如何撰写现有的Linq表达式

Question

我想编写两个Linq表达式的结果.它们以形式存在

Expression<Func<T, bool>>

所以我要编写的两个本质上是一个参数(类型为T)的委托,它们都返回一个布尔值.我想要的结果是对布尔值的逻辑评价.我可能会将它作为扩展方法实现,所以我的语法将是这样的:

Expression<Func<User, bool>> expression1 = t => t.Name == "steve";
Expression<Func<User, bool>> expression2 = t => t.Age == 28;
Expression<Func<User, bool>> composedExpression = expression1.And(expression2);

后来在我的代码中我想评估组合表达式

var user = new User();
bool evaluated = composedExpression.Compile().Invoke(user);

我用了一些不同的想法,但我担心它比我希望的更复杂.这是怎么做到的？

Answer 1

这是一个例子:

var user1 = new User {Name = "steve", Age = 28};
var user2 = new User {Name = "foobar", Age = 28};

Expression<Func<User, bool>> expression1 = t => t.Name == "steve";
Expression<Func<User, bool>> expression2 = t => t.Age == 28;

var invokedExpression = Expression.Invoke(expression2, expression1.Parameters.Cast<Expression>());

var result = Expression.Lambda<Func<User, bool>>(Expression.And(expression1.Body, invokedExpression), expression1.Parameters);

Console.WriteLine(result.Compile().Invoke(user1)); // true
Console.WriteLine(result.Compile().Invoke(user2)); // false

您可以通过扩展方法重用此代码:

class User
{
  public string Name { get; set; }
  public int Age { get; set; }
}

public static class PredicateExtensions
{
  public static Expression<Func<T, bool>> And<T>(this Expression<Func<T, bool>> expression1,Expression<Func<T, bool>> expression2)
  {
    InvocationExpression invokedExpression = Expression.Invoke(expression2, expression1.Parameters.Cast<Expression>());

    return Expression.Lambda<Func<T, bool>>(Expression.And(expression1.Body, invokedExpression), expression1.Parameters);
  }
}

class Program
{
  static void Main(string[] args)
  {
    var user1 = new User {Name = "steve", Age = 28};
    var user2 = new User {Name = "foobar", Age = 28};

    Expression<Func<User, bool>> expression1 = t => t.Name == "steve";
    Expression<Func<User, bool>> expression2 = t => t.Age == 28;

    var result = expression1.And(expression2);

    Console.WriteLine(result.Compile().Invoke(user1));
    Console.WriteLine(result.Compile().Invoke(user2));
  }
}