database filled select form issue.

[Anon]

Greetings!

Back again with another one of my brain buster questions.

I am currently propagating a select form drop down list with database driven elements like so:

<?
$sql = "SELECT item FROM types ORDER BY item ASC";
$connection = mysql_connect("localhost", "user", "pass")
or die ("Couldn't connect to server.");
$db = mysql_select_db("db", $connection)
or die ("Couldn't select database.");
$sql_result = mysql_query($sql)
or die("Couldn't execute query.");
if (!$sql_result) {

echo "<P>Could not generate list!";
} else {
echo "
<select name=\"sel_record\">
<option value=\"\"> -- Select Item -- </option>
";
while ($row = mysql_fetch_array($sql_result)) {
$item = $row["item"];
echo "<option value=\"$item\">$item</option>";
}
echo "</select>";
}
?>

While all that seems fine and dandy, if "item" has two of the same things listed on different rows, then a duplicate entry is presented. Is there any way around this to where when the query is returned, it filters out duplicate entries before sending them to the drop down list?

[Anon]

Try:
$sql = "SELECT DISTINCT item FROM types ORDER BY item ASC";

[Anon]

sometimes the easiest of solutions get over looked. I actually knew about distinct, just forgot about using it in this instance. I appreciate your help! It works like a charm now!

[Anon]

Aside from using DISTINCT in your query, you can also solve the problem by taking the array of items and using the array_unique() function, like so:

$array = array_unique($array);