch_assoc($result)) 
    { 
$test2 = $row['test_name'];?>
    <option><?php echo $test2 ?></option>
    <?php } 
 
    } 
    else {?>
    <option>No Names Present</option>"; 
   <?php } 
    ?>
    </select>
    </td>
    </tr>






IN THIS CODE I AM GETTIN AN ERROR Undefined index: testseries in ..addques.php on line 155 WHICH IS THIS ONE $testseries = $_POST['testseries'];

I have attempted to get a php reporting page built for our off site managers to run reports.  Because of the number of employees I think it ouwld be best to sort by foreman, then be able to pick from the resulting drop down list of current employees under that foreman.   My issue I think really comes from having the first drop down be dynamic in that you would see a list of foremen which comes from a query of the main employees table and then picks up only active employees and their foreman.  The tutorials I have seen all reference an 'id" field but since I am pulling the info for the drop down from a query, should I have some sort of rowID?  If so how?  The tutorial I have found checks to see if you pick a number which I seem to have gotten around but when I do the query and look for $GF (general foreman) then the query comes back empty (this is bolded and underlined below).  Should this be some sort of rowID for the general foreman drop down?   Below is what I have been able to sort-of get working (adding in the URL handlers I think will be useful for people in the future to be able to bookmark it/send out emails with links already in it etc)   Here is the tutorial http://www.plus2net....p_down_list.php   Not sure if i understand why he used the $nocitia tags either   I know Im close on this but am starting to go crosseyed!  Thanks for pointing me in the right direction.

<?php
require 'config.php';  // Database connection
//////// End of connecting to database ////////
?>

<!doctype html public "-//w3c//dtd html 3.2//en">

<html>

<head>
<title>Multiple drop down list box</title>
<SCRIPT language=JavaScript>
function reload(form)
{
var val=form.GF.options[form.GF.options.selectedIndex].value;
var val2=form.EMPLOYEE_ID.options[form.EMPLOYEE_ID.options.selectedIndex].value;
var val3=form.SEPARATED.options[form.SEPARATED.options.selectedIndex].value;

self.location='dd3.php?GF=' + val + '&EMPLOYEE_ID=' + val2 + '&SEPARATED=' + val3 ;
}

</script>
</head>

<body>
<?Php

@$GF=$_GET['GF']; // Use this line or below line if register_global is off
@$EMPLOYEE_ID=$_GET['EMPLOYEE_ID']; // Use this line or below line if register_global is off
@$SEPARATED=$_GET['SEPARATED']; // Use this line or below line if register_global is off


///////// Getting the data from Mysql table for first list box//////////
$quer2="SELECT tbl_Employee_Master.Current_GF as GF FROM tbl_Employee_Master GROUP BY tbl_Employee_Master.Current_GF, tbl_Employee_Master.Separated HAVING (((tbl_Employee_Master.Separated)=0))";
///////////// End of query for first list box////////////

/////// for second drop down list we will check if category is selected else we will display all the subcategory/////
if(strlen($GF) > 0){
$quer="SELECT Employee_ID tbl_Employee_Master WHERE Current_GF=$GF";
}else{$quer="SELECT DISTINCT Employee_ID, Last_Name, First_Name FROM tbl_Employee_Master order by Last_Name"; }
////////// end of query for second subcategory drop down list box ///////////////////////////

echo "<form method=post name=f1 action='dd-check.php'>";
/// Add your form processing page address to action in above line. Example  action=dd-check.php////
//////////        Starting of first drop downlist /////////
echo "<select name='GF' onchange=\"reload(this.form)\"><option value=''>Select one</option>";
foreach ($dbo->query($quer2) as $noticia2) {
if($noticia2['GF']==@$GF){echo "<option selected value='$noticia2[GF]'>$noticia2[GF]</option>"."<BR>";}
else{echo  "<option value='$noticia2[GF]'>$noticia2[GF]</option>";}
}
echo "</select>";
//////////////////  This will end the first drop down list ///////////

//////////        Starting of second drop downlist /////////
echo "<select name='EMPLOYEE_ID'><option value=''>Select one</option>";
foreach ($dbo->query($quer) as $noticia) {
echo  "<option value='$noticia[Employee_ID]'>$noticia[Employee_ID] $noticia[Last_Name] $noticia[First_Name]</option>";
}
echo "</select>";
//////////////////  This will end the second drop down list ///////////
//// Add your other form fields as needed here/////
echo "<input type=submit value=Submit>";
echo "</form>";
?>
<br><br>
<a href=dd.php>Reset and start again</a>
<br><br>
<center><a href='http://www.plus2net.com' rel="nofollow">PHP SQL HTML free tutorials and scripts</a></center>
</body>

</html>

i have created a seating system.
based on the number of seats booked that value is passed to a page that dynamicly produces a drop down box for each seat booked so the type of seat price can be assigned.

this part works fine.
the problem i have is how do i code the recieving page not knowing how many values are going to be passed?

this is the drop down menu page

Thanks

Code: [Select]
<form action="payment.php?seatco=<?php echo $s_count; ?>" method="post" name="f1">
<div id="tick_select">
<?php
for($i=1; $i<=$s_count; $i++){
$tick= mysql_query("select * from ticket_list"); 
?>
<select name="tick_com<?php echo $i; ?>" id="tick_com" onChange="" title="Choose A Ticket">
<option value="">Select Ticket type</option>
<?php 
while($result= mysql_fetch_assoc($tick))
{ 
?>
<?php echo '<option value="'.$result[ticket_id].'" name="'.$result[price].'">'.$result[type].'</option>';?>
<?php }}?>
</select>
</div>
<INPUT TYPE="submit" VALUE="Select">
</form>

Hi guys, I'm still not having any luck with coming up with an idea of how to make my multi-level drop down menu work with items from a database. I don't necessarily need code, but an idea of how to make it work, so any ideas are welcome.

The table:

Column Descriptions:
id       - The unique id
label    - The text to display
link     - The link to be used
relation - I cannot remember why I wanted that... just disregard it
parent   - The parent item's id (0 for no parent)
sort     - A sort number for the item (ascending)
active   - 0 for false, 1 for true (not implemented yet, will add a WHERE `active` = 1 into the sql later)
depth    - How many parent items there are

Now, here's what it has to do (the class):
Grab items from the database (sorted by depth, parent, and sort) Place the children in a child array of the parents (or however it can be sorted) Implode the array and turn it into an html menu with the below structure
Menu HTML Structure
Code: [Select]
<div id="menu">
  <ul>
    <li><h2>Level 1 A</h2>
      <ul>
        <li><a href="?r=level2/a">Level 2 A</a>
          <ul>
            <li><a href="?r=level3/a">Level 3 A</a></li>
            <li><a href="?r=level3/b">Level 3 B</a></li>
          </ul>
        </li>
      </ul>
    </li>
    <li><a href="?r=level1/b"><h2>Level 1 B</h2></a></li>
  </ul>
</div>
And here is the code I have so far, but I can't figure out how to get the children into the parent array, or at least sort it so that I can achieve the above HTML structure.

Thank you in advance for your help, and as I said above, I don't necessarily need code, but an idea of how to make it work, so any ideas are welcome.

Hello,, I have looked for numerous examples, and have only got part of the answer.
I have successfully populated a drop down list by reading the database.
It populates with a list of quest titles. Once the user selects the quest.
I want a textbox to populate with the Quest Description.

I found an example that uses AJAX/PHP to do this, but it doesn't access a db to read/populate.
The main code is in index.html
The original example you select a country, and it populates a text field with the currency code.
It passes a variable called $country to another file called find_ccode.php.
I changed the first case in find_ccode.php to access my database, and populate the text box with quest description.
Ideally I will like to make this case dynamic based off QuestID, would populate textbox with corresponding Quest Description (QDescrip).
I then tried the other way in index.php
I dynamically populated the dropdown list, and gave the option values the QuestIDs to be passed to $country, but something isn't working.
I have a write files command to let me know what the $country variable is passing, and it never writes. It works in index.html though.
I apologize in advance for my amateurish coding. I still have a lot to learn. I have included the code below, but if you know a more elegant way of doing it. I am creating a role-playing game, as you probably have figured out by now. I hope to one day put it on the web, but right now using it as a way of learning html, and PHP.

I really want to know how to catch the value of the dropdown box. I would like to know what it is before sending over.
also what is this.value. I assume this is what the value of the dropdown box is when you have selected something.

I have been beating my head for days trying to figure this out. Any help is very much appreciated!

Index.html
<html>
<head>
<title>Changing textbox value based on dropdown list using Ajax and PHP</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<script>
//  Developed by Roshan Bhattarai
//  Visit http://roshanbh.com.np for this script and more.
//  This notice MUST stay intact for legal use

//fuction to return the xml http object
function getXMLHTTP() {
      var xmlhttp=false;   
      try{
         xmlhttp=new XMLHttpRequest();
      }
      catch(e)   {      
         try{         
            xmlhttp= new ActiveXObject("Microsoft.XMLHTTP");
         }
         catch(e){
            try{
            xmlhttp = new ActiveXObject("Msxml2.XMLHTTP");
            }
            catch(e1){
               xmlhttp=false;
            }
         }
      }
          
      return xmlhttp;
   }
   
   
   
function getCurrencyCode(strURL)
{      
   var req = getXMLHTTP();      
   if (req)
   {
      //function to be called when state is changed
      req.onreadystatechange = function()
      {
         //when state is completed i.e 4
         if (req.readyState == 4)
         {         
            // only if http status is "OK"
            if (req.status == 200)
            {                  
               document.getElementById('cur_code').value=req.responseText;                  
            }
            else
            {
               alert("There was a problem while using XMLHTTP:\n" + req.statusText);
            }
         }            
       }         
       req.open("GET", strURL, true);
       req.send(null);
   }         
}
</script>




</head>

<body style="font: 12px Verdana, Arial, Helvetica, sans-serif;">

<form style="text-align:center" method="post" action="" name="form1">
<p style="color:#000099 ">When you change the dropdown list, the respective currency code of the country will be displayed in the textbox which is fetched from PHP using Ajax. </p>
<p>Country : <select name="country" onChange="getCurrencyCode('find_ccode.php?country='+this.value)"

<?php

mysql_connect('localhost', '$user', '$password');

mysql_select_db('sok');

$result = mysql_query('select QuestID,QTitle, QDescrip from QuestList');

$options="";

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

    $QuestID=$row["QuestID"];
    $QTitle=$row["QTitle"];
    $options.="<OPTION VALUE=\"$QuestID\">".$QTitle;
}
?>
   <option value="">Select Country</option>
   <option value="1">USA</option>
   <option value="2">UK</option>
   <option value="3">Nepal</option>   
        </select><br/><br/>
 Currency :   <input type="text" name="cur_code" id="cur_code" ></p>
</form>
</body>
</html>

Index.php
<html>
<head>
<title>Changing textbox value based on dropdown list using Ajax and PHP</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<script>
//  Developed by Roshan Bhattarai
//  Visit http://roshanbh.com.np for this script and more.
//  This notice MUST stay intact for legal use

//fuction to return the xml http object
function getXMLHTTP() {
      var xmlhttp=false;   
      try{
         xmlhttp=new XMLHttpRequest();
      }
      catch(e)   {      
         try{         
            xmlhttp= new ActiveXObject("Microsoft.XMLHTTP");
         }
         catch(e){
            try{
            xmlhttp = new ActiveXObject("Msxml2.XMLHTTP");
            }
            catch(e1){
               xmlhttp=false;
            }
         }
      }
          
      return xmlhttp;
   }
   
   
   
function getCurrencyCode(strURL)
{      
   var req = getXMLHTTP();      
   if (req)
   {
      //function to be called when state is changed
      req.onreadystatechange = function()
      {
         //when state is completed i.e 4
         if (req.readyState == 4)
         {         
            // only if http status is "OK"
            if (req.status == 200)
            {                  
               document.getElementById('cur_code').value=req.responseText;                  
            }
            else
            {
               alert("There was a problem while using XMLHTTP:\n" + req.statusText);
            }
         }            
       }         
       req.open("GET", strURL, true);
       req.send(null);
   }         
}
</script>

</head>

<body style="font: 12px Verdana, Arial, Helvetica, sans-serif;">

<form style="text-align:center" method="post" action="" name="form1">
<p style="color:#000099 ">When you change the dropdown list, the respective currency code of the country will be displayed in the textbox which is fetched from PHP using Ajax. </p>
<p>Quest : <select name="country" getCurrencyCode('find_ccode.php?country='+this.value)">'

<?php

mysql_connect('localhost', '$user', '$password');

mysql_select_db('sok');

$result = mysql_query('select QuestID,QTitle, QDescrip from QuestList');

$options="";

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

    $QuestID=$row["QuestID"];
    $QTitle=$row["QTitle"];
    $options.="<OPTION VALUE=\"$QuestID\">".$QTitle;
   
}

$myFile = "test_catchoption.txt";
$fh = fopen($myFile, 'w') or die("can't open file");   
fwrite($fh, $options);
fclose($fh);
?>

Choose your Quest
<?=$options?>
</SELECT>
 <br/><br/>
 Quest Description :   <input type="text" name="cur_code" id="cur_code" ></p>
</form>
</body>
</html>


find_ccode.php
<?php
// Developed by Roshan Bhattarai
// Visit http://roshanbh.com.np for this script and more.
// This notice MUST stay intact for legal use
$country=$_REQUEST['country'];
$myFile = "testFilefindcc.txt";
$fh = fopen($myFile, 'w') or die("can't open file");   
fwrite($fh, $country);
fclose($fh);
switch($country)
{
   case "1" :
   mysql_connect('localhost', '$user', '$password');
    mysql_select_db('sok');

$result = mysql_query("select * from QuestList where QuestID = '1'");
$row = mysql_fetch_assoc($result);
        echo $row['QDescrip'];
      break;
   case "2" :   
      echo "GBP";
      break;
   case "3" :   
      echo "NPR";
      break;
}
?>

I am having a lot of trouble with this code, and I have no clue how to fix it.  Right now, I have a GUI for a fictitious car dealership that has 5 populated drop down menus called Make, Model, Year, Color, and Mileage.  What I want the code to do is read the selections made by the user with the drop down menus once the user hits the submit button and then filter the tables that I have in a mysql database to meets the choice requirements of the user.

The code will bring up the GUI, but once I hit the submit button, I get the following errors

Please if anyone can help me that would be fantastic.  I really have no clue

Hi there.

Im a noob to sql and php not sure if this is right place to post,

Im trying to get a dynamic drop down menu to show the 1st column in my sql database the column is called cat and holds category info ie audio, internet, music ect. ( i have no idea how to do lol )

it has taken me 2 days to find and edit this  the bold and underline'd bit is what im trying to change with the dropdown menu.

Or thinking bout it is there a way to do it with the URL. IE.. page name.php?cat=audio ?
would that be easer ? is there any security issues with doing it that way ?


Code==================================
$db_host = '*******';
$db_user = '******';
$db_pwd = '*****;

$database = 'nbbcj_co_uk';
$table = 'penapps';

if (!mysql_connect($db_host, $db_user, $db_pwd))
    die("Can't connect to database");

if (!mysql_select_db($database))
    die("Can't select database");

// sending query
$result = mysql_query("SELECT * FROM {$table} WHERE `cat` = 'audio' LIMIT 10 ");
if (!$result) {
    die("Query to show fields from table failed");
}

$fields_num = mysql_num_fields($result);

//echo "<h1>Table: {$table}</h1>";
echo "<table border='1' width='100%'><tr>";
// printing table headers
for($i=0; $i<$fields_num; $i++)
{
    $field = mysql_fetch_field($result);
    //echo "<td>{$field->name}</td>";
}
echo "</tr>\n";
// printing table rows
while($row = mysql_fetch_row($result))
{
    echo "<tr>";

    // $row is array... foreach( .. ) puts every element
    // of $row to $cell variable
    foreach($row as $cell)
        echo "<td>$cell</td>";

    echo "</tr>\n";
}
mysql_free_result($result);
?>
code end ======================

the test page can been seen here http://www.nbbcj.co.uk/testd/1/test1.php

Any more questions let me know

Thanks for any help you can give, a we all have to start some ware lol

thanks kaine.

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