具有条件渲染的 Vue 组件功能

Question

I have a vue component called <PlanView/> , and I'm rendering this component conditionally:

<div v-if="show_plan" id="mainplan">
  <PlanView/>
</div>
<div class="icon" v-else>
  <font-awesome-icon icon="fa-solid fa-angles-right" @click="openPlan"/>
</div>

openPlan() {
    this.show_plan = true;
},

but I want the functionality to be called even if the component is not rendered, can you please advise me how can I do that? thanks in advance.

Answer 1

If you want the component to be renedered and not displayed, you can hide the visibility of the template inside the component rather than hiding the complete compoenent.

Pass a prop to PlanView to decide the template is to be rendered or not

<PlanView :showPlan="show_plan"/>

Accept the prop inside PlanView component like

defineProps({
    showPlan: {
        type: Boolean,
        required: false,
        default: false
    }
})

Render the template of PlanView only if the prop is satisfied. So the template of PlanView will be looking like

<template>
    <!-- Div or some wraper element -->
    <div v-if="showPlan"> 
        <!-- Rest of your template here -->
    </div>
</template>

OR

Simply use v-show on the wrapper so that the element will be loaded, but will not be displayed in the UI when the condition is false

<PlanView v-show="show_plan"/>

Answer 2

You can simply use v-show instead of v-if to provide the same functionality as the answer from @Nitheesh suggests.

<div v-show="show_plan" id="mainplan">
  <PlanView/>
</div>
<div v-show="!show_plan" class="icon">
  <font-awesome-icon icon="fa-solid fa-angles-right" @click="openPlan"/>
</div>

But I am not sure this is what you means by by calling the functionality .