Conditionally open a menu entry in a new tab

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1

Hi,

How to do this conditionally though ?

I tried adding an is_external boolean, but the IDE says Schema validation: Property 'is_external' is not allowed and the terminal command outputs can't evaluate field is_external in type *navigation.MenuEntry :thinking:

Thanks

3

There are a couple of ways to do this.

1) Open all external URLs in new tab

{{ with site.Menus.main }}
  <nav class="menu">
    {{ range . }}
      {{ if (urls.Parse .URL).IsAbs }}
        <a href="{{ .URL }}" rel="external" target="_blank">{{ .Name }}</a>
      {{ else }}
        <a href="{{ .URL }}">{{ .Name }}</a>
      {{ end }}
    {{ end }}
  </nav>
{{ end }}

2) Configure each menu entry as needed

menus:
  main:
    - name: Home
      pageRef: /
      weight: 10
    - name: Posts
      pageRef: /posts
      weight: 20
    - name: Hugo
      url: https://gohugo.io/
      weight: 30
      params:
        external: true

{{ with site.Menus.main }}
  <nav class="menu">
    {{ range . }}
      {{ if .Params.external }}
        <a href="{{ .URL }}" rel="external" target="_blank">{{ .Name }}</a>
      {{ else }}
        <a href="{{ .URL }}">{{ .Name }}</a>
      {{ end }}
    {{ end }}
  </nav>
{{ end }}