[英]MySQL join same table twice on same column with different value returning most recent row only
我被困於嘗試解決否則是復雜的JOIN的一小部分。
我們有一個“指令”表和一個“估計”表。 在“估計”中,對於給定指令,我們有多行用於不同類型的estimates
。
說明表
id | address | status
1 | 27 TAYLOR ROAD, ALBION PARK NSW 2527 | InProgress
估算表
id | instruction_id | basis | basis_date | basis_value
1 | 1 | ContractPrice | 2012-04-05 | 124000
2 | 1 | CAMV | 2012-02-01 | 120000
3 | 1 | CustomerEstimate | 2012-06-07 | 132000
4 | 1 | ContractPrice | 2013-01-03 | 140000
5 | 1 | CustomerEstimate | 2013-02-09 | 145000
我們想要的實際上是基於structions.id = Estimates.instruction_id和estimate.basis的“ estimates”上的“ instructions”的2個聯接,用於1)最新的“ CustomerEstimate”(別名為basic_date和basic_value的estimate_date和estimate_value)和2)最近的“合同價格”(再次,將base_date和base_value別名為contact_date和contract_value)。
預期結果如下:
id | address | status | contract_price | contract_date | estimate_date | estimate_value
1 | 27 TAYLOR ROAD, ALBION PARK NSW 2527 | InProgress | 2013-01-03 | 140000 | 2013-02-09 | 145000
我真的很感謝來自SQL專家的幫助。
非常感謝,特倫特。
嘗試
SELECT i.id,
i.address,
i.status,
p.max_date contract_date,
p.basis_value contract_price,
e.max_date estimate_date,
e.basis_value estimate_value
FROM Instructions i LEFT JOIN
(
SELECT q1.instruction_id, max_date, basis_value
FROM Estimates e JOIN
(
SELECT instruction_id, MAX(basis_date) max_date
FROM Estimates
WHERE basis = 'CustomerEstimate'
GROUP BY instruction_id
) q1 ON e.instruction_id = q1.instruction_id AND e.basis_date = q1.max_date
) e ON i.id = e.instruction_id LEFT JOIN
(
SELECT q2.instruction_id, max_date, basis_value
FROM Estimates e JOIN
(
SELECT instruction_id, MAX(basis_date) max_date
FROM Estimates
WHERE basis = 'ContractPrice'
GROUP BY instruction_id
) q2 ON e.instruction_id = q2.instruction_id AND e.basis_date = q2.max_date
) p ON i.id = p.instruction_id
輸出:
| ID | ADDRESS | STATUS | CONTRACT_PRICE | CONTRACT_DATE | ESTIMATE_VALUE | ESTIMATE_DATE | ---------------------------------------------------------------------------------------------------------------------------- | 1 | 27 TAYLOR ROAD, ALBION PARK NSW 2527 | InProgress | 140000 | 2013-01-03 | 145000 | 2013-02-09 |
這是SQLFiddle演示。
這里的合同價格是多少? 您可以嘗試以下
select inst.id
, inst.address
, inst.status
, est.basis_value as estimate_value
, est.basis_date as estimate_date
from instructions inst
, estimates est
where inst.id=est.instruction_id
and (est.basis='CustomerEstimate' or est.basis='ContractPrice')
order
by est.basis
, est.basis_date desc;
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