[Vuex] Lazy Load a Vuex Module at Runtime using TypeScript

Sometimes we need to create modules at runtime, for example depending on a condition. We could even want to lazy load that module by using Webpack’s code splitting feature.

 

For example, in current appliation, we are loading login and todos modules at the same time, now what we want is lazy load login module:

import Vue from 'vue';
import Vuex from 'vuex';

import {todos} from './todos';
import {login} from './login';

Vue.use(Vuex);

export default new Vuex.Store({
  modules: {
    todos,
    login,
  }
});

 

Go to the main.ts file, set login module to be lazy load:

import './hooks';

import Vue from 'vue';
import App from './App.vue';
import router from '@/router';
import store from '@/store/index';
import '@/registerServiceWorker';
Vue.config.productionTip = false;

const load = true;
if (load) {
  import('./store/login').then(({login}) => {
    store.registerModule('login', login);
  });
}

new Vue({
  router,
  store,
  render: (h) => h(App),
}).$mount('#app');

We can use dynamic import syntax to lazy load login store.

To support the syntax, we need to change "compilerOptions.modules='esnext'".

 

To code safty. in login component, we add v-if to check the login module is loaded or not:

<template>
    <section v-if="login">
        <button v-if="!login.isLoggedIn" @click="loginMutation">Login</button>
        <p v-else>Hello {{login.user}}</p>
    </section>
</template>

<script lang="ts">
import {Component, Vue} from 'vue-property-decorator';
import {State, Mutation} from 'vuex-class';
import {LoginState} from '../types';

@Component()
export default class Login extends Vue{
    @State login: LoginState;
    @Mutation('login') loginMutation;
}
</script>