css类基于加载的模板 [英] css class based on loaded template

问题描述

我在我的_base.html模板中有这个引导导航：

I've got this bootstrap nav in my _base.html template like this:

<ul class="nav navbar-nav">
   <li><a href="/" class="">Home</a></li>
   <li><a href="/blog/">Blog</a></li>
</ul>

使用Golang我要添加

Using Golang I want to add a

class="active"

。

我已阅读 HTML /模板文档和 thisone 等文章，但我看来我必须写一个golang函数添加

I've read the html/template docs and articles like thisone, but it appears to me that I have to write a golang function that adds

class="active"

到每个相应的列表项。但不知何故仍然我认为如果我可以只添加像

to every correspondending corresponding list-item. But somehow still I think it would be cleaner if I could just add something like

<ul>
    <li{{ if .template = "index.html" }} class="active"{{ end }}><a href="/">Home</a></li>
    <li{{ if .template = "blog.html" }} class="active"{{ end }}><a href="/blog/">Blog</a></li>
</ul>

或类似的东西。我记得Rob Pike说Golang应该为你做所有的计算，但为什么在html / template-package中有一个if语句？

or something like that. I remember Rob Pike saying Golang should be doing all the calculations for you, but why is there an "if" statement in the html/template-package?

推荐答案

我个人经常为这样的任务实施一个小 eq 助手：

I personally often implement a small eq helper for tasks like that:

var tmpl = template.Must(template.New("").Funcs(template.FuncMap{
    "eq": func(a, b interface{}) bool {
        return a == b
    },
}).ParseGlob("templates/*.html")

示例用法：

<li{{if eq .Active "index"}} class="active"{{end}}><a href="/">Home</a></li>

但是它只用于显示逻辑本身，保持显示逻辑和实际计算分开是一个很好的做法。

But use it only for the display logic itself. It's a good practice to keep the display logic and the real computation apart.

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