How to put the id instead of year? - php

Question

Everyone please I really need help, I have a problem on how to put id's under the current_year. For an example in the image that's my data structure sample image. What are the ways to save it's id not the current year I am using $YearNow=Date('Y'); for the current year, I would something that if the current year is 2015 after if I submit it to the database, it will save the id under the current year like 2000
=======================
if the year of my pc is 2016 then 2016 = 2001(syearid). In my school_year table I have (syearid(pri), from_year, to_year). And my studentvotes(studeid(pri)autoincremtn ,candid,idno,syearid(foreign key)). Therefore if the year of my pc is 2016 as you can see that2016(from_year)is under of thesyearid 2001`
here's my code:
<?php
require_once('auth.php');
include('connection/connect.php');
$idnum=$_POST['idnum'];
$stat='voted';
$sqla = "UPDATE student
SET status=?
WHERE idno=?";
$qa = $db->prepare($sqla);
$qa->execute(array($stat,$idnum));
$edittable=$_POST['votes'];
$a=1;
$N = count($edittable);
$YearNow=Date('Y');
for($i=0; $i < $N; $i++)
{
$sql = "UPDATE candidates
SET votes=votes+?
WHERE candid =? ";
//don't know it's correct?
$years = array();
$sq12="select * from studentvotes,school_year where studentvotes.syearid = school_year.syearid AND school_year.from_year = $YearNow";
$result1= mysql_fetch_assoc($sq12);
for($i=0; $row = $result->fetch(); $i++){
$years[$row['syearid']] = $row['from_year'];
}
$q = $db->prepare($sql);
$q->execute(array($a,$edittable[$i]));
//$this part will have something to update
$sqlas =
"INSERT INTO studentvotes(candid,idno,syearid) VALUES (:m,:n,:o)";
$qs = $db->prepare($sqlas);
$qs->execute(array(':m'=>$edittable[$i],':n'=>$idnum,':o'=>$YearNow ));
}
header("location: notification.php?". http_build_query($query));
mysql_close($con);
?>

Answer 1

You are mixing the mysql and mysqli extension together and also using the wrong variable $result1. You can use like that for getting year id.
$sq12="select studentvotes.syearid from studentvotes,school_year where studentvotes.syearid = school_year.syearid AND school_year.from_year = $YearNow";
$result = $db->prepare($sq12);
$result->execute();
$result_final = $result->fetchAll(PDO::FETCH_COLUMN, 0);
$yearid = $result_final[0]; // get year id
Now you can insert it in your Query:
$sqlas = "INSERT INTO studentvotes(candid,idno,syearid) VALUES (:m,:n,:o)";
$qs = $db->prepare($sqlas);
$qs->execute(array(':m'=>$edittable[$i],':n'=>$idnum,':o'=>$yearid));