动态where子句的最简单方法

Question

动态where子句的最简单方法

c#entity-frameworklambdalinq-to-entities

3

在使用EF生成动态where子句时，我感到信息过载。一些解决方案似乎非常陈旧（我正在使用.NET 4.5和EF 5）。

这是我的代码：

public enum PersonTypes
{
    Lazy = 1,
    Awesome = 2,
    SuperHero = 3
}

public bool IncludeLazyPeople { get; set; }
public bool IncludeAwesomePeople { get; set; }
public bool IncludeSuperHeroPeople { get; set; }

使用EF，我需要查询与提供的布尔值匹配的人员类型。

我研究了这个链接: http://www.albahari.com/nutshell/predicatebuilder.aspx

以下是我得出的结果:

// create an expression that would include none of the person types
var personTypeExpression = PredicateBuilder.False<DAL.Models.Person>();

if (IncludeLazyPeople)
    personTypeExpression = personTypeExpression.Or(person => person.PersonType == (int)PersonTypes.Lazy);

if (IncludeAwesomePeople)
    personTypeExpression = personTypeExpression.Or(person => person.PersonType == (int)PersonTypes.Awesome);

if (IncludeSuperHeroPeople)
    personTypeExpression = personTypeExpression.Or(person => person.PersonType == (int)PersonTypes.SuperHero);

// filter the people for all included types
var filteredPeople = ctx.People.Where(personTypeExpression);

由于我没有包含 AsExpandable 扩展，所以那个方法失败了。在尝试之前，我想知道是否有更简单的方法？

- mbursill

2个回答

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- user1914530 · Answer 1

var query = context.People.AsQueryable();

if (IncludeLazyPeople)
    query = query.Where(person => person.PersonType == (int)PersonTypes.Lazy);

if (IncludeAwesomePeople)
    query = query.Where(person => person.PersonType == (int)PersonTypes.Awesome);

if (IncludeSuperHeroPeople)
    query = query.Where(person => person.PersonType == (int)PersonTypes.SuperHero);

如果你只需要 AND 逻辑，那么你可以在查询中调用额外的 Where() 方法。否则，如果你需要更强大的选项，那么就使用谓词构建器或 System.Linq.Expressions。

- TDaver · Answer 2

ParameterExpression p = Expression.Parameter(typeof(Person));
Expression personType = Expression.MakeMemberAccess(p, typeof(Person).GetProperty("PersonType"));
Expression personTypeExpression = Expression.Constant(false);
if (IncludeLazyPeople)
    personTypeExpression = Expression.OrElse(personTypeExpression, Expression.Equal(personType, Expression.Constant((int)PersonTypes.Lazy)));

//... same as above only with different Constant values

// filter the people for all included types
var filteredPeople = ctx.People.Where(Expression.Lambda<Func<Person,bool>>(personTypeExpression,p));

这应该大致符合您的要求。没有在编译器中检查，希望没有太多的拼写错误。