Is it possible to escape the code within the setup block using the Vue-Composition API?

<script setup lang="ts">
import router from '@/router';
import { useMainStore } from '@/stores/main';
import { ref } from 'vue';

const mainStore = useMainStore();
const x = ref<object| undefined>();

if (!mainStore.x) {
    router.replace('/');
} else {
    x.value = mainStore.x;
}

if (!x.value) {
    router.replace('/');
} else {
    console.log(x.value.location);
}

function fun() {
    if (!x.value) {
        router.replace('/');
        return 5;
    }
    return x.value.location;
}
</script>

I am looking for a way to exit this single file component if mainStore.x is undefined. Once I have confirmed that mainStore.x is not undefined, I don't need to keep checking if x.value is defined every time I use it.

By using router.replace to navigate to a different single file component, I ensure that x.value will never be undefined in this component.

Answer №1

To tackle this issue, consider encapsulating the remainder of your code within an else block and verifying for non-null values at the beginning of each function to return if null.

Furthermore, should you require that variable in your template, try using v-if="!variable", as :if doesn't seem to function correctly in my case.

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