I am working with a vue component called <PlanView/>. In my code, I am 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;
},
However, I want the functionality to be triggered even when the component is not rendered. Can you please provide guidance on how I can achieve that? Thank you in advance.
To ensure that the component is rendered but not displayed, you can manipulate the visibility of the template within the component instead of hiding the entire component.
Include a prop in PlanView to determine whether the template should be rendered or not
<PlanView :showPlan="show_plan"/>
Access the prop within the PlanView component like this:
defineProps({
showPlan: {
type: Boolean,
required: false,
default: false
}
})
Render the PlanView template only if the prop condition is met. In this case, the template will appear as follows:
<template>
<!-- Container element -->
<div v-if="showPlan">
<!-- The rest of your template goes here -->
</div>
</template>
Alternatively
You can simply use v-show on the wrapper element so that it loads but remains hidden on the UI when the condition is false
<PlanView v-show="show_plan"/>
To achieve the desired outcome, consider utilizing v-show in place of v-if, as suggested by @Nitheesh.
<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>
It's important to clarify if this approach aligns with your concept of "calling the functionality."
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