Guide on how to link a prop to a ref in Vue 3.0 without modifying the prop value

Question

Guide on how to link a prop to a ref in Vue 3.0 without modifying the prop value

When passing a prop user from the parent component to the child component, I want to create a separate copy of it without altering the original prop value.

I attempted to achieve this with the following code:

export default defineComponent({
  props: {
    apiUser: {
      required: true,
      type: Object
    }
  },
  setup(props) {
    const user = ref(props.apiUser);

    return { user };
  }
});

However, modifying the user object also ends up changing the apiUser prop. I considered using Object.assign, but then the reactivity of the ref is lost.

In Vue 2.0, I would approach it like this:

export default {
  props: {
     apiUser: {
       required: true,
       type: Object
     }
  },
  data() {
    return {
      user: {}
    }
  },
  mounted() {
    this.user = this.apiUser;
    // Now I can use this.user without altering this.apiUser's value.
  }
};

Credit goes to @butttons for providing the helpful suggestion that led to the solution.

const user = reactive({ ...props.apiUser });

javascript vue.js vue-composition-api

Answer 1

Answer №1

data: {
 apiUser: {
   required: true,
   type: Object
 }
},
created() {
   const userDuplicate = Vue.reactive(this.apiUser)
}

Utilizing the composition API, we can employ the Vue.reactive method to create a duplicate of any source reactive object. By ensuring that the apiUser is a reactive object, using Vue.reactive() will prevent mutations from occurring in your data.

Answer 2

data: {
 apiUser: {
   required: true,
   type: Object
 }
},
created() {
   const userDuplicate = Vue.reactive(this.apiUser)
}

Utilizing the composition API, we can employ the Vue.reactive method to create a duplicate of any source reactive object. By ensuring that the apiUser is a reactive object, using Vue.reactive() will prevent mutations from occurring in your data.

Answer 3

Answer №2

If you're searching for it, this is what you need: https://vuejs.org/guide/components/props.html#one-way-data-flow

Set up data to include the prop:

export default {
  props: ['apiUser'],
  data() {
    return {
      // user only uses this.apiUser as the initial value;
      // it is disconnected from future prop updates.
      user: this.apiUser
    }
  }
}

Alternatively, if you are using api composition:

import {ref} from "vue";
const props = defineProps(['apiUser']);
const user = ref(props.apiUser);

You may also want to explore computed methods (refer to linked documentation section above) or v-model.

Keep in mind that the solution marked here, , does not work. Trying to update user will result in a readonly error in the console. If you don't plan on modifying user, it's best to just use the prop directly.

Answer 4

If you're searching for it, this is what you need: https://vuejs.org/guide/components/props.html#one-way-data-flow

Set up data to include the prop:

export default {
  props: ['apiUser'],
  data() {
    return {
      // user only uses this.apiUser as the initial value;
      // it is disconnected from future prop updates.
      user: this.apiUser
    }
  }
}

Alternatively, if you are using api composition:

import {ref} from "vue";
const props = defineProps(['apiUser']);
const user = ref(props.apiUser);

You may also want to explore computed methods (refer to linked documentation section above) or v-model.

Keep in mind that the solution marked here, , does not work. Trying to update user will result in a readonly error in the console. If you don't plan on modifying user, it's best to just use the prop directly.

Answer 5

Answer №3

When implementing this method

const copiedUser = toRef(props, 'apiUser')

the ref created will attempt to change the prop whenever there is a modification, resulting in a warning.

The proper technique for linking a ref to a prop without causing any prop mutations is as follows:

const copiedUser = ref();

watchEffect(() => (copiedUser.value = props.apiUser));

Answer 6

When implementing this method

const copiedUser = toRef(props, 'apiUser')

the ref created will attempt to change the prop whenever there is a modification, resulting in a warning.

The proper technique for linking a ref to a prop without causing any prop mutations is as follows:

const copiedUser = ref();

watchEffect(() => (copiedUser.value = props.apiUser));

Answer 7

Answer №4

The optimal approach, in my opinion, involves cloning the prop when initializing the local variable.

Here is how I would tackle it:

Parent component:

<script setup>
import { ref } from 'vue';
import Child from './Child.vue';

const user = ref({
  age: 20,
});
</script>

<template>
  Parent: {{ user.age }}
  <br />
  <child :user="user" />
</template>

Child component:

<script setup>
import { ref, toRaw } from 'vue';
const props = defineProps({
  user: {
    type: Object,
    required: true,
  },
});

const localUser = ref(structuredClone(toRaw(props.user)));
localUser.value.age = 40;
</script>

<template>Child: {{ localUser.age }}</template>

Please note: simply referencing like this will not suffice:

const localUser = ref(props.user);

This method will not prevent changes in the parent component when altering the age of the local reference.

Personally, I opt for using the native structuredClone for deep clonings. Although it does have limitations with types like Proxy, utilizing toRaw overcomes that obstacle by converting the prop into a plain Object compatible with structuredClone.

Feel free to explore various deep clone methods available online.

In my observation, structuredClone outperforms alternatives like lodash in terms of speed and efficiency. It also comes built-in, eliminating the need for additional code snippets and is suitable for most scenarios.
Benchmark comparison available [here]
For more information on supported functionalities of native clone, check out this article [here] or visit the official documentation.

Avoid using the spread operator (...) or Object.assign for deep copying as they only perform shallow copies, failing to replicate nested objects accurately.

Explore a functional Stack Blitz example here: https://stackblitz.com/edit/vitejs-vite-wycfiv?file=src%2Fcomponents%2FChild.vue

Answer 8

The optimal approach, in my opinion, involves cloning the prop when initializing the local variable.

Here is how I would tackle it:

Parent component:

<script setup>
import { ref } from 'vue';
import Child from './Child.vue';

const user = ref({
  age: 20,
});
</script>

<template>
  Parent: {{ user.age }}
  <br />
  <child :user="user" />
</template>

Child component:

<script setup>
import { ref, toRaw } from 'vue';
const props = defineProps({
  user: {
    type: Object,
    required: true,
  },
});

const localUser = ref(structuredClone(toRaw(props.user)));
localUser.value.age = 40;
</script>

<template>Child: {{ localUser.age }}</template>

Please note: simply referencing like this will not suffice:

const localUser = ref(props.user);

This method will not prevent changes in the parent component when altering the age of the local reference.

Personally, I opt for using the native structuredClone for deep clonings. Although it does have limitations with types like Proxy, utilizing toRaw overcomes that obstacle by converting the prop into a plain Object compatible with structuredClone.

Feel free to explore various deep clone methods available online.

In my observation, structuredClone outperforms alternatives like lodash in terms of speed and efficiency. It also comes built-in, eliminating the need for additional code snippets and is suitable for most scenarios.
Benchmark comparison available [here]
For more information on supported functionalities of native clone, check out this article [here] or visit the official documentation.

Avoid using the spread operator (...) or Object.assign for deep copying as they only perform shallow copies, failing to replicate nested objects accurately.

Explore a functional Stack Blitz example here: https://stackblitz.com/edit/vitejs-vite-wycfiv?file=src%2Fcomponents%2FChild.vue

Answer 9

Answer №5

During the discussion in the comments, I came across a Vue 2 method that I find particularly useful in these scenarios. It essentially creates a roundtrip when updating a model.

Here is how it works:

Parent

<template>
   <div class="parent-root"
      <child :apiUser="apiUser" @setUserData="setUserData" />
   </div>
</template>

export default {
   data() {
      return {
         apiUser: {
            id: 'e134',
            age: 27
         }
      };
   },

   methods: {
      setUserData(payload) {
         this.$set(this.apiUser, 'age', payload);
      }
   }
}

Child

<template>
   <div class="child-root"
      {{ apiUser }}
   </div>
</template>

export default {
   props: {
      apiUser: {
         required: true,
         type: Object
      }
   },

   data() {
      return {
         user: null
      };
   },

   watch: {
      apiUser: {
         deep: true,
         handler() {
            this.user = cloneDeep(this.apiUser);
         }
      }
   },

   mounted() {
      this.user = cloneDeep(this.apiUser);
   },

   methods: {
      setUserData(payload) {
         this.$emit('setUserData', this.user);
      }
   }
}

I apologize for any typos or mistakes made in the content above.

Answer 10

During the discussion in the comments, I came across a Vue 2 method that I find particularly useful in these scenarios. It essentially creates a roundtrip when updating a model.

Here is how it works:

Parent

<template>
   <div class="parent-root"
      <child :apiUser="apiUser" @setUserData="setUserData" />
   </div>
</template>

export default {
   data() {
      return {
         apiUser: {
            id: 'e134',
            age: 27
         }
      };
   },

   methods: {
      setUserData(payload) {
         this.$set(this.apiUser, 'age', payload);
      }
   }
}

Child

<template>
   <div class="child-root"
      {{ apiUser }}
   </div>
</template>

export default {
   props: {
      apiUser: {
         required: true,
         type: Object
      }
   },

   data() {
      return {
         user: null
      };
   },

   watch: {
      apiUser: {
         deep: true,
         handler() {
            this.user = cloneDeep(this.apiUser);
         }
      }
   },

   mounted() {
      this.user = cloneDeep(this.apiUser);
   },

   methods: {
      setUserData(payload) {
         this.$emit('setUserData', this.user);
      }
   }
}

I apologize for any typos or mistakes made in the content above.

Guide on how to link a prop to a ref in Vue 3.0 without modifying the prop value

Answer №1

Answer №2

Answer №3

Answer №4

The optimal approach, in my opinion, involves cloning the prop when initializing the local variable.

Answer №5

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