CODEWITHSUNDEEP
reactjs
myusuf
myusuf

Reputation: 12240

setState vs replaceState in React.js

I am new to React.js Library and I was going over some of the tutorials and I came across:

The Description given is not very clear (IMO).

setState is done to 'set' the state of a value, even if its already set 
in the 'getInitialState' function.

Similarly,

The replaceState() method is for when you want to clear out the values 
already in state, and add new ones.

I tried this.setState({data: someArray}); followed by this.replaceState({test: someArray}); and then console.logged them and I found that state now had both data and test.

Then, I tried this.setState({data: someArray}); followed by this.setState({test: someArray}); and then console.logged them and I found that state again had both data and test.

So, what exactly is the difference between the two ?

Upvotes: 103

Views: 89027

Answers (4)

stormpie
stormpie

Reputation: 25

Since replaceState is now deprecated, here is a very simple workaround. Though it is probably quite seldom that you would / should resort to needing this.

To remove the state:

for (const old_key of Object.keys(this.state))
    this.setState({ [old_key]: undefined });

Or, alternatively, if you don't want to have multiple calls to setState

const new_state = {};
for (const old_key of Object.keys(this.state))
    new_state[old_key] = undefined;
this.setState(new_state);

Essentially, all previous keys in the state now return undefined, which can very easily be filtered with an if statement: 

if (this.state.some_old_key) {
    // work with key
} else {
    // key is undefined (has been removed)
}

if ( ! this.state.some_old_key) {
    // key is undefined (has been removed)
} else {
    // work with key
}

In JSX:

render() {
    return <div>
        { this.state.some_old_key ? "The state remains" : "The state is gone" }
    </div>
}

Finally, to "replace" the state, simply combine the new state object with a copy of the old state that has been undefined, and set it as state: 

const new_state = {new_key1: "value1", new_key2: "value2"};
const state = this.state;

for (const old_key of Object.keys(state))
    state[old_key] = undefined;

for (const new_key of Object.keys(new_state))
    state[new_key] = new_state[new_key];

this.setState(state);

Upvotes: 2

Liran Brimer
Liran Brimer

Reputation: 3556

According to the docs, replaceState might get deprecated:

This method is not available on ES6 class components that extend React.Component. It may be removed entirely in a future version of React.

Upvotes: 14

Grav
Grav

Reputation: 1714

Definition by example:

// let's say that this.state is {foo: 42}

this.setState({bar: 117})

// this.state is now {foo: 42, bar: 117}

this.setState({foo: 43})

// this.state is now {foo: 43, bar: 117}

this.replaceState({baz: "hello"})

// this.state. is now {baz: "hello"}

Take note of this from the docs, though:

setState() does not immediately mutate this.state but creates a pending state transition. Accessing this.state after calling this method can potentially return the existing value.

Same goes for replaceState()

Upvotes: 49

Brigand
Brigand

Reputation: 86220

With setState the current and previous states are merged. With replaceState, it throws out the current state, and replaces it with only what you provide. Usually setState is used unless you really need to remove keys for some reason; but setting them to false/null is usually a more explicit tactic.

While it's possible it could change; replaceState currently uses the object passed as the state, i.e. replaceState(x), and once it's set this.state === x. This is a little lighter than setState, so it could be used as an optimization if thousands of components are setting their states frequently.
I asserted this with this test case.

If your current state is {a: 1}, and you call this.setState({b: 2}); when the state is applied, it will be {a: 1, b: 2}. If you called this.replaceState({b: 2}) your state would be {b: 2}.

Side note: State isn't set instantly, so don't do this.setState({b: 1}); console.log(this.state) when testing.

Upvotes: 153

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