2桌:
Employees
- EmployeeID
- LeadCount
Leads
- leadID
- employeeID
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我想Employees.LeadCount
通过计算Leads
表中具有相同的引导数来更新列EmployeeID
.
注意:可能有超过1个具有相同employeeID的潜在客户,所以我必须做一个DISTINCT(SUM(employeeID))
.
Jon*_*nan 12
UPDATE
Employees E
SET
E.LeadCount = (
SELECT COUNT(L.EmployeeID)
FROM Leads L
WHERE L.EmployeeID = E.EmployeeID
)
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您正在为数据同步问题做好准备.在插入,更新或删除Leads表中的行时,您需要不断更新Employees.LeadCount列.
最好的解决方案是根本不存储LeadCount列,而是在需要值时使用SQL聚合查询重新计算潜在客户数.这样它总是正确的.
SELECT employeeID, COUNT(leadId) AS LeadCount
FROM Leads
GROUP BY employeeID;
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另一种解决方案是在Leads表上为INSERT,UPDATE和DELETE创建触发器,以便始终使Employees.LeadCount列保持最新.例如,使用MySQL触发器语法:
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
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如果您使用MySQL,另一个选择是使用多表UPDATE语法.这是SQL的MySQL扩展,它不能移植到其他品牌的RDBMS.首先,将所有行中的LeadCount重置为零,然后对Leads表进行连接,并在连接生成的每一行中增加LeadCount.
UPDATE Employees SET LeadCount = 0;
UPDATE Employees AS e JOIN Leads AS l USING (employeeID)
SET e.LeadCount = e.LeadCount+1;
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联接对更新(和删除)的作用相同,就像对选择(编辑:在某些流行的RDBMS中,至少*)一样:
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
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SUM(DISTINCT EmployeeId)没有意义-您只需要一个COUNT(*)。
UPDATE Employees SET LeadCount = (
SELECT Distinct(SUM(employeeID)) FROM Leads WHERE Leads.employeeId = Employees.employeeId
)
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