How to group data by year with paid and unpaid values

i have a MYSQL database where i want to sort the totals of both paind and unpaid amounts. The Query i used was :

SELECT DISTINCT 
YEAR( app_payments.created_at ) AS YEARS,
SUM( app_payments.amount ) AS Total,
app_users.serial,
app_payments.`status` AS payment_state 
FROM
app_payments
INNER JOIN app_users ON app_payments.created_by = app_users.serial 
WHERE
app_payments.created_by = 'd88faa' 
GROUP BY
YEAR ( app_payments.created_at ),
app_payments.status

i got the results as:

2017    1995    d88faa  1
2018    1200    d88faa  1
2019    1250    d88faa  0
2019    4990    d88faa  1

Where 1 stands for PAID and 0 stand for UNPAID

in my php code, i tried to group the data into years

$Stats = array ();
while(!$this->EndofSeek()){
$result = $this->Row();
if($result->payment_state == 0 ){
 if(in_array($result->YEARS,$Stats)){
 array_replace($Stats,['y'=>$result->YEARS , 'b'=>$result->Total ]);
}else{ array_push($Stats,['y'=>$result->YEARS , 'a'=>0 , 'b'=>$result->Total ]);}
}else if($result->payment_state == 1){
 array_push($Stats,['y'=>$result->YEARS , 'a'=>$result->Total , 'b'=>0 ]);
}
 }
  return json_encode($Stats)

This returns the output:

[{"y":"2017","a":"1995","b":0},
{"y":"2018","a":"1200","b":0},
{"y":"2019","a":"4990","b":"1450"},
{"y":"2019","a":"4990","b":0}]

Where y is YEARS , a is PAID and b is UNPAID

What i seek to achieve is to group all the data to a particular year where i would have [{"y":"2017","a":"1995","b":0}, {"y":"2018","a":"1200","b":0}, {"y":"2019","a":"4990","b":"1450"}]

Without it duplicating the year but rather merging them into a single unit.

What do i need to do, and which code do i need to implement to achieve this.

Answer

Do you just want conditional aggregation?

SELECT YEAR(p.created_at) AS YEAR,
       SUM( CASE WHEN p.status = 0 THEN p.amount END) AS Total_0,
       SUM( CASE WHEN p.status = 1 THEN p.amount END) AS Total_1
FROM app_payments p INNER JOIN
     app_users u
     ON p.created_by = u.serial 
WHERE p.created_by = 'd88faa' 
GROUP BY YEAR(p.created_at);

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Answer