Show same item only once in drop down list? - php

Question

I have the following drop down list, that has a common item 'a'. This item is shown two times in output, but it want to have it only once. Can anybody help me?
<select name="">
<option value="a">a</option>
<option value="a">a</option>
<option value="b">b</option>
</select>
My code is
<select name="brand" id="brand" class="txtfld" >
<option value="">Select</option>
<?php
$country_sql="SELECT DISTINCT(brand) FROM customer where status='A' and brand<>''";
$result_country=executequery($country_sql);
while($country_array=ms_stripslashes(mysql_fetch_array($result_country)))
{
$sel_con=($country_array['brand']==$_REQUEST['brand'])? " selected='selected'" : " ";
$brand=$country_array['brand'];
$brand2=explode('|',$brand);
if(count($brand2)<2)
{
?><option value="<?=$brand;?>" ><?=$brand;?></option><?php
}
else
{
for($i=0;$i<count($brand2);$i++)
{
?><option value="<?=$brand2[$i];?>" ><?=$brand2[$i];?></option><?php
}
}
}
?>
</select>

Answer 1

You can use array_unique() which will keep only unique values of your array :
$brand2 = explode('|',$brand);
$brand2 = array_unique($brand2);

Answer 2

EDIT
Try :
$new_array = array();
while($country_array=ms_stripslashes(mysql_fetch_array($result_country)))
{
$brand=$country_array['brand'];
$brand2=explode('|',$brand);
$NewArray = array_unique (array_merge ($NewArray, $brand2));
}
Now you have all unique value in array, loop through to show the list.
NOTE: It might be a overhead if you have big data.