PHP - Using Php To Sleect Value From Drop Down Menu Not Working

trying to get a drop-down menu (in a form) to work  that needs to get populated from a table
table called  :   CarCategory
in this table are 2 columns  called
CarCategoryID    (unique number from 1-18)
CarCategoryDesc  (actual name of category.. e.g. 'convertible')

link :   http://98.131.37.90/postPart.php

I'm a total novice.. and can't figure out why the first is not working and the second (country) is working


code :

 <!--NOT WORKING ... why not ?? should show drop down menu-->   
    <label>
      <div align="center">
          <select name="carcategory" id="CarCategory" class="validate[required]" style="width: 200px;">

          <option value="">Select Car Category from list...</option>

             <?php while($obj_queryCarCategory = mysql_fetch_object($result_queryCarCategory)) { ?>
                <option value="<?php echo $obj_queryCarCategory->CarCategoryID;?>"
                <?php if($obj_queryCarCategory->CarCategoryID == $UserCountry) { echo 'selected="selected"'; } ?> >
                <?php echo $obj_queryCarCategory->CarCategoryDesc;?></option>
            <?php } ?>

        </select>
      </div>
    </label>

<!-- working ! -->
    <label>
      <div align="center">
          <select name="country" id="Country" class="validate[required]" style="width: 200px;">

          <option value="">Select...</option>

            <?php while($obj_queryCountry = mysql_fetch_object($result_queryCountry)) { ?>
                <option value="<?php echo $obj_queryCountry->CountryID;?>"
                <?php if($obj_queryCountry->CountryID == $UserCountry) { echo 'selected="selected"'; } ?> >
                <?php echo $obj_queryCountry->CountryName;?></option>
            <?php } ?>

        </select>
      </div>
    </label>

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

Okay, so, here's the scenario.

I have a form that is editing an item that is already in the database. The text fields fill in just fine with that info. However, the drop down menus don't retrieve that info, rather resorting to the defaults, which can be a problem if you don't remember what you originally had.

Is there anyway to make the dropdown menus pull the info from the table and use that rather than resorting back to the default?

I tried using this:

Code: [Select]
<tr><td width="20%">Bonus:</td><td><select name="bonus" value="{{bonus}}">
<option value="Attack" {{bonus1select}}>Add to the attack power of weapon</option>
<option value="Defense" {{bonus2select}}>Add to the defensive power of armor</option>
<option value="None" {{bonus3select}}>No effect</option>
</select><br /></td></tr>

So, it's obvious the "value" portion not working. Any help would be great!!

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>

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>