Question

如果我有一个包含列表的预期对象图，如下所示：

        var expectedExperiment = new Experiment
        {
            Number= "12345",
            AllocatedInstrument = "Instrument 1",
            Experimenters = new List<Experimenter>
            {
                new Experimenter
                {
                    Name = "Sue"
                    Role = "Scientist",
                    Id = 1,
                    Initials = "S"
                },
                new Experimenter()
                {
                    Name = "Mark",
                    Role = "Technician",
                    Id = 2,
                    Initials = "M"
                },
            }
        };

当我只想在子对象列表中包含某些属性时，如何将它与我的实际对象进行比较。

例如，我想写这样的东西来比较所有父对象属性和一些子对象属性：

        actualExperiment.ShouldBeEquivalentTo(expectedExperiment, options => options
            .Including(o => o.Number)
            .Including(o => o.AllocatedInstrument)
            .Including(o => o.Experimenters.Select(e => e.Role))
            .Including(o => o.Experimenters.Select(e => e.Name)));

但我得到一个例外：

System.ArgumentException : Expression <o.Experimenters.Select(e => e.Role)> cannot be used to select a member.

在这种情况下，我不介意子项目的顺序，但我想断言我关心的属性符合预期。

Answer 1

您可以通过检查当前对象的运行时类型和选定成员路径来执行此操作：

.Including(subjectInfo => subjectInfo.RuntimeType == typeof(Experimenter) &&
                          subjectInfo.SelectedMemberPath.EndsWith("Role"))

这可以提取到一种方法以便进一步重用：

private Expression<Func<ISubjectInfo, bool>> BuildMemberExpression<TSource, TProperty>(Expression<Func<TSource,TProperty>> propertySelector)
    {
        var memberExpression = propertySelector.Body as MemberExpression;
        if (memberExpression == null)
        {
            throw new NotSupportedException();
        }

        return subjectInfo => subjectInfo.RuntimeType == typeof(TSource) && 
                              subjectInfo.SelectedMemberPath.EndsWith(memberExpression.Member.Name);
    }

现在以这种方式使用它：

.Including(BuildMemberExpression((Experimenter e) => e.Name))
.Including(BuildMemberExpression((Experimenter e) => e.Role))

如何使用流畅的断言比较对象图中的嵌套列表

1 个答案: