Skips the first repeated Value and fetch the 2nd Value - PHP - php

Question

PHP and MYSQLi
I am trying to get the below output: for a Project ID (PJ006)
I put the "date" column in Descending order first and then i want to skip all the Repeated value of the first value (Skip all entries of EM207) and display the immedate 2nd value (Output expected: EM206). This output is needed against the for a Project ID (PJ006)
Database:
SL No. Badge ID Date ProjectID
1 EM207 2015-05-12 00:00:00 PJ006
2 EM207 2015-05-11 00:00:00 PJ006
3 EM207 2015-05-09 00:00:00 PJ006
4 EM207 2015-05-09 00:00:00 PJ001
5 EM206 2015-05-08 00:00:00 PJ001
6 EM206 2015-09-05 00:00:00 PJ006
7 EM206 2015-09-03 00:00:00 PJ006
8 EM205 2015-09-01 00:00:00 PJ006
Expected output: For Project ID: PJ006,
Badge ID: EM206
PHP Code for: Using the First Badge ID (EM207), fetch the data of the Project ID PJ006 and display
$mysqli=mysqli_connect('serverIP','username','password','databasename');
//This data i passed by Ajax
$someID = trim($_POST["storedValue"]);
//This query puts the date column in Desending Order
$query2 ="SELECT * FROM scores1 WHERE ProjectID='$someID' order by ScoreDate desc limit 1";
$result = mysqli_query($mysqli,$query2)or die(mysqli_error());
$num_row = mysqli_num_rows($result);
while($row=mysqli_fetch_array($result))
{
//the output is displayed here
echo $row[1];
}
What modification is required in the above code "to skip all the Repeated value of the first value (Skip all entries of EM207) and display the immediate 2nd value (Output expected: EM206)". This output is needed against the for a Project ID (PJ006)

Answer 1

$query2 ="SELECT * FROM scores1 WHERE ProjectID='$someID' order by ScoreDate desc limit 1";
$result = mysqli_query($mysqli,$query2)or die(mysqli_error());
$num_row = mysqli_num_rows($result);
$badgeid=0; $proid=0;
while($row= $database->fetch_array($result))
{
if($badgeid!=$row['batchid'] || $proid!=$row['proid']; ){
echo "$badgeid and $proid "
$b=$row['batchid'];
$p=$row['proid'];
}
}

Answer 2

In this answer i am thanking Mr.Thamaraiselvam for helping me with the exact required output. Thanks You.
The Solution which worked:
$mysqli=mysqli_connect('serverIP','username','password','databasename');
$query2 ="SELECT * FROM scores1 where ProjectID='12345' ORDER BY ScoreDate DESC";
$result = mysqli_query($mysqli,$query2)or die(mysqli_error());
$num_row = mysqli_num_rows($result);
$badgeid=0; $proid=0 ; $date=0;
$flag=0;
while($row=mysqli_fetch_array($result))
{
if($badgeid!=$row['BadgeID'] || $proid!=$row['ProjectID'] ){
if($flag==2){
echo "$badgeid and $date<br>";
}
$flag++;
$badgeid =$row['BadgeID'];
$proid =$row['ProjectID'];
$date =$row['ScoreDate'];
}
}
?>

Answer 3

Okay, as discussed, what you're actually after is the most recent badge id for a particular project - the fact that this is the second entry is a bit of a red herring and may not hold true in other scenarios.
I'd use a subquery to find the most recent entries for each project:
select
badgeId,
projectId
from
scores1,
(
select
max(date) as max,
projectId
from
scores1
group by
projectId
) as sub
where
scores1.date = sub.max
and scores1.projectId = sub.projectId
You can then add a check for whichever project id you're interested in. Don't forget to add some indexes for better performance.