Why can't I change my input value in React even with the onChange listener

Question

I am quite new to React and after going through some tutorials, I was trying the below code of mine.

I made one component, passed props to it from a store, on componentWillMount I make a new state for component. Rendering is fine till now.

Next I bound my state to value of an input box and I have onChange listener as well. Still, I can't change my values in the field.

Since, I am from Angular background, I am assuming binding input's value to state like below will automatically update the property name in state object. Am I wrong here?

componentWillMount(){
    this.setState({
        updatable : false,
        name : this.props.name,
        status : this.props.status
    });
}

//relevant DOM from component's render function
<input className="form-control" type="text" value={this.state.name} id={'todoName' + this.props.id} onChange={this.onTodoChange.bind(this)}/>

onTodoChange(){
    console.log(this);
    //consoling 'this' here, shows old values only.
    //not sure how and even if I need to update state here.
    // Do I need to pass new state to this function from DOM
    //TODO: send new data to store
}

My onTodoChange function console the value of this which has same value of state as while initializing. How do I go about changing state by typing in input boxes, so that I can send them to the stores?

Answer 1

Unlike in the case of Angular, in React.js you need to update the state manually. You can do something like this:

<input
    className="form-control"
    type="text" value={this.state.name}
    id={'todoName' + this.props.id}
    onChange={e => this.onTodoChange(e.target.value)}
/>

And then in the function:

onTodoChange(value){
        this.setState({
             name: value
        });
    }

Also, you can set the initial state in the constructor of the component:

  constructor (props) {
    super(props);
    this.state = {
        updatable: false,
        name: props.name,
        status: props.status
    };
  }