如何根据订购排序

Mar*_*gus 10 .net c# lambda

可以说我有物品

items : [{id:1,...}, {id:2,...}, {id:3,...}]
Run Code Online (Sandbox Code Playgroud)

命令:[ 2,3,1 ]得到一个可枚举的

items : [{id:2,...}, {id:3,...}, {id:1,...}]
Run Code Online (Sandbox Code Playgroud)

我希望它能成为符合标准的东西

items.Select(o => new {key = ordering[i++], value = o})
     .OrderBy(k => k.key)
     .Select(o => o.value)
Run Code Online (Sandbox Code Playgroud)

但有更清洁的解决方案吗?


以下我已经确认了这项工作(HimBromBeere,Domysee,qxg)

var expectedOrder = ordering.Select(x => result.First(o => o.Id == x));
var expectedOrder = result.OrderBy(item => Array.FindIndex(ordering,i => i == item.Id));
var expectedOrder = result.OrderBy(item => ordering.ToList().FindIndex(i => i == item.Id));
var expectedOrder = 
  from o in ordering
  join i in result 
    on o equals i.Id
  select i;
Run Code Online (Sandbox Code Playgroud)

Fwi,这是用于验证测试:

    [Test]
    [TestCase(1, 2, 3)]
    [TestCase(1, 3, 2)]
    [TestCase(2, 1, 3)]
    [TestCase(2, 3, 1)]
    [TestCase(3, 1, 2)]
    public void Test_Should_Fail_If_GetMessages_Does_Not_Return_Sorted_By_Sent_Then_By_Id_Result(params int[] ordering)
    {
        var questions = GetQuestionsData();                     
        Mock.Get(_questionService)
            .Setup(o => o.GetQuestions())
            .Returns(questions);
        var result = _mailboxService.GetMessages();    
        var expectedOrder = ordering.Select(x => result.First(o => o.Id == x));

        // Act
        Action sortOrder = () => expectedOrder.Should()
            .BeInDescendingOrder(o => o.Sent)
            .And.BeInDescendingOrder(o => o.Id);

        // Assert
        sortOrder.ShouldThrow<AssertionException>();
    }
Run Code Online (Sandbox Code Playgroud)

Him*_*ere 3

我猜是这样的:

var result = ordering.Select(x => items.First(y => y.id == x.id));
Run Code Online (Sandbox Code Playgroud)

工作示例:

var items = new[] { new { id = 1, name = "1" }, new { id = 2, name = "2" }, new { id = 3, name = "3" }, new { id = 4, name = "4" } };
var result = new[] { 2, 3, 1 }.Select(x => items.First(y => y.id == x));
Run Code Online (Sandbox Code Playgroud)

这也会过滤items掉那些索引不包含在 中的索引ordering