[转]SQL,LINQ,Lambda语法对照图
如果你熟悉SQL语句,当使用LINQ时,会有似曾相识的感觉.但又略有不同.下面是SQL和LINQ,Lambda语法对照图
SELECT * FROM HumanResources.Employee |
from e in Employees select e |
Employees |
SELECT e.LoginID, e.JobTitle FROM HumanResources.Employee AS e |
from e in Employees select new {e.LoginID, e.JobTitle} |
Employees |
SELECT e.LoginID AS ID, e.JobTitle AS Title FROM HumanResources.Employee AS e |
from e in Employees select new {ID = e.LoginID, Title = e.JobTitle} |
Employees |
SELECT DISTINCT e.JobTitle FROM HumanResources.Employee AS e |
(from e in Employees select e.JobTitle).Distinct() |
Employees |
SELECT e.* FROM HumanResources.Employee AS e WHERE e.LoginID = 'test' |
from e in Employees where e.LoginID == "test" select e |
Employees |
SELECT e.* FROM HumanResources.Employee AS e WHERE e.LoginID = 'test' AND e.SalariedFlag = 1 |
from e in Employees where e.LoginID == "test" && e.SalariedFlag select e |
Employees |
SELECT e.* WHERE e.VacationHours >= 2 AND e.VacationHours <= 10 |
from e in Employees where e.VacationHours >= 2 && e.VacationHours <= 10 select e |
Employees |
SELECT e.* FROM HumanResources.Employee AS e |
from e in Employees orderby e.NationalIDNumber select e |
Employees |
SELECT e.* FROM HumanResources.Employee AS e ORDER BY e.HireDate DESC, e.NationalIDNumber |
from e in Employees orderby e.HireDate descending, e.NationalIDNumber select e |
Employees |
SELECT e.* WHERE e.JobTitle LIKE 'Vice%' OR SUBSTRING(e.JobTitle, 0, 3) = 'Pro' |
from e in Employees where e.JobTitle.StartsWith("Vice") || e.JobTitle.Substring(0, 3) == "Pro" select e |
Employees |
SELECT SUM(e.VacationHours) FROM HumanResources.Employee AS e |
Employees.Sum(e => e.VacationHours); |
|
SELECT COUNT(*) FROM HumanResources.Employee AS e |
Employees.Count(); |
|
SELECT SUM(e.VacationHours) AS TotalVacations, e.JobTitle FROM HumanResources.Employee AS e GROUP BY e.JobTitle |
from e in Employees group e by e.JobTitle into g select new {JobTitle = g.Key, TotalVacations = g.Sum(e => e.VacationHours)} |
Employees |
SELECT e.JobTitle, SUM(e.VacationHours) AS TotalVacations FROM HumanResources.Employee AS e GROUP BY e.JobTitle HAVING e.COUNT(*) > 2 |
from e in Employees group e by e.JobTitle into g where g.Count() > 2 select new {JobTitle = g.Key, TotalVacations = g.Sum(e => e.VacationHours)} |
Employees |
SELECT * FROM Production.Product AS p, Production.ProductReview AS pr |
from p in Products from pr in ProductReviews select new {p, pr} |
Products |
SELECT * FROM Production.Product AS p INNER JOIN Production.ProductReview AS pr ON p.ProductID = pr.ProductID |
from p in Products join pr in ProductReviews on p.ProductID equals pr.ProductID select new {p, pr} |
Products |
SELECT * FROM Production.Product AS p INNER JOIN Production.ProductCostHistory AS pch ON p.ProductID = pch.ProductID AND p.SellStartDate = pch.StartDate |
from p in Products join pch in ProductCostHistories on new {p.ProductID, StartDate = p.SellStartDate} equals new {pch.ProductID, StartDate = pch.StartDate} select new {p, pch} |
Products |
SELECT * FROM Production.Product AS p LEFT OUTER JOIN Production.ProductReview AS pr ON p.ProductID = pr.ProductID |
from p in Products join pr in ProductReviews on p.ProductID equals pr.ProductID into prodrev select new {p, prodrev} |
Products |
SELECT p.ProductID AS ID FROM Production.Product AS p UNION SELECT pr.ProductReviewID FROM Production.ProductReview AS pr |
(from p in Products select new {ID = p.ProductID}).Union( from pr in ProductReviews select new {ID = pr.ProductReviewID}) |
Products |
SELECT TOP (10) * FROM Production.Product AS p WHERE p.StandardCost < 100 |
(from p in Products where p.StandardCost < 100 select p).Take(10) |
Products |
SELECT * FROM [Production].[Product] AS p WHERE p.ProductID IN( SELECT pr.ProductID FROM [Production].[ProductReview] AS [pr] WHERE pr.[Rating] = 5 ) |
from p in Products where (from pr in ProductReviews where pr.Rating == 5 select pr.ProductID).Contains(p.ProductID) select p |
Products |
转自:http://hi.baidu.com/ren_long_/blog/item/4c574455a37e570e3b293518.html
转载于:https://www.cnblogs.com/rohan/archive/2011/01/18/1937925.html
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