How can I access the $route object in a split Vuex getters.js file?

In my vuex setup, I have split files for getters, mutations, actions, and state. I recently switched to using Quasar which suggests organizing the Vuex module in this way. Although I used to have them all in one file before, I decided to give this a try. Here is my folder structure:

└── store
      └─ routes
          ├── state.js
          ├── actions.js
          ├── mutations.js
          ├── getters.js
          └── index.js

In the index.js file, everything is imported and bound together like this:

import state from './state'
import * as getters from './getters'
import * as mutations from './mutations'
import * as actions from './actions'

export default {
  namespaced: true,
  getters,
  mutations,
  actions,
  state
}

Within getters.js, I export my previously used functions. However, I encountered an issue where I needed to access the $route plugin within one of them.

export const current_route = state => {
    return this.$route;
}

While I am aware that I can directly use this.$route without going through the getter, I wanted to understand if it's possible to access these variables within the splitted files.

Answer №1

Here is the provided source code for vue-router:

  Vue.mixin({
    beforeCreate () {
      if (isDef(this.$options.router)) {
        /* additional logic */
        Vue.util.defineReactive(this, '_route', this._router.history.current)
      } else {
        this._routerRoot = (this.$parent && this.$parent._routerRoot) || this
      }
      /* additional logic */
    }
  })

  Object.defineProperty(Vue.prototype, '$route', {
    get () { return this._routerRoot._route }
  })

In the code snippet above, it demonstrates that $route serves as an alias for $router.history.current. Therefore, you have the option to directly import router like so:

import router from 'path/to/your/router'
export const current_route = state => {
  return router.history.current
}

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