如何按已付费和未付费值对数据进行分组
How to group data by year with paid and unpaid values
我有一个 MYSQL 数据库,我想在其中对已支付金额和未支付金额的总计进行排序。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)
这个returns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}]
其中 y
是 YEARS
,a
是 PAID
并且 b
是 UNPAID
我想要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"}]
没有重复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才能实现。
您只需要条件聚合吗?
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);
我有一个 MYSQL 数据库,我想在其中对已支付金额和未支付金额的总计进行排序。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)
这个returns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}]
其中 y
是 YEARS
,a
是 PAID
并且 b
是 UNPAID
我想要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"}]
没有重复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才能实现。
您只需要条件聚合吗?
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);