how to display the two different columns from the selected year - php

Question

I do not know how will I display the position and names of the officers from year selected by the user.
Edit:
I used ajax now, but I'm still having a problem. When I click the combobox and select a year, the officers still don't show up. Only the table with the header Position and Name that shows up, but no data from my database under those columns.
getyear.php
<?php
$q = strval($_GET['q']);
$con = mysqli_connect('localhost','root','','test');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"ajax_demo");
$sql="SELECT * FROM officers WHERE year = '".$q."'";
$result = mysqli_query($con,$sql);
echo "<table>
<tr>
<th>Position</th>
<th>Name</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['position'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
main.php
<form>
<?php
mysql_connect("localhost","root","");
mysql_select_db("test");
$sql = "SELECT DISTINCT year FROM officers ORDER BY year DESC";
$result = mysql_query($sql);
/* assign an onchange event handler */
echo "<select name='year' onchange='showofficers(this.value)'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['year'] ."'>" . $row['year'];
}
echo "</select> <br>";
?>
<div id="txtHint">
</div>
</form>
<script>
/* event handler ~ no ajax function shown */
function showofficers(str){
if (str == "") {
document.getElementById("txtHint").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
}
};
xmlhttp.open("GET","getyear.php?q="+str,true);
xmlhttp.send();
}
}
</script>

Answer 1

some pseudo-code o give you an idea of how you could achieve the desired goal.
<?php
mysql_connect("localhost","root","");
mysql_select_db("test");
$sql = "SELECT DISTINCT year FROM officers ORDER BY year DESC";
$result = mysql_query($sql);
/* assign an onchange event handler */
echo "<select name='year' onchange='showofficers(this.value)'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['year'] ."'>" . $row['year'];
}
echo "</select> <br>";
?>
<script>
/* event handler ~ no ajax function shown */
function showofficers( value ){
/*
use ajax to send a request that fetches the officers details
based upon the year selected. Preferred method=POST for ajax query
*/
alert( 'send '+value+' via ajax, build the sql query and use the ajax callback to generate the new html content' );
}
</script>
<?php
if( $_SERVER['REQUEST_METHOD']=='POST' ){
/* Intercept and process ajax request */
/* the year is POSTed by ajax */
$year = $_POST['year'];
$sql='select * from table where year='.$year;
$res=$db->query( $sql );
if( $res ){
/* process recordset and send back response */
}
}
?>