Filter results in an array

brett

Hi, I am having a bit of trouble. I have a query that returns state codes from my database in a list dropdown.

<td><select name="state" id="state">
<?php

print "<option selected>All</option>";
for ($y = 0; $y < count($state); $y++)
{
	print "<option>" . $state[$y] . "</option>";
}

?>
</select></td>

The problem I am having is that I have multiple records in the database that are the same so the drop down lists something like CA, UT, UT, CO, HI, CO, ID etc.

How do I filter the array so that only one record is displayed for those that have duplicates? I want to only list one state on those that have multiple entries in the database. So I get something like CA,UT,CO,HI, ID (based on example above)
Thanks
Brett

swon

You can do something like that:

<td><select name="state" id="state">
<?php
$arr = array();
print "<option selected>All</option>";
for ($y = 0; $y < count($state); $y++)
{

if(!in_array($state[$y],$arr))
{
print "<option>" . $state[$y] . "</option>";
array_push($arr,$state[$y]);
}
}

?>
</select></td>

wbriggs

Well... if you're pulling the array from a database, is it a SQL database?

If so, just use SELECT DISTINCT.

If not, you can use the array_unique function.

-Will

brett

MySQL

wbriggs

If you're using MySQL, then you should definitely just use a SELECT DISTINCT statement.

However, if you've already gone to all the trouble of building the array you've got now, the easiest and quickest fix is to just make a call to the array_unique function.