List Rendering
v-for
We can use the v-for
directive to render a list of items based on an array. The v-for
directive requires a special syntax in the form of item in items
, where items
is the source data array and item
is an alias for the array element being iterated on:
js
const items = ref([{ message: 'Foo' }, { message: 'Bar' }])
template
<li v-for="item in items">
{{ item.message }}
</li>
Inside the v-for
scope, template expressions have access to all parent scope properties. In addition, v-for
also supports an optional second alias for the index of the current item:
js
const parentMessage = ref('Parent')
const items = ref([{ message: 'Foo' }, { message: 'Bar' }])
template
<li v-for="(item, index) in items">
{{ parentMessage }} - {{ index }} - {{ item.message }}
</li>
The variable scoping of v-for
is similar to the following JavaScript:
js
const parentMessage = 'Parent'
const items = [
/* ... */
]
items.forEach((item, index) => {
// has access to outer scope `parentMessage`
// but `item` and `index` are only available in here
console.log(parentMessage, item.message, index)
})
Notice how the v-for
value matches the function signature of the forEach
callback. In fact, you can use destructuring on the v-for
item alias similar to destructuring function arguments:
template
<li v-for="{ message } in items">
{{ message }}
</li>
<!-- with index alias -->
<li v-for="({ message }, index) in items">
{{ message }} {{ index }}
</li>
For nested v-for
, scoping also works similar to nested functions. Each v-for
scope has access to parent scopes:
template
<li v-for="item in items">
<span v-for="childItem in item.children">
{{ item.message }} {{ childItem }}
</span>
</li>
You can also use of
as the delimiter instead of in
, so that it is closer to JavaScript's syntax for iterators:
template
<div v-for="item of items"></div>
v-for
with an Object
You can also use v-for
to iterate through the properties of an object. The iteration order will be based on the result of calling Object.keys()
on the object:
js
const myObject = reactive({
title: 'How to do lists in Vue',
author: 'Jane Doe',
publishedAt: '2016-04-10'
})
template
<ul>
<li v-for="value in myObject">
{{ value }}
</li>
</ul>
You can also provide a second alias for the property's name (a.k.a. key):
template
<li v-for="(value, key) in myObject">
{{ key }}: {{ value }}
</li>
And another for the index:
template
<li v-for="(value, key, index) in myObject">
{{ index }}. {{ key }}: {{ value }}
</li>
v-for
with a Range
v-for
can also take an integer. In this case it will repeat the template that many times, based on a range of 1...n
.
template
<span v-for="n in 10">{{ n }}</span>
Note here n
starts with an initial value of 1
instead of 0
.
v-for
on <template>
Similar to template v-if
, you can also use a <template>
tag with v-for
to render a block of multiple elements. For example:
template
<ul>
<template v-for="item in items">
<li>{{ item.msg }}</li>
<li class="divider" role="presentation"></li>
</template>
</ul>
v-for
with v-if
Note
It's not recommended to use v-if
and v-for
on the same element due to implicit precedence. Refer to style guide for details.
When they exist on the same node, v-if
has a higher priority than v-for
. That means the v-if
condition will not have access to variables from the scope of the v-for
:
template
<!--
This will throw an error because property "todo"
is not defined on instance.
-->
<li v-for="todo in todos" v-if="!todo.isComplete">
{{ todo.name }}
</li>
This can be fixed by moving v-for
to a wrapping <template>
tag (which is also more explicit):
template
<template v-for="todo in todos">
<li v-if="!todo.isComplete">
{{ todo.name }}
</li>
</template>
Maintaining State with key
When Vue is updating a list of elements rendered with v-for
, by default it uses an "in-place patch" strategy. If the order of the data items has changed, instead of moving the DOM elements to match the order of the items, Vue will patch each element in-place and make sure it reflects what should be rendered at that particular index.
This default mode is efficient, but only suitable when your list render output does not rely on child component state or temporary DOM state (e.g. form input values).
To give Vue a hint so that it can track each node's identity, and thus reuse and reorder existing elements, you need to provide a unique key
attribute for each item:
template
<div v-for="item in items" :key="item.id">
<!-- content -->
</div>
When using <template v-for>
, the key
should be placed on the <template>
container:
template
<template v-for="todo in todos" :key="todo.name">
<li>{{ todo.name }}</li>
</template>
Note
key
here is a special attribute being bound with v-bind
. It should not be confused with the property key variable when using v-for
with an object.
It is recommended to provide a key
attribute with v-for
whenever possible, unless the iterated DOM content is simple (i.e. contains no components or stateful DOM elements), or you are intentionally relying on the default behavior for performance gains.
The key
binding expects primitive values - i.e. strings and numbers. Do not use objects as v-for
keys. For detailed usage of the key
attribute, please see the key
API documentation.
v-for
with a Component
This section assumes knowledge of Components. Feel free to skip it and come back later.
You can directly use v-for
on a component, like any normal element (don't forget to provide a key
):
template
<MyComponent v-for="item in items" :key="item.id" />
However, this won't automatically pass any data to the component, because components have isolated scopes of their own. In order to pass the iterated data into the component, we should also use props:
template
<MyComponent
v-for="(item, index) in items"
:item="item"
:index="index"
:key="item.id"
/>
The reason for not automatically injecting item
into the component is because that makes the component tightly coupled to how v-for
works. Being explicit about where its data comes from makes the component reusable in other situations.
Check out this example of a simple todo list to see how to render a list of components using v-for
, passing different data to each instance.
Array Change Detection
Mutation Methods
Vue is able to detect when a reactive array's mutation methods are called and trigger necessary updates. These mutation methods are:
push()
pop()
shift()
unshift()
splice()
sort()
reverse()
Replacing an Array
Mutation methods, as the name suggests, mutate the original array they are called on. In comparison, there are also non-mutating methods, e.g. filter()
, concat()
and slice()
, which do not mutate the original array but always return a new array. When working with non-mutating methods, we should replace the old array with the new one:
js
// `items` is a ref with array value
items.value = items.value.filter((item) => item.message.match(/Foo/))
You might think this will cause Vue to throw away the existing DOM and re-render the entire list - luckily, that is not the case. Vue implements some smart heuristics to maximize DOM element reuse, so replacing an array with another array containing overlapping objects is a very efficient operation.
Displaying Filtered/Sorted Results
Sometimes we want to display a filtered or sorted version of an array without actually mutating or resetting the original data. In this case, you can create a computed property that returns the filtered or sorted array.
For example:
js
const numbers = ref([1, 2, 3, 4, 5])
const evenNumbers = computed(() => {
return numbers.value.filter((n) => n % 2 === 0)
})
template
<li v-for="n in evenNumbers">{{ n }}</li>
In situations where computed properties are not feasible (e.g. inside nested v-for
loops), you can use a method:
js
const sets = ref([
[1, 2, 3, 4, 5],
[6, 7, 8, 9, 10]
])
function even(numbers) {
return numbers.filter((number) => number % 2 === 0)
}
template
<ul v-for="numbers in sets">
<li v-for="n in even(numbers)">{{ n }}</li>
</ul>
Be careful with reverse()
and sort()
in a computed property! These two methods will mutate the original array, which should be avoided in computed getters. Create a copy of the original array before calling these methods:
diff
- return numbers.reverse()
+ return [...numbers].reverse()