獲取從屬性值的詳細信息視圖

Question

我有我的部分觀點Preview.cshtml

@model MonitoreoIntegrado.Models.Figuras 

    @Html.DisplayFor(model => model.tipo) 

    @Html.DisplayFor(model => model.nombre) 

    @Html.DisplayFor(model => model.size) 

    @Html.DisplayFor(model => model.color) 

    @Html.DisplayFor(model => model.borde) 

    @Html.DisplayFor(model => model.colorBorde) 

    @Html.DisplayFor(model => model.rotacion) 

的model.size數據類型爲短。我需要得到model.size的價值做這樣的事情提前這個

<div style="width:@(model.size*2); height:@(model.size*3)"></div> 

感謝。

Answer 1

就像@Pricey說的。它只是寫Model.attribute，而不是model.attribute。謝謝！ 這是實現代碼:)

@{  
int largo = (Model.size-Model.borde).Value; 
double centro = Model.size*1.42/2; 
short x1, x2, x3, x4, y1, y2, y3, y4, deg; 
deg = Model.rotacion.Value; 
x1=(short)Math.Round(centro + (largo*1.42/2) * Funciones.coseno(225 - deg)); 
y1 = (short)Math.Round(centro + (largo*1.42/2) * Funciones.seno(225 - deg)); 
x2 = (short)Math.Round(x1 + largo * Funciones.coseno(90 - deg)); 
y2 = (short)Math.Round(y1 + largo * Funciones.seno(90 - deg)); 
x3 = (short)Math.Round(x2 + largo * Funciones.coseno(0 - deg)); 
y3 = (short)Math.Round(y2 + largo * Funciones.seno(0 - deg)); 
x4 = (short)Math.Round(x3 + largo * Funciones.coseno(270 - deg)); 
y4 = (short)Math.Round(y3 + largo * Funciones.seno(270 - deg)); 
<svg xmlns="http://www.w3.org/2000/svg" version="1.1" width="@(Model.size*1*42)" height="@(Model.size*1.42)"> 
    <polygon points="@x1,@y1 @x2,@y2 @x3,@y3 @x4,@y4" 
     fill="#@Model.color" stroke-width="@Model.borde" stroke="@Model.colorBorde" /> 
</svg> 
} 

Funciones.coseno就像Math.cos，但它與度而非弧度的作品。