How do I translate an expression tree of one type to a different expression type?

Here's a visitor that does the job it makes a copy of the parameter (since we'll need to create a new parameter and substitute all references of the old parameter for the new one) it walks the Body of the tree, substituting the parameter, and switching any member-access against the old type to a like-named member on the new type it re-assembles a lambda using the parameter we invented earler Code: class TypeChangeVisitor : ExpressionVisitor { private readonly Type from, to; private readonly Dictionary substitutions; public TypeChangeVisitor(Type from, Type to, Dictionary substitutions) { this. From = from; this. To = to; this.

Substitutions = substitutions; } public override Expression Visit(Expression node) { // general substitutions (for example, parameter swaps) Expression found; if(substitutions! = null && substitutions. TryGetValue(node, out found)) { return found; } return base.

Visit(node); } protected override Expression VisitMember(MemberExpression node) { // if we see x. Name on the old type, substitute for new type if (node.Member. DeclaringType == from) { return Expression.

MakeMemberAccess(Visit(node. Expression), to. GetMember(node.Member.Name, node.Member.

MemberType, BindingFlags. Instance | BindingFlags. Static | BindingFlags.

Public | BindingFlags. NonPublic).Single()); } return base. VisitMember(node); } } public class Program { public static void Main() { Expression> predicate = x => x.

Name == "abc"; var switched = Translate(predicate); } public static Expression> Translate(Expression> expression) { var param = Expression. Parameter(typeof(TTo), expression. Parameters0.

Name); var subst = new Dictionary { { expression. Parameters0, param } }; var visitor = new TypeChangeVisitor(typeof(TFrom), typeof(TTo), subst); return Expression. Lambda>(visitor.

Visit(expression. Body), param); } } Note that if you have x.Something.Name you might need to be a bit more careful, but this should get you a reasonable way.

Here's a visitor that does the job. It makes a copy of the parameter (since we'll need to create a new parameter and substitute all references of the old parameter for the new one) it walks the . Body of the tree, substituting the parameter, and switching any member-access against the old type to a like-named member on the new type it re-assembles a lambda using the parameter we invented earler Code: class TypeChangeVisitor : ExpressionVisitor { private readonly Type from, to; private readonly Dictionary substitutions; public TypeChangeVisitor(Type from, Type to, Dictionary substitutions) { this.

From = from; this. To = to; this. Substitutions = substitutions; } public override Expression Visit(Expression node) { // general substitutions (for example, parameter swaps) Expression found; if(substitutions!

= null && substitutions. TryGetValue(node, out found)) { return found; } return base. Visit(node); } protected override Expression VisitMember(MemberExpression node) { // if we see x.Name on the old type, substitute for new type if (node.Member.

DeclaringType == from) { return Expression. MakeMemberAccess(Visit(node. Expression), to.

GetMember(node.Member. Name, node.Member. MemberType, BindingFlags.

Instance | BindingFlags. Static | BindingFlags. Public | BindingFlags.

NonPublic).Single()); } return base. VisitMember(node); } } public class Program { public static void Main() { Expression> predicate = x => x.Name == "abc"; var switched = Translate(predicate); } public static Expression> Translate(Expression> expression) { var param = Expression. Parameter(typeof(TTo), expression.

Parameters0.Name); var subst = new Dictionary { { expression. Parameters0, param } }; var visitor = new TypeChangeVisitor(typeof(TFrom), typeof(TTo), subst); return Expression. Lambda>(visitor.

Visit(expression. Body), param); } } Note that if you have x.Something. Name you might need to be a bit more careful, but this should get you a reasonable way.

Jul 15 at 22:38 Ain't help though – Agzam Jul 15 at 23:02 check for yourself - if you have Int Id in Animal and also AnimalViewModel derived from BaseViewModel which holds Id. And if you try to use Id in predicate wouldn't work. Don't know why – Agzam Jul 15 at 23:22.

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