比较C＃ASP.NET MVC LINQ实体框架中的两个列表

debugcn 发表于 Dev

夸脱学

如标题所示，即时通讯试图比较带时间戳的日期时间和带时间戳的日期时间。

我有一个flex变量，它将获取一个工作日（8h）这两个日期之间的差值。

“如果我在08:00戳记并在17:00戳记，则我的弹性为+1（h）”

模型：

public class FlexModel
{
    public List<User> Users { get; set; }
    public List<Stamping> Stampings { get; set; } 
    public decimal FlexTime { get; set; }
}

public partial class Stamping
{
    [Key]
    [DatabaseGenerated(DatabaseGeneratedOption.Identity)]
    public int Id { get; set; }

    [Required]
    public int UserId { get; set; }

    [Required]
    [DataType(DataType.DateTime)]
    public DateTime Timestamp { get; set; }

    [Required]
    [StringLength(3)]
    public string StampingType { get; set; }

    public virtual User User { get; set; }
}

看法：

@Html.LabelFor(model => model.FlexTime, htmlAttributes: new { @class = "control-label col-md-2" })
<div class="col-md-10">
    @Html.DisplayFor(model => model.FlexTime, new { htmlAttributes = new { @class = "form-control" } })
    @Html.ValidationMessageFor(model => model.FlexTime, "", new { @class = "text-danger" })
</div>

控制器：

public ActionResult Info()
{
    var flexModel = new FlexModel();
    var userId = (int)Session["userId"];
    var user = _db.Users.Find(userId);
    var stampIn = _db.Stampings
                     .Where(i => i.StampingType == "in")
                     .Where(i => i.User == user)
                     .ToList();

    var stampOut = _db.Stampings
                      .Where(i => i.StampingType == "out")
                      .Where(i => i.User == user)
                      .ToList();

    var workDay = 8;

    if (stampIn.Count == 0)
    {
        return View();
    }

    foreach (var itemIn in stampIn)
    {
        //Dont know what to do here
    }

    foreach (var itemOut in stampOut)
    {
        //Dont know what to do here either
    }

    return View();
}

请帮忙。

戴维

您可以像这样关联列表：

var attendance = from sin in stampIn
                 select new 
                 {
                    StampIn = sin,
                    StampOut = stampOut.FirstOrDefault(sout => 
                                    sout.Timestamp > sin.Timestamp)
                 };

这为您提供了一个踩踏与踩踏事件的列表（尽管踩踏可能为空）。然后，您需要像这样计算弹性时间：

var flexitime = from att in attendance
                select new
                {
                    TimeIn = att.StampIn.Timestamp,
                    TimeOut = att.StampOut == null ? (DateTime?)null : att.StampOut.Timestamp,
                    TotalTime = att.StampOut == null ? 0 :  
                        att.StampOut.Timestamp.Subtract(att.StampIn.Timestamp).TotalHours
                };

您现在可以将其转换为您的FlexModel对象（我仅填写该FlexTime属性，因为不确定如何/为什么需要其他属性）：

var workDay = 8;
var flexModel = new FlexModel 
{
    FlexTime = Convert.ToDecimal(flexitime
                   .Sum(f => f.TotalTime - workDay))
};
return View(flexModel);

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