小编Dan*_* T.的帖子

更新实体框架中的嵌套对象

最近我发现 EF 不更新嵌套对象。几天来,我试图弄清楚如何做到这一点,但不幸的是我被这个问题困住了。

我有对象

public class ProjectEntity : AuditableEntity<int>
{

    public string CustumerCompany { get; set; }
    public string CustomerRepresentative { get; set; }

    public string ProjectTitle { get; set; }
    public string WwsNumber { get; set; }

    [ForeignKey("ParentProjectId")]
    public virtual ProjectEntity ParentProject { get; set; }

    public int? ParentProjectId { get; set; }

    public virtual ICollection<ProjectServicesEntity> Service { get; set; }
}
Run Code Online (Sandbox Code Playgroud)

然后服务对象

public class ProjectServicesEntity : AuditableEntity<int>
{
    /// <summary>
    /// Service Number
    /// </summary>
    public int Number …
Run Code Online (Sandbox Code Playgroud)

.net c# entity-framework entity-framework-6

7
推荐指数
1
解决办法
8172
查看次数

使用Knockout.js发布集合

我正在编写一个asp.net MVC应用程序,并决定尝试使用Knockout.js来获取动态UI内容.这是一个很棒的框架,它迄今为止帮助了我很多.

但我遇到了两个我无法解决的问题,并找到了有用的信息.我将从代码开始向您展示我所拥有的内容然后我将尝试解释我想要实现的目标.

C#ViewModel

项目ViewModel


ProjectServicesViewModel


ProjectPositionsViewModel

我的HTML/Razor和淘汰模块

var Project = function (project) {
	var self = this;
	self.Id = ko.observable(project ? project.Id : 0);
	self.CustumerCompany = ko.observable(project ? project.CustumerCompany : "");
	self.CustomerRepresentative = ko.observable(project ? project.CustomerRepresentative : "");
	self.ProjectTitle = ko.observable(project ? project.ProjectTitle : "");
	self.WWSNumber = ko.observable(project ? project.WWSNumber : "");
	self.AqStatus = ko.observable(project ? project.AqStatus : "");
	self.Completed = ko.observable(project ? project.Completed : "");
	self.StartDate = ko.observable(project ? project.StartDate : "");
	self.EndDate = ko.observable(project ? project.EndDate : "");
	self.ProjectLeader = …
Run Code Online (Sandbox Code Playgroud)

javascript c# asp.net asp.net-mvc knockout.js

6
推荐指数
1
解决办法
214
查看次数

离开网站时弹出

我遇到了JavaScript问题.我想要一个脚本,在退出整个网站时弹出一条带有问题的消息,如果访问者回答"否"网页关闭,如果他回答"是",他将被重定向到另一个页面.我在http://www.pgrs.net/2008/01/30/popup-when-leaving-website/找到了一个例子,但似乎它对我不起作用.我找不到任何解决方案.请检查我的代码并告诉我也许我做错了什么?这是我的源代码.

也许有人会看到一个问题.

 <!DOCTYPE html>
<html lang="lt">
<head>
    <meta charset="utf-8">
    <title>PUA.LT</title>
    <meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1">
    <meta name="description" content="">
    <meta name="author" content="Perfect WEB Solutions">
    <link rel="stylesheet" href="<?php echo base_url("additional/style.css") ?>">
    <script src='<?php echo base_url("additional/prototype.js")?>' type='text/javascript' ></script>
</head>
<body>
<script type="text/javascript">


Event.observe(document.body, 'click', function(event) {
  if (Event.element(event).tagName == 'A') {
    staying_in_site = true;
  }
});

window.onunload = popup;

function popup() {
  if(staying_in_site) {
    return;
  }
  alert('I see you are leaving the site');
}
</script>

</body>
</html>
Run Code Online (Sandbox Code Playgroud)

javascript popup function-exit

2
推荐指数
1
解决办法
2万
查看次数