我试图追加 where 谓词,我的目标是创建与以下相同的表达式:
Services.Where(s => s.Name == "Modules" && s.Namespace == "Namespace");
我有以下代码:
Expression<Func<Service,string>> sel1 = s => s.Name;
Expression<Func<Service,string>> sel2 = s => s.Namespace;
var val1 = Expression.Constant("Modules");
var val2 = Expression.Constant("Namespace");
Expression e1 = Expression.Equal(sel1.Body, val1);
Expression e2 = Expression.Equal(sel2.Body, val2);
var andExp = Expression.AndAlso(e1, e2);
ParameterExpression argParam = Expression.Parameter(typeof(string), "s");
var lambda = Expression.Lambda<Func<string, bool>>(andExp, argParam);
这将创建以下输出:
s => ((s.Name == "Modules") AndAlso (s.Namespace == "Namespace"))
但是,这是错误的,因为Name和Namespace的参数不一样。如果我将表达式选择器之一更改为:
Expression<Func<Service,string>> sel2 = srv => srv.Namespace;
输出将是:
s => ((s.Name == "Modules") AndAlso (srv.Namespace == "Namespace"))
如何使用sel1和sel2创建有效的表达式?
更新(2011 年 2 月 28 日)
我通过创建调用表达式解决了这个问题:Expression.Invoke
所以 lambda 表达式 sel1 和 sel2 不需要是 MemberExpression:
Expression<Func<Service,string>> sel1 = s => s.Name;
Expression<Func<Service,string>> sel2 = srv => srv.Namespace;
var val1 = Expression.Constant("Modules");
var val2 = Expression.Constant("Namespace");
Expression<Func<Service, bool>> lambda = m => true;
var modelParameter = lambda.Parameters.First();
// sel1 predicate
{
var invokedExpr = Expression.Invoke(sel1, modelParameter);
var binaryExpression = Expression.Equal(invokedExpr, val1);
lambda = Expression.Lambda<Func<Service, bool>>(Expression.AndAlso(binaryExpression, lambda.Body), lambda.Parameters);
}
// sel2 predicate
{
var invokedExpr = Expression.Invoke(sel2, modelParameter);
var binaryExpression = Expression.Equal(invokedExpr, val2);
lambda = Expression.Lambda<Func<Service, bool>>(Expression.AndAlso(binaryExpression, lambda.Body), lambda.Parameters);
}