我有两个表达式的类型表达式<Func<T, bool>>和我想采取或，与或不是这些，并获得一个相同类型的新表达式
Expression<Func<T, bool>> expr1;
Expression<Func<T, bool>> expr2;
...
//how to do this (the code below will obviously not work)
Expression<Func<T, bool>> andExpression = expr AND expr2
当前回答
如果您的提供程序不支持Invoke，而您需要组合两个表达式，则可以使用ExpressionVisitor将第二个表达式中的参数替换为第一个表达式中的参数。
class ParameterUpdateVisitor : ExpressionVisitor
{
private ParameterExpression _oldParameter;
private ParameterExpression _newParameter;
public ParameterUpdateVisitor(ParameterExpression oldParameter, ParameterExpression newParameter)
{
_oldParameter = oldParameter;
_newParameter = newParameter;
}
protected override Expression VisitParameter(ParameterExpression node)
{
if (object.ReferenceEquals(node, _oldParameter))
return _newParameter;
return base.VisitParameter(node);
}
}
static Expression<Func<T, bool>> UpdateParameter<T>(
Expression<Func<T, bool>> expr,
ParameterExpression newParameter)
{
var visitor = new ParameterUpdateVisitor(expr.Parameters[0], newParameter);
var body = visitor.Visit(expr.Body);
return Expression.Lambda<Func<T, bool>>(body, newParameter);
}
[TestMethod]
public void ExpressionText()
{
string text = "test";
Expression<Func<Coco, bool>> expr1 = p => p.Item1.Contains(text);
Expression<Func<Coco, bool>> expr2 = q => q.Item2.Contains(text);
Expression<Func<Coco, bool>> expr3 = UpdateParameter(expr2, expr1.Parameters[0]);
var expr4 = Expression.Lambda<Func<Recording, bool>>(
Expression.OrElse(expr1.Body, expr3.Body), expr1.Parameters[0]);
var func = expr4.Compile();
Assert.IsTrue(func(new Coco { Item1 = "caca", Item2 = "test pipi" }));
}
其他回答
我需要实现相同的结果，但是使用一些更通用的东西(因为不知道类型)。多亏了marc的回答，我终于明白了我想要达到的目标:
public static LambdaExpression CombineOr(Type sourceType, LambdaExpression exp, LambdaExpression newExp)
{
var parameter = Expression.Parameter(sourceType);
var leftVisitor = new ReplaceExpressionVisitor(exp.Parameters[0], parameter);
var left = leftVisitor.Visit(exp.Body);
var rightVisitor = new ReplaceExpressionVisitor(newExp.Parameters[0], parameter);
var right = rightVisitor.Visit(newExp.Body);
var delegateType = typeof(Func<,>).MakeGenericType(sourceType, typeof(bool));
return Expression.Lambda(delegateType, Expression.Or(left, right), parameter);
}
我觉得这样挺好的，不是吗?
Func<T, bool> expr1 = (x => x.Att1 == "a");
Func<T, bool> expr2 = (x => x.Att2 == "b");
Func<T, bool> expr1ANDexpr2 = (x => expr1(x) && expr2(x));
Func<T, bool> expr1ORexpr2 = (x => expr1(x) || expr2(x));
Func<T, bool> NOTexpr1 = (x => !expr1(x));
如果您的提供程序不支持Invoke，而您需要组合两个表达式，则可以使用ExpressionVisitor将第二个表达式中的参数替换为第一个表达式中的参数。
class ParameterUpdateVisitor : ExpressionVisitor
{
private ParameterExpression _oldParameter;
private ParameterExpression _newParameter;
public ParameterUpdateVisitor(ParameterExpression oldParameter, ParameterExpression newParameter)
{
_oldParameter = oldParameter;
_newParameter = newParameter;
}
protected override Expression VisitParameter(ParameterExpression node)
{
if (object.ReferenceEquals(node, _oldParameter))
return _newParameter;
return base.VisitParameter(node);
}
}
static Expression<Func<T, bool>> UpdateParameter<T>(
Expression<Func<T, bool>> expr,
ParameterExpression newParameter)
{
var visitor = new ParameterUpdateVisitor(expr.Parameters[0], newParameter);
var body = visitor.Visit(expr.Body);
return Expression.Lambda<Func<T, bool>>(body, newParameter);
}
[TestMethod]
public void ExpressionText()
{
string text = "test";
Expression<Func<Coco, bool>> expr1 = p => p.Item1.Contains(text);
Expression<Func<Coco, bool>> expr2 = q => q.Item2.Contains(text);
Expression<Func<Coco, bool>> expr3 = UpdateParameter(expr2, expr1.Parameters[0]);
var expr4 = Expression.Lambda<Func<Recording, bool>>(
Expression.OrElse(expr1.Body, expr3.Body), expr1.Parameters[0]);
var func = expr4.Compile();
Assert.IsTrue(func(new Coco { Item1 = "caca", Item2 = "test pipi" }));
}
你可以用表达式。和also / OrElse等结合逻辑表达式，但问题是参数;你在expr1和expr2中使用相同的参数表达式吗?如果是这样，那么:
var body = Expression.AndAlso(expr1.Body, expr2.Body);
var lambda = Expression.Lambda<Func<T,bool>>(body, expr1.Parameters[0]);
这也适用于对单个操作求反:
static Expression<Func<T, bool>> Not<T>(
this Expression<Func<T, bool>> expr)
{
return Expression.Lambda<Func<T, bool>>(
Expression.Not(expr.Body), expr.Parameters[0]);
}
否则，取决于LINQ提供程序，你可以将它们与Invoke结合使用:
// OrElse is very similar...
static Expression<Func<T, bool>> AndAlso<T>(
this Expression<Func<T, bool>> left,
Expression<Func<T, bool>> right)
{
var param = Expression.Parameter(typeof(T), "x");
var body = Expression.AndAlso(
Expression.Invoke(left, param),
Expression.Invoke(right, param)
);
var lambda = Expression.Lambda<Func<T, bool>>(body, param);
return lambda;
}
在某个地方，我有一些代码重写了一个表达式树，替换节点以消除对Invoke的需求，但它相当长(我不记得我把它放在哪里了……)
选择最简单路线的通用版本:
static Expression<Func<T, bool>> AndAlso<T>(
this Expression<Func<T, bool>> expr1,
Expression<Func<T, bool>> expr2)
{
// need to detect whether they use the same
// parameter instance; if not, they need fixing
ParameterExpression param = expr1.Parameters[0];
if (ReferenceEquals(param, expr2.Parameters[0]))
{
// simple version
return Expression.Lambda<Func<T, bool>>(
Expression.AndAlso(expr1.Body, expr2.Body), param);
}
// otherwise, keep expr1 "as is" and invoke expr2
return Expression.Lambda<Func<T, bool>>(
Expression.AndAlso(
expr1.Body,
Expression.Invoke(expr2, param)), param);
}
从。net 4.0开始，有了ExpressionVisitor类，它允许你构建EF安全的表达式。
public static Expression<Func<T, bool>> AndAlso<T>(
this Expression<Func<T, bool>> expr1,
Expression<Func<T, bool>> expr2)
{
var parameter = Expression.Parameter(typeof (T));
var leftVisitor = new ReplaceExpressionVisitor(expr1.Parameters[0], parameter);
var left = leftVisitor.Visit(expr1.Body);
var rightVisitor = new ReplaceExpressionVisitor(expr2.Parameters[0], parameter);
var right = rightVisitor.Visit(expr2.Body);
return Expression.Lambda<Func<T, bool>>(
Expression.AndAlso(left, right), parameter);
}
private class ReplaceExpressionVisitor
: ExpressionVisitor
{
private readonly Expression _oldValue;
private readonly Expression _newValue;
public ReplaceExpressionVisitor(Expression oldValue, Expression newValue)
{
_oldValue = oldValue;
_newValue = newValue;
}
public override Expression Visit(Expression node)
{
if (node == _oldValue)
return _newValue;
return base.Visit(node);
}
}
using System;
using System.Linq.Expressions;
namespace Extensions
{
public class Example
{
//How to use it
public static void Main()
{
Expression<Func<string, bool>> expression1 = exp => true;
Expression<Func<string, bool>> expression2 = exp => false;
Expression<Func<string, bool>> expression3 = ExpressionExtensions.AndAlso(expression1, expression2);
Expression<Func<string, bool>> expression4 = ExpressionExtensions.OrElse(expression1, expression2);
Expression<Func<string, bool>> expression = ExpressionExtensions.AndAlso(expression3, expression4);
}
}
public static class ExpressionExtensions
{
public static Expression<Func<T, bool>> AndAlso<T>(this Expression<Func<T, bool>> expr1, Expression<Func<T, bool>> expr2)
{
ParameterExpression parameter1 = expr1.Parameters[0];
var visitor = new ReplaceParameterVisitor(expr2.Parameters[0], parameter1);
var body2WithParam1 = visitor.Visit(expr2.Body);
return Expression.Lambda<Func<T, bool>>(Expression.AndAlso(expr1.Body, body2WithParam1), parameter1);
}
public static Expression<Func<T, bool>> OrElse<T>(this Expression<Func<T, bool>> expr1, Expression<Func<T, bool>> expr2)
{
ParameterExpression parameter1 = expr1.Parameters[0];
var visitor = new ReplaceParameterVisitor(expr2.Parameters[0], parameter1);
var body2WithParam1 = visitor.Visit(expr2.Body);
return Expression.Lambda<Func<T, bool>>(Expression.OrElse(expr1.Body, body2WithParam1), parameter1);
}
private class ReplaceParameterVisitor : ExpressionVisitor
{
private readonly ParameterExpression _oldParameter;
private readonly ParameterExpression _newParameter;
public ReplaceParameterVisitor(ParameterExpression oldParameter, ParameterExpression newParameter)
{
_oldParameter = oldParameter;
_newParameter = newParameter;
}
protected override Expression VisitParameter(ParameterExpression node)
{
if (ReferenceEquals(node, _oldParameter))
return _newParameter;
return base.VisitParameter(node);
}
}
}
}
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