这是我的选择:
SELECT
SUM(t.amount + c.designFeeValue) as cashReceived,
ROUND(SUM(i.value) * (m.percentOurs / 100)) as adValue,
m.managementFee as managementFee,
m.productionCost as productionCost,
5 as emailAddress,
(
(
SELECT value
FROM commission_transactions
WHERE isDebit IS
TRUE
) -
(
SELECT value
FROM commission_transactions
WHERE isDebit IS
FALSE
)
) as miscExpenses,
(managementFee + productionCost + emailAddress + miscExpenses) as totalExpenses
这是轰炸,因为下面的行,我添加了一些别名。
(managementFee + productionCost + emailAddress + miscExpenses) as totalExpenses
别名是未知字段。
有没有办法可以保留此算法的别名,还是我需要重新执行所有生成每个别名的数学运算来计算totalExpenses?这似乎是一种非常难看的方式。
更新:
根据您的建议,我现在使用派生表。
SELECT
cashReceived,
adValue,
managementFee,
productionCost,
emailAddress,
miscExpenses,
adValue + managementFee + productionCost + emailAddress + miscExpenses as totalExpenses
FROM (
SELECT
SUM(t.amount + c.designFeeValue) as cashReceived,
ROUND(SUM(i.value) * (m.percentOurs / 100)) as adValue,
m.managementFee as managementFee,
m.productionCost as productionCost,
5 as emailAddress,
(
(
SELECT value
FROM commission_transactions
WHERE isDebit IS TRUE
) -
(
SELECT value
FROM commission_transactions
WHERE isDebit IS FALSE
)
) as miscExpenses
FROM magazines m
JOIN insertions i ON i.magazineId = m.id
JOIN transactions t ON t.insertionId = i.id
JOIN contracts c ON i.contractId = c.id
JOIN commission_transactions ct ON m.id = ct.magazineId
WHERE m.id = 17
AND t.isChargedBack IS FALSE
AND t.`timestamp` >= '2013-08-01 00:00:00'
AND t.`timestamp` < '2013-09-01 00:00:00'
AND ct.createdDate >= '2013-08-01 00:00:00'
AND ct.createdDate < '2013-09-01 00:00:00'
) sub;
答案 0 :(得分:4)
别名不能在字段列表中重复使用,例如:
mysql> select 5 as five, five + 1 as six;
ERROR 1054 (42S22): Unknown column 'five' in 'field list'
您必须将select包装为另一个,然后在该包装中进行别名数学运算,例如
select *, managementFee + productionCost + emailAddress + miscExpenses) as totalExpenses
FROM (
... your above query here ...
)