How to generate an employee id with year of registration in PHP? - php

Question

I need to generate an employee id for eg: E13001, the number should be auto incremented, and when the next year comes, it should be E14001(year is 2014).It should again start from 001 for every year. I have tried with yy format but the sequence number continues from the previous year. How i suppose to my change the sequence number generation ? help me. Thanks in Advance.

Answer 1

Well, the auto-increment on a column in MySQL will not work except for integers. Also it will increment with each field.
To do what you want, you need to separate the id and year in tow (for the mysql table). So you get a table with id,emp_id(should not be managed by the db, no AUTO-INCREMENT) and year. Then select max(emp_id) where year ='2014', so when adding a new employee you will increment that value. To get the employe id the way you need it you should merge the tow columns into one and dump the first tow chars from the year.
You should know that this then becomes a distributed e_id, and every time you need to work with it you will need to do extra work. But you keep the format you need.
Hope this helps you get an idea of how it can be done.

Answer 2

Hi you can try in the following way
<?php
$connection = mysqli_connect("localhost","root","")
or die("no connection");
$db_select=mysql_select_db("test",$connection)
or die("no connection to db");
$query3 = "SELECT * FROM simple";
$result3 = mysqli_query($query3);
$count = mysql_num_rows($result3);
if($count == 0)
{
$year = date('y');
$seq = 1;
$a = sprintf("%04d", $seq);
$emp_id = E.$year.$a;
$query= "INSERT INTO simple (emp_id) VALUES ('$emp_id')";
$result=mysqli_query($query)or die(mysql_error());
}
else
{
$query2 = "SELECT * FROM simple ORDER BY id DESC LIMIT 1";
$result2 = mysqli_query($query2);
$row = mysql_fetch_array($result2);
$last_id = $row['emp_id'];
$rest = substr("$last_id", -4);
$insert_id = "$rest" + 1;
echo $ars = sprintf("%04d", $insert_id);
$year = date('y');
$emp_id = E.$year.$ars;
$query= "INSERT INTO simple (emp_id) VALUES ('$emp_id')";
$result=mysqli_query($query)or die(mysql_error());
}
Hope this code helps you.