1084

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 doesn't work.

(Parse Error: Unexpected identifier)

function MyInput({isRequired}) {
  return <input classname="foo" {isRequired ? "required" : ""} />
}
1
  • 3
    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. Commented Feb 13, 2019 at 9:46

29 Answers 29

916

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.

10
  • 80
    Generally null means "act like I didn't specify it at all". For boolean dom attributes true/false is preferred over repeating the attribute name/false, e.g. <input disabled> compiles to React.createElement('input', {disabled: true})
    – Brigand
    Commented Jul 1, 2015 at 14:31
  • 13
    readonly never gets added because React is expecting the attribute readOnly (with a capital O).
    – Max
    Commented Feb 22, 2016 at 13:54
  • 17
    Thanks! Make sure the value is not just an empty string or zero, these may not get removed. For example, you could pass a value like this, and it should make sure it is removed if it evaluates to false: alt={props.alt || null}.
    – Jake
    Commented Jul 27, 2016 at 5:35
  • 11
    Thanks, @Jake. I had tried setting the attribute to false, but only null made sure the attribute was actually removed. Commented Jun 15, 2017 at 14:07
  • 3
    I am getting Warning: Received `false` for a non-boolean attribute `active`... Commented Apr 24, 2019 at 18:41
640

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' : null} />
);

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 null.

16
  • 6
    @Punit, The spread operator has a lower precedence than the conditional operator, so the condition is evaluated first, (either {bsStyle: 'success'} or {} results from it), then that object is spread. Commented Oct 17, 2017 at 13:21
  • 13
    Would the following do the same <Button bsStyle={ condition ? 'success': undefined } /> I find the syntax slightly easier, passing undefined will omit the property. Commented Jan 5, 2018 at 18:00
  • 4
    @AndyPolhill looks good to me and much easier to read the code, the only small difference is that in your code example inner component <Button/>'s props object will have a key bsStyle with value of undefined. Commented Jan 6, 2018 at 2:39
  • 4
    This was the only solution that allowed a radio button to not throw a warning on a checked value being set without an onChange set (even though checked was being set to false). So {...(checked ? {checked} : {})}. Thanks for the solution!
    – ScottS
    Commented Jan 23, 2021 at 18:27
  • 3
    Solved my problem! Thanks. Here is an example of how I used it, spreading React Native style classes conditionally: style={{...(isReadOnly ? styles.readOnly : {}), ...styles.baseStyle }} Then, inside the component, you can mix it into whatever styles the component is using: <Text style={...props.style, ...localStyleClass} >Hello World<Text>
    – C.T. Bell
    Commented Dec 23, 2021 at 14:23
619

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}
 />
1
  • 2
    If anyone cares about the precise semantics of this "sugar," you can look at the script that your .jsx is transpiled into you'll see that a function _extends has been added to it, which will (under normal circumstances) take the props constructed from the "normal attributes" and apply Object.assign(props, inputProps). Commented Jan 31, 2020 at 13:12
237

Here is an alternative.

var condition = true;

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

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

Or its inline version

var condition = true;

var component = (
  <div value="foo" {...(condition && { disabled: true })} /> 
);
7
  • 14
    I like this approach, it makes me cool among my workmates. Kidding aside, from the looks of it, the props are just passed as a key-value pair after all, is that correct?
    – JohnnyQ
    Commented Jan 31, 2017 at 9:30
  • 3
    If condition is false, this will try to expand/iterate over false, which I don't think is correct. Commented Feb 14, 2017 at 13:14
  • 4
    @LarsNyström, That makes sense. The spread operator accepts only iterables, where false is not one. Just check with Babel. This works with it when condition evaluates to false since the way Babel implements the operator. Though a trivial workaround could be ...( condition ? { disabled: true } : {} ), it becomes a bit verbose then. Thanks for this nice input!
    – Season
    Commented Feb 14, 2017 at 14:23
  • 1
    +1 This approach is required if you want to conditionally output data-* or aria-* attributes, as they're a special case in JSX, so data-my-conditional-attr={ false } will output data-my-conditional-attr="false" rather than omitting the attribute. facebook.github.io/react/docs/dom-elements.html
    – ptim
    Commented Jul 26, 2017 at 7:36
  • Any idea how to do this with an attribute that takes an array or jsx expression? columns={{ xs: 4, sm: 8, md: 12 }} say if I wanted to change the numbers based on a boolean?
    – Jared Rice
    Commented Jul 13, 2022 at 19:42
59

Here's a way I do it.

With a conditional:

<Label
    {...{
      text: label,
      type,
      ...(tooltip && { 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} />
2
  • Could you pls explain how this part works - ...(tooltip && { tooltip }),? It does work on component but when I try to use something like this in the code I get an error meaning that I try to spread non-iterable value
    – skwisgaar
    Commented Mar 16, 2020 at 20:17
  • 1
    probably because falseyValue && {} will return false, so its likely you are spreading on false eg ...(false). much better to use full expression so the spread continues to behave ...(condition ? {foo: 'bar'} : {}) Commented Apr 20, 2020 at 13:17
37

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

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

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

{...this.props.isTrue && {style : {color: 'red'}}}
23

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 && someValue} />
    );
  }
}
2
  • 3
    If someCondition is true but someValue is falsy (e.g. false itself, or 0, etc.), does the attribute still get included? This is important if it is necessary to explicitly include a falsy value, e.g. a 0 for a coordinate attribute, etc. Commented Dec 14, 2016 at 13:25
  • Normally, the attribute is omitted, but not in the case of data-* and aria-*, see my comment above. If you quote the value, or cast it as a String, the attribute will display: eg someAttr={ `${falsyValue}` } could render someAttr="false"
    – ptim
    Commented Jul 26, 2017 at 7:41
15

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}} />
2
13

Using undefined works for most properties:

const name = "someName";

return (
    <input name={name ? name : undefined} />
);
8

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>
    );
}
0
7

Given a local variable isRequired You can do the following in your render method (if using a class) or return statement (if using a function component):

 <MyComponent required={isRequired ? 'true' : undefined} />

In this case, the attribute will not be added if isRequired is undefined, false, or null (which is different from adding the attribute but setting it to 'false'.) Also note that I am using strings instead of booleans in order to avoid a warning message from react (Boolean value received on non-boolean attribute).

1
  • This one is solved the issue for me.
    – Or Assayag
    Commented Apr 20, 2023 at 5:16
4

For example using property styles for custom container

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

  return (
    <Container {...otherProps} >
4
  1. For some boolean attributes listed by React [1]:
<input disabled={disabled} />

// renders either `<input>` or `<input disabled>` 
  1. For other attributes:
<div aria-selected= {selected ? "" : undefined} />

// renders either `<div aria-selected></div>` or `<div></div>`

[1] The list of boolean attributes: https://github.com/facebook/react/blob/3f9480f0f5ceb5a32a3751066f0b8e9eae5f1b10/packages/react-dom/src/shared/DOMProperty.js#L318-L345

0
3

Indeed, you may conditionally add an attribute depending on a certain circumstance by using a ternary expression. On the other hand, you must use caution while using the proper JSX syntax.

  return <input className="foo" { ...(isRequired ? { required: true } : {}) } />
}

In this instance, the decision of whether or not to add the "required" property is made using a ternary expression. The "required" property will be applied to the element if isRequired is true. Nothing will be added in the event that isRequired is false. This is accomplished by inserting the attribute into JSX using the spread syntax (...).

As a result, properties will be added or removed based on the set criteria. In your scenario, the code will execute and add the "required" element if isRequired is true; otherwise, it will not add the attribute.

2

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>

2
  • 1
    This can get really messy with lots of attributes. I like the spread variant better.
    – Remi Sture
    Commented Jan 19, 2018 at 7:06
  • Yes Your are right but this is for someone who need to get overview I will make another example. But want keep things simple.
    – Juraj
    Commented Jan 19, 2018 at 17:08
2

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.

0
2

One way to do this is to use the spread operator (...) to pass a dynamic object of props to your component.

function MyInput({isRequired}) {
 const inputProps = { className: 'foo' };

 if (isRequired) {
  inputProps.required = true;
 }
 return <input {...inputProps} />;
}

In this code, we first create an object inputProps with the className property. Then, if the isRequired prop is true, we add the required property to the inputProps object. Finally, we pass the inputProps object to the input component using the spread operator (...), which will dynamically add the className and required attributes to the input element.

Note that we're not adding the readOnly attribute in this example, but you could add it in a similar way by checking another prop and conditionally adding it to the inputProps object.

1
<input checked={true} type="checkbox"  />
1
  • Some explanation would make this a better answer.
    – isherwood
    Commented Jan 6, 2022 at 14:42
1

In react functional component you can try something like this to omit unnecessary tag property.

<div className="something" ref={someCondition ? dummyRef : null} />

This works for me if I need to omit tags like ref, class, etc. But I don't know if that's work for every tag property

1
<Button {...(isWeb3Enabled ? {} : { isExternal: true })}>
    Metamask
</Button>
1

You must set as undefined the value for when you do not need the attribute Example:

<a data-tooltip={sidebarCollapsed?'Show details':undefined}></a>
1

The accepted answer did not work for me when using Apollo's MockedProvider component.

What worked:

const props = {
    mocks: mockedApolloResponse ?? undefined, // simplified logic to make this example shorter
};

return <MockedProvider {...props}>{children}</MockedProvider>

Extra context:

I think the reason is that while mocks is optional on MockedProviderProps, it has to be of type ReadonlyArray<MockedResponse> when present.

In my IDE, writing this line of code caused resulted in a warning:

<MockedProvider mocks={mockedApolloResponse ?? undefined}</MockedProvider>

The warning was: Type ReadonlyArray<MockedResponse> | undefined is not assignable to type ReadonlyArray<MockedResponse>.

1

Use the conditional (ternary) operator from JavaScript within the JSX expression to conditionally add attributes to a React component based on a certain circumstance. However, in order to add the attribute dynamically, you must use the proper syntax.

Your example has a syntax error since you are attempting to add the "required" property via string interpolation. Instead, you must encapsulate the attribute and its value in curly brackets '{}'.

This modified version of your code appropriately adds the "required" property to the input component in a conditional manner and adds it to the input component:function MyInput({ isRequired }) { return <input className="foo" {isRequired ? 'required' : null} />; } The "required" property is conditionally added in this code using the isRequired prop. The property is inserted with the value "required" if isRequired is true. Otherwise, the property is skipped if isRequired is false or null.

Keep in mind that classname, not class, is used to specify the element's CSS class.

If you need to conditionally add it depending on a different circumstance, be sure to handle the readOnly property in a manner that is comparable as well.

1

The answer by gian-marco-toso is technically correct in that the disabled attribute will not be rendered in HTML, but if you're using a framework like material-ui, the actual MUI components will still render in a disabled state since they are just looking for the existence of the disabled attribute on the JSX component rather than in the DOM :(

To get around this, I followed the advice from a comment by brigand. Simply set the disabled attribute to null instead of false.

export const MyInput = (props) => {
    ...
    const disableField = useMemo<boolean>(() => {
        // TODO : add complex logic here ...
        return true;
    }, []);
    ...
    return (
        <input
            disabled={disableField ? true : null}
        >
        </input>
    );
    ...
};
0

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" />
);
0

I think this may be useful for those who would like attribute's value to be a function:

import { RNCamera } from 'react-native-camera';
[...]

export default class MyView extends React.Component {

    _myFunction = (myObject) => {
        console.log(myObject.type); //
    }

    render() {

        var scannerProps = Platform.OS === 'ios' ? 
        {
            onBarCodeRead : this._myFunction
        } 
        : 
        { 
            // here you can add attribute(s) for other platforms
        }

        return (
            // it is just a part of code for MyView's layout
            <RNCamera 
                ref={ref => { this.camera = ref; }}
                style={{ flex: 1, justifyContent: 'flex-end', alignItems: 'center', }}
                type={RNCamera.Constants.Type.back}
                flashMode={RNCamera.Constants.FlashMode.on}
                {...scannerProps}
            />
        );
    }
}
0

in an easy way

const InputText= ({required = false , disabled = false, ...props}) => 
         (<input type="text" disabled={disabled} required={required} {...props} />);

and use it just like this

<InputText required disabled/>
0

If it is for a limited number of properties this will do


    function MyInput({isRequired}) {
        if (isRequired) {
            return <input classname="foo" isRequired={isRequired} />
        }
        return <input classname="foo" />
    }

If you have a large number of properties, it will be difficult to write if else conditions for every property and return accordingly. For that, you can push those properties in an object and use the spread operator in the returned element.

    function MyInput({ prop1, prop2, ...propN }) {
        const props = {};
        if (prop1) props.prop1 = prop1;
        .
        .
        .
        if (propN) props.propN = propN;
        return <input classname="foo" {...props} />
    }
-1

you can following thiss code :

interface InputComponentProps {
required?: boolean;
}

    const InputComponent = function(props:InputComponentProps ) {
    const {required} = props

    return (
         <input
                      required={required === true ? true : false }
                      type="text"
                      {...props}
                    />
    );
}

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