How to update data when input is drop down? - php

Question

I am so new in html, php, javascript, mysql. i created a drop down menu using java script that contains a list of 2011 to 2510. During update i can not display the stored value that is generated using java script. I searched but found nothing. A portion of code.....
<select name="eca">
<option id="eca" required="required" class="ecadetail" value="NSS" <?php if ($eca == 'NSS') echo 'selected="selected"';?>">NSS</option>
<option id="eca" required="required" class="ecadetail" value="NCC" <?php if ($eca == 'NCC') echo 'selected="selected"';?>">NCC</option>
<option id="eca" required="required" class="ecadetail" value="Cultural" <?php if ($eca == 'Cultural') echo 'selected="selected"';?>">Cultural</option>
</select>
</div>
<div class="label">Year</div>
<div class="inputyear">
<select name="years" >
<script language="JavaScript">
// loop to create the list
var year = 2010
for (var i=1; i <=500; i++)
{
year++;
document.write("<option>" + year + "</option>");
}
// end JS code hide -->
<option value="<?php if ($year == 'year') echo 'selected="selected"';?>"></script>
</script>
</select>
</div>
</div> <!-- end of 7th row -->
the value for eca is working fine. The value year is stored in $year.
help please...

Answer 1

This loop can purely done in php then why involving the javascript
<select name="years" >
<?php
$matchyear = 2010;
for ($i=2010; $i <=2510; $i++)
{ ?>
<option value="<?php echo $i;?>" <?php if ($i== $matchyear) echo 'selected="selected"';?>><?php echo $i;?> </option>
<?php }
?>
</select>
Make things easier for you not a complex programming so that other person can understand easily and call you by good names