如何创建比较子对象中的属性的表达式树？

Question

如何创建比较子对象中的属性的表达式树？

例如,我已经可以创建一个lambda表达式树来比较对象的直接属性.像这样:

var propertyName = "JobNumber";
var propertyValue = "1005";
Type entityType = typeof(ParentObject);
OperatorDelegate comparisonMethod = Expression.Equal;

var parameter = Expression.Parameter(entityType, "entity");
var lambda =
    Expression.Lambda<Func<ParentObject, bool>>(
        comparisonMethod(Expression.Property(parameter, propertyName), Expression.Constant(propertyValue)),
        parameter);

这是(我相信)相当于:

entity => entity.JobNumber == "1005";

我被挂起的地方是试图弄清楚如何比较子对象属性中的值.例如,给定:

public class ParentObject
{
    public ChildObject Child { get; set; }

    // more properties of ParentObject
}

public class ChildObject
{
    public string JobNumber { get; set; }
}

我如何构建一个等价于的表达式:

parentEntity => parentEntity.Child.JobNumber == "1005"

谢谢你的帮助.

Answer 1

您只需要获取返回的属性值的属性:

var child = Expression.Property(parameter, "Child");
var jobNumber = Expression.Property(child, propertyName);

var lambda =
    Expression.Lambda<Func<ParentObject, bool>>(
        comparisonMethod(jobNumber, Expression.Constant(propertyValue)),
        parameter);