I have got the following form:
<div id="Quick_find_container">
<form action="" method="post" target="_self">
<div id="qcategory_1">Product</div>
<div id="qcategory">
<select name="Category" class="dropmenu" id="Category"onchange="this.form.submit()">
<option value="">Any</option>
<option value="Keyboard"<?php if ($_POST['Category']=="Keyboard") {echo "selected='selected'"; } ?>>Keyboard</option>
<option value="Piano"<?php if ($_POST['Category']=="Piano") {echo "selected='selected'"; } ?>>Piano</option>
</select>
</div>
<div id="qbrand_1">Brand</div>
<div id="qbrand"><select name='Manufacturer' onchange="this.form.submit()">
<?php
echo '<option value="">Any</option>';
$value = $_POST['Manufacturer'];
while ($row = mysql_fetch_array($RS_Search)) {
echo "<option value=" . $row['Manufacturer'] . " ' . (selected == $value ? ' selected' : '') . '>" . $row['Manufacturer'] . "</option>";
}
?>
</select>
</div>
<div id="qsubmit">
<input name="Search2" type="submit" id="Search2" value="Submit">
</div>
</form>
</div>
<?php echo $_POST['Category']; ?>
<?php echo $_POST['Manufacturer']; ?>
Echo post Category and Manufacturer are purely to see what it is submitting.
My problem is the second drop down menu. I would like to display after something is selected what the selected value was. At the moment it just jumps back to the default value Any even though the out put of POST_[Manufacturer'] is correct. Is there a way to display the selected value like in the first drop down menu? I still would like to keep the values to choose from the database as well. But just display the selected value.
Any help welcome