Employees
- EmployeeID
- LeadCount
Leads
- leadID
- employeeID
我想通過計數Leads
表具有相同EmployeeID
引線#更新Employees.LeadCount
列。
注意:可能有超過1名領導具有相同的employeeID,所以我必須做一個DISTINCT(SUM(employeeID))
。
Employees
- EmployeeID
- LeadCount
Leads
- leadID
- employeeID
我想通過計數Leads
表具有相同EmployeeID
引線#更新Employees.LeadCount
列。
注意:可能有超過1名領導具有相同的employeeID,所以我必須做一個DISTINCT(SUM(employeeID))
。
UPDATE Employees SET LeadCount = (
SELECT Distinct(SUM(employeeID)) FROM Leads WHERE Leads.employeeId = Employees.employeeId
)
UPDATE
Employees E
SET
E.LeadCount = (
SELECT COUNT(L.EmployeeID)
FROM Leads L
WHERE L.EmployeeID = E.EmployeeID
)
從上方鋼化和除去從屬子查詢。
// create tmp -> TBL (EmpID, count)
insert into TBL
SELECT employeeID COUNT(employeeID) Di
FROM Leads WHERE Leads.employeeId = Employees.employeeId GROUP BY EmployeeId
UPDATE Employees SET LeadCount = (
SELECT count FROM TBL WHERE TBL.EmpID = Employees.employeeId
)
// drop TBL
編輯它的「分組依據」不「不同」:B(感謝馬克·布拉克特)
連接的工作原理相同的更新(和刪除),就像他們的選擇(編輯做的:在一些流行的RDBMS」,至少*):
UPDATE Employees SET
LeadCount = Leads.LeadCount
FROM Employee
JOIN (
SELECT EmployeeId, COUNT(*) as LeadCount
FROM Leads
GROUP BY EmployeeId
) as Leads ON
Employee.EmployeeId = Leads.EmployeeId
的SUM(DISTINCT僱員)是沒有意義的 - 你只需要一個COUNT(*)。
你有鏈接嗎?我想了解更多。 – BCS 2008-09-26 00:31:03
您正在設置自己的數據同步問題。隨着Leads表中的行被插入,更新或刪除,您需要不斷更新Employees.LeadCount列。
最好的解決方案不是存儲LeadCount列,而是根據需要重新計算帶有SQL聚合查詢的銷售線索數。這樣它總是正確的。
SELECT employeeID, COUNT(leadId) AS LeadCount
FROM Leads
GROUP BY employeeID;
另一個解決辦法是對信息表的INSERT,UPDATE創建觸發器和DELETE,讓你保持Employees.LeadCount列當前所有的時間。例如,使用MySQL觸發語法:
CREATE TRIGGER leadIns AFTER INSERT ON Leads
FOR EACH ROW BEGIN
UPDATE Employees SET LeadCount = LeadCount + 1 WHERE employeeID = NEW.employeeID;
END
CREATE TRIGGER leadIns AFTER UPDATE ON Leads
FOR EACH ROW BEGIN
UPDATE Employees SET LeadCount = LeadCount - 1 WHERE employeeID = OLD.employeeID;
UPDATE Employees SET LeadCount = LeadCount + 1 WHERE employeeID = NEW.employeeID;
END
CREATE TRIGGER leadIns AFTER DELETE ON Leads
FOR EACH ROW BEGIN
UPDATE Employees SET LeadCount = LeadCount - 1 WHERE employeeID = OLD.employeeID;
END
如果您使用的是MySQL,另一種選擇是使用多表UPDATE語法。這是SQL的MySQL擴展,它不能移植到其他品牌的RDBMS。首先,將所有行中的LeadCount重置爲零,然後對Leads表執行聯接,並在聯接所產生的每行中增加LeadCount。
UPDATE Employees SET LeadCount = 0;
UPDATE Employees AS e JOIN Leads AS l USING (employeeID)
SET e.LeadCount = e.LeadCount+1;
sum(employeeid)沒有任何意義,並且圍繞單個值的不同將始終返回相同的值,因此它是多餘的。 – 2008-09-26 00:08:36
IIRC會很慢(MySQL是我用過的) – BCS 2008-09-26 00:13:35