javascript – How do I conditionally add attributes to React components?

The Question :

648 people think this question is useful

Is there a way to only add attributes to a React component if a certain condition is met?

I’m supposed to add required and readOnly attributes to form elements based on an Ajax call after render, but I can’t see how to solve this since readOnly=”false” is not the same as omitting the attribute completely.

The example below should explain what I want, but it won’t work (Parse Error: Unexpected identifier).

var React = require('React');

var MyOwnInput = React.createClass({
    render: function () {
        return (
            <div>
                <input type="text" onChange={this.changeValue} value={this.getValue()} name={this.props.name}/>
            </div>
        );
    }
});

module.exports = React.createClass({
    getInitialState: function () {
        return {
            isRequired: false
        }
    },
    componentDidMount: function () {
        this.setState({
            isRequired: true
        });
    },
    render: function () {
        var isRequired = this.state.isRequired;

        return (
            <MyOwnInput name="test" {isRequired ? "required" : ""} />
        );
    }
});

The Question Comments :
  • May be one comment help someone, i found out React 16.7 doesnt rerenders and update the component’s html attributes if you changed only them in a store (f.e. redux) and tied to component. This means the component has f.e.aria-modal=true, you push the changes (to false) to the store of aria/data attributes, but nothing else is changed (such as component’s content or class or variables in there) as the result ReactJs will not update aria/data attrs in that components. I’ve been messing around about whole day to realise that.

The Answer 1

609 people think this answer is useful

Apparently, for certain attributes, React is intelligent enough to omit the attribute if the value you pass to it is not truthy. For example:

const InputComponent = function() {
    const required = true;
    const disabled = false;

    return (
        <input type="text" disabled={disabled} required={required} />
    );
}

will result in:

<input type="text" required>

Update: if anyone is curious as to how/why this happens, you can find details in ReactDOM’s source code, specifically at lines 30 and 167 of the DOMProperty.js file.

The Answer 2

441 people think this answer is useful

juandemarco’s answer is usually correct, but here is another option.

Build an object how you like:

var inputProps = {
  value: 'foo',
  onChange: this.handleChange
};

if (condition)
  inputProps.disabled = true;

Render with spread, optionally passing other props also.

<input
    value="this is overridden by inputProps"
    {...inputProps}
    onChange={overridesInputProps}
 />

The Answer 3

352 people think this answer is useful

Here is an example of using Bootstrap‘s Button via React-Bootstrap (version 0.32.4):

var condition = true;

return (
  <Button {...(condition ? {bsStyle: 'success'} : {})} />
);

Depending on the condition, either {bsStyle: 'success'} or {} will be returned. The spread operator will then spread the properties of the returned object to Button component. In the falsy case, since no properties exist on the returned object, nothing will be passed to the component.


An alternative way based on Andy Polhill’s comment:

var condition = true;

return (
  <Button bsStyle={condition ? 'success' : undefined} />
);

The only small difference is that in the second example the inner component <Button/>‘s props object will have a key bsStyle with a value of undefined.

The Answer 4

105 people think this answer is useful

Here is an alternative.

var condition = true;

var props = {
  value: 'foo',
  ...( condition &amp;&amp; { disabled: true } )
};

var component = <div { ...props } />;

Or its inline version

var condition = true;

var component = (
  <div
    value="foo"
    { ...( condition &amp;&amp; { disabled: true } ) } />
);

The Answer 5

44 people think this answer is useful

Here’s a way I do it.

With a conditional:

<Label
    {...{
      text: label,
      type,
      ...(tooltip &amp;&amp; { tooltip }),
      isRequired: required
    }}
/>

I still prefer using the regular way of passing props down, because it is more readable (in my opinion) in the case of not have any conditionals.

Without a conditional:

<Label text={label} type={type} tooltip={tooltip} isRequired={required} />

The Answer 6

26 people think this answer is useful

Let’s say we want to add a custom property (using aria-* or data-*) if a condition is true:

{...this.props.isTrue &amp;&amp; {'aria-name' : 'something here'}}

Let’s say we want to add a style property if a condition is true:

{...this.props.isTrue &amp;&amp; {style : {color: 'red'}}}

The Answer 7

20 people think this answer is useful

You can use the same shortcut, which is used to add/remove (parts of) components ({isVisible && <SomeComponent />}).

class MyComponent extends React.Component {
  render() {
    return (
      <div someAttribute={someCondition &amp;&amp; someValue} />
    );
  }
}

The Answer 8

13 people think this answer is useful

If you use ECMAScript 6, you can simply write like this.

// First, create a wrap object.
const wrap = {
    [variableName]: true
}
// Then, use it
<SomeComponent {...{wrap}} />

The Answer 9

8 people think this answer is useful

This should work, since your state will change after the Ajax call, and the parent component will re-render.

render : function () {
    var item;
    if (this.state.isRequired) {
        item = <MyOwnInput attribute={'whatever'} />
    } else {
        item = <MyOwnInput />
    }
    return (
        <div>
            {item}
        </div>
    );
}

The Answer 10

3 people think this answer is useful

In React you can conditionally render Components, but also their attributes, like props, className, id, and more.

In React it’s very good practice to use the ternary operator which can help you conditionally render Components.

An example also shows how to conditionally render Component and its style attribute.

Here is a simple example:

class App extends React.Component {
  state = {
    isTrue: true
  };

  render() {
    return (
      <div>
        {this.state.isTrue ? (
          <button style={{ color: this.state.isTrue ? "red" : "blue" }}>
            I am rendered if TRUE
          </button>
        ) : (
          <button>I am rendered if FALSE</button>
        )}
      </div>
    );
  }
}

ReactDOM.render(<App />, document.getElementById("root"));
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/15.1.0/react.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/15.1.0/react-dom.min.js"></script>

<div id="root"></div>

The Answer 11

1 people think this answer is useful

For example using property styles for custom container

const DriverSelector = props => {
  const Container = props.container;
  const otherProps = {
    ...( props.containerStyles &amp;&amp; { style: props.containerStyles } )
  };

  return (
    <Container {...otherProps} >

The Answer 12

0 people think this answer is useful

Considering the post JSX In Depth, you can solve your problem this way:

if (isRequired) {
  return (
    <MyOwnInput name="test" required='required' />
  );
}
return (
    <MyOwnInput name="test" />
);

The Answer 13

0 people think this answer is useful

From my point of view the best way to manage multiple conditional props is the props object approach from @brigand. But it can be improved in order to avoid adding one if block for each conditional prop.

The ifVal helper

rename it as you like (iv, condVal, cv, _, …)

You can define a helper function to return a value, or another, if a condition is met:

// components-helpers.js
export const ifVal = (cond, trueValue=true, falseValue=null) => {
  return cond ? trueValue : falseValue
}

If cond is true (or truthy), the trueValue is returned – or true. If cond is false (or falsy), the falseValue is returned – or null.

These defaults (true and null) are, usually the right values to allow a prop to be passed or not to a React component. You can think to this function as an “improved React ternary operator”. Please improve it if you need more control over the returned values.

Let’s use it with many props.

Build the (complex) props object

// your-code.js
import { ifVal } from './components-helpers.js'

// BE SURE to replace all true/false with a real condition in you code
// this is just an example

const inputProps = {
  value: 'foo',
  enabled: ifVal(true), // true
  noProp: ifVal(false), // null - ignored by React
  aProp: ifVal(true, 'my value'), // 'my value'
  bProp: ifVal(false, 'the true text', 'the false text') // 'my false value',
  onAction: ifVal(isGuest, handleGuest, handleUser) // it depends on isGuest value
};

 <MyComponent {...inputProps} />

This approach is something similar to the popular way to conditionally manage classes using the classnames utility, but adapted to props.

Why you should use this approach

You’ll have a clean and readable syntax, even with many conditional props: every new prop just add a line of code inside the object declaration.

In this way you replace the syntax noise of repeated operators (..., &&, ? :, …), that can be very annoying when you have many props, with a plain function call.

Our top priority, as developers, is to write the most obvious code that solve a problem. Too many times we solve problems for our ego, adding complexity where it’s not required. Our code should be straightforward, for us today, for us tomorrow and for our mates.

just because we can do something doesn’t mean we should

I hope this late reply will help.

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