如何按年份将数据与付费和未付费值分组

Question

我有一个 MYSQL 数据库，我想在其中对已付金额和未付金额的总数进行排序。我使用的查询是：

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 

我得到的结果是：

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

其中1 代表PAID，0 代表UNPAID

在我的 php 代码中，我尝试将数据分组为年份

$Stats = array ();
while(!$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)

这将返回输出：

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

y 是 YEARS ，a 是 PAID 和 b 是 UNPAID

我寻求实现的是将所有数据分组到我将拥有的特定年份 [{"y":"2017","a":"1995","b":0}, {"y":"2018","a":"1200","b":0}, {"y":"2019","a":"4990","b":"1450"}]

没有复制年份，而是将它们合并为一个单元。

我需要做什么，我需要实现哪些代码来实现这一点。

Answer 1

你只是想要条件聚合吗？

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);