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Asp.net mvc 3 如何使用Razor在mvc3.0中编码链接按钮单击事件打开模式弹出窗口?_Asp.net Mvc 3 - Fatal编程技术网

Asp.net mvc 3 如何使用Razor在mvc3.0中编码链接按钮单击事件打开模式弹出窗口?

Asp.net mvc 3 如何使用Razor在mvc3.0中编码链接按钮单击事件打开模式弹出窗口?,asp.net-mvc-3,Asp.net Mvc 3,我有如下webgrid @grid.GetHtml(columns: grid.Columns(grid.Column( "FirstName",format: @<text>@Html.ActionLink((string)item.FirstName, "Details", "product", new { id = item.FirstName }, null)</text>), grid.Column("LastName","Last") ))

我有如下webgrid

@grid.GetHtml(columns: grid.Columns(grid.Column(
    "FirstName",format: @<text>@Html.ActionLink((string)item.FirstName, "Details", "product", new { id = item.FirstName }, null)</text>),
    grid.Column("LastName","Last")
))
@grid.GetHtml(列:grid.columns(grid.Column(
“FirstName”,格式:@@Html.ActionLink((字符串)item.FirstName,“详细信息”,“产品”,新的{id=item.FirstName},null)),
grid.Column(“LastName”、“Last”)
))
如果我点击Firstname链接,模态弹出窗口会打开吗?如何在mvc3.0中做到这一点?

推荐

为fancybox添加document.ready。(查看示例和文档)

然后更改actionlink,使其具有fancybox可以附加到的类

@Html.ActionLink((string)item.FirstName, "Details", "product"
    , new { id = item.FirstName }, new {Class = "fancybox"})
推荐

为fancybox添加document.ready。(查看示例和文档)

然后更改actionlink,使其具有fancybox可以附加到的类

@Html.ActionLink((string)item.FirstName, "Details", "product"
    , new { id = item.FirstName }, new {Class = "fancybox"})