比较C#ASP.NET MVC LINQ实体框架中的两个列表
本文关键字:列表 两个 NET C#ASP MVC LINQ 框架 实体 比较 | 更新日期: 2023-09-27 18:25:24
正如标题所说,我试图比较戳入日期时间和戳出日期时间。
我有一个弹性变量,它将得到一个工作日(8小时)中这两个日期之间的差值。
"如果我在08:00盖章,在17:00盖章,我的弹性应为+1(h)"
型号:
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; }
}
视图:
@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。
您可以关联如下列表:
var attendance = from sin in stampIn
select new
{
StampIn = sin,
StampOut = stampOut.FirstOrDefault(sout =>
sout.Timestamp > sin.Timestamp)
};
这会给你一个stampin与stampout事件的列表(尽管stampout可能为null)。然后你需要这样计算弹性时间:
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FlexModel
对象(我只填写FlexTime
属性,因为我不确定您如何/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);