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Is there any way to pull out the properties, the operator and matching value from an Expression<Func<T>,bool>? Given the following example:

var customers = GetCustomers();
var customerQuery = customers.Where(x=> x.CustomerID == 1 
    && x.CustomerName == "Bob"); // The query is for illustration only

I need to be able to get out something like the following:

Property: CustomerID
Operator: Equals
Value:    1

Property: CustomerName
Operator: Equals
Value:    Bob

I've already written something that can pull out the property name of an Expression, but I cannot seem to find out where the value and operator are held, although it's quite clearly visible in the Expression's DebugView property.

djdd87
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2 Answers2

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The operator will be on the BinaryExpression's Method that is the Equals node. You should also look at the expressions .NodeType, which reveals much (it should be Equal).

The values will typically be in a ConstantExpression in the .Right of that BinaryExpression, or in the case of a captured variable: the capture-context will be the ConstantExpression, so the value will be a MemberExpression over a ConstantExpression (you will need to investigate whether the member is a FieldInfo vs PropertyInfo, and fetch the value via .GetValue(...) on that).

Marc Gravell
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  • Excellent, thanks. Just had to cast `predicate` using `(BinaryExpression)predicate.Body` and now everything's rather obvious. Cheers Marc. – djdd87 Nov 07 '11 at 09:08
3

In addition to Marc Gravells answer (+1 there) I'll just add that it's worth taking a look at the ExpressionVisitor class (out of the box in .Net 4; MSDN has an example that you can copy/paste for 3.5). It makes writing code to extract certain types of expression very very easy.

In your case you would be looking to override it's VisitBinary method.

I typically use the class to push the expression(s) I'm interested in into a read-only list, for example, which I then make available publicly on my implementation of the class. You're not using it to rewrite the expression.

Andras Zoltan
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  • +1, thanks. It certainly makes it easier to find the Binary Expressions, but I'll leave Marc's answer as accepted as it says how to pull out the information I require. – djdd87 Nov 07 '11 at 09:17
  • Yes I agree - Marc's answer gives you the nuts and bolts for this question and deserves the tick; mine is only offering a simple way to achieve what he's suggested :) – Andras Zoltan Nov 07 '11 at 14:09