stuck on a PHP program. Need some idea - php

Question

Want to make a php program, where there will be a drop down which will contain some name of brands .. after selecting the " first drop down/ brands" products of the selected brand will show on another drop down.. need help . anyone ?

Answer 1

What you looking for is called a dependent select. It have barely nothing to do with php (except populating select options). I've found a demo for your case. You will need to install jquery to implement it in your code.
var $city = $(".city").on('change', function() {
$city.not(this).get(0).selectedIndex = this.selectedIndex;
});

Answer 2

You need to read about jQuery or CSS.
Look at this example (jQuery): http://dev7studios.com/dropit/

Answer 3

so you have to use ajax to do this
$(document).on("change","first select box",function(){
var id = $("first select box").val();
$.ajax({
url: "path to your file where you should write db code",
type: "POST",
dataType: "HTML",
async: false,
data: {"id": id},
success: function(data) {
$("second select box").html(data);
// here directly manipulate the data in controller or get the data in success function and manipulate .
}
});
})
in the file where you write db code
$a = "";
foreach(rows fro db as $a){
$a .= "<select value='db id'><?= name ?></select>";
}
echo $a;
we are capturing $a to out normal file add making that as the value for our second select box.
Hope it hlps

Answer 4

Use javascript function onchange select element and fetch records according to selected first select element value.
<form name="product" method="post" >
<select id="category" name="category" onChange="relodme()">
<option value=''></option>
<?php
$qry = "select * from category order by name";
$res = mysql_query($qry) or die ("MYSQL ERROR:".mysql_error());
while ($arr = mysql_fetch_array($res))
{
?>
<option value="<?=$arr['category_id']?>" <? if($_POST['category'] == $arr['category_id']) { ?> selected="selected" <? } ?> ><?=$arr['name']?></option>
<?
}
?>
</select>
<select id="Type" name="Type" >
<option value=''></option>
<?php
$qry = "select * from subcategory where category_id = '".$_POST['category']."' order by name";
$res = mysql_query($qry) or die ("MYSQL ERROR:".mysql_error());
while ($arr = mysql_fetch_array($res))
{
?>
<option value="<?=$arr['sub_category_id']?>" <? if($_POST['Type'] == $arr['sub_category_id']) { ?> selected="selected" <? } ?> ><?=$arr['name']?></option>
<?
}
?>
</select>
</form>
Javascript function:
function relodme()
{
document.forms[0].action="test1.php"; //your page name give here....
document.forms[0].submit();
}