mysql dropdown list when selected - php

Question

I have a table with this structure and trying to use a dropdown to select a language from the list and then display the contents of respective language_id .. I am using mysql and php 5.4
My code is as follows, got stuck with looping and not sure how to get it
$sql=mysql_query("SELECT lang_id, lang_desc FROM languages ");
if(mysql_num_rows($sql))
{
$select= '<select lang_desc="select">';
while($rows=mysql_fetch_array($sql))
{
$select.='<option value="'.$rows['lang_id'].'">'.$rows['lang_desc'].'</option>';
}
}
$select.='</select>';
echo $select ;
$result = mysql_query("SELECT * FROM contents LEFT JOIN languages ON contents.lang_id = languages.lang_id WHERE contents.page_name = 'index' AND contents.lang_id = '$select'");
while ($row = mysql_fetch_array($result))
{
}
Please help

Answer 1

I think this want minimum 2 PHP pages 1st for show data and 2nd for creat data in hear I didn't create DB connection please add it
in 1st I use jquery for getting data from 2nd PHP
show.php
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
<select id="lang" onchange="setlang(this.value)">
<?PHP
$sql=mysql_query("SELECT lang_id, lang_desc FROM languages");
while($rows=mysql_fetch_array($sqL,MYSQL_ASSOC)){
echo '<option value="'.$rows['lang_id'].'">'.$rows['lang_desc'].'</option>'
}
?>
</select>
<div id='dis'></div>
<script type="text/javascript">
$( document ).ready(function() {
setlang($('#lang').val());
});
function setlang(val) {
$.ajax({
url: "data.php",
type: "POST",
data:{val:val},
success: function(result){
$("#dis").html(result);
}
});
}
</script>
data.php
<?php
$result = mysql_query("SELECT * FROM contents LEFT JOIN languages ON contents.lang_id = languages.lang_id WHERE contents.page_name = 'index' AND contents.lang_id = '".$_POST['val']."'");
$data="";
while ($row = mysql_fetch_array($result))
$data.="your row data for dis";
//
}
echo $data;
?>