Correct way to handle conditional styling in React

Question

I'm doing some React right now and I was wondering if there is a "correct" way to do conditional styling. In the tutorial they use

style={{
  textDecoration: completed ? 'line-through' : 'none'
}}

I prefer not to use inline styling so I want to instead use a class to control conditional styling. How would one approach this in the React way of thinking? Or should I just use this inline styling way?

Answer 1

 <div style={{ visibility: this.state.driverDetails.firstName != undefined? 'visible': 'hidden'}}></div>

Checkout the above code. That will do the trick.

Answer 2

If you need to conditionally apply inline styles (apply all or nothing) then this notation also works:

style={ someCondition ? { textAlign:'center', paddingTop: '50%'} : {}}

In case 'someCondition' not fulfilled then you pass empty object.

Answer 3

instead of this:

style={{
  textDecoration: completed ? 'line-through' : 'none'
}}

you could try the following using short circuiting:

style={{
  textDecoration: completed && 'line-through'
}}

https://codeburst.io/javascript-short-circuit-conditionals-bbc13ac3e9eb

key bit of information from the link:

Short circuiting means that in JavaScript when we are evaluating an AND expression (&&), if the first operand is false, JavaScript will short-circuit and not even look at the second operand.

It's worth noting that this would return false if the first operand is false, so might have to consider how this would affect your style.

The other solutions might be more best practice, but thought it would be worth sharing.

Answer 4

inline style handling

style={{backgroundColor: selected ? 'red':'green'}}

using Css

in js

className={`section ${selected && 'section_selected'}`}

in css

.section {
  display: flex;
  align-items: center;
} 
.section_selected{
  background-color: whitesmoke;
  border-width: 3px !important;
}

same can be done with Js stylesheets

Answer 5

Another way, using inline style and the spread operator

style={{
  ...completed ? { textDecoration: completed } : {}
}}

That way be useful in some situations where you want to add a bunch of properties at the same time base on the condition.

Answer 6

First, I agree with you as a matter of style - I would also (and do also) conditionally apply classes rather than inline styles. But you can use the same technique:

<div className={{completed ? "completed" : ""}}></div>

For more complex sets of state, accumulate an array of classes and apply them:

var classes = [];

if (completed) classes.push("completed");
if (foo) classes.push("foo");
if (someComplicatedCondition) classes.push("bar");

return <div className={{classes.join(" ")}}></div>;

Answer 7

 style={{
              whiteSpace: "unset",
              wordBreak: "break-all",
              backgroundColor: one.read == false && "#e1f4f3",
              borderBottom:'0.8px solid #fefefe'
            }}

Answer 8

If you want assign styles based on condition, its better you use a class name for styles. For this assignment, there are different ways. These are two of them.

1.

    <div className={`another-class ${condition ? 'active' : ''}`} />

    <div className={`another-class ${condition && 'active'}`} />

Answer 9

I came across this question while trying to answer the same question. McCrohan's approach with the classes array & join is solid.

Through my experience, I have been working with a lot of legacy ruby code that is being converted to React and as we build the component(s) up I find myself reaching out for both existing css classes and inline styles.

example snippet inside a component:

// if failed, progress bar is red, otherwise green 
<div
    className={`progress-bar ${failed ? 'failed' : ''}`}
    style={{ width: this.getPercentage() }} 
/>

Again, I find myself reaching out to legacy css code, "packaging" it with the component and moving on.

So, I really feel that it is a bit in the air as to what is "best" as that label will vary greatly depending on your project.

Answer 10

The best way to handle styling is by using classes with set of css properties.

example:

<Component className={this.getColor()} />

getColor() {
    let class = "badge m2";
    class += this.state.count===0 ? "warning" : danger;
    return class;
}

Answer 11

You can use somthing like this.

render () {
    var btnClass = 'btn';
    if (this.state.isPressed) btnClass += ' btn-pressed';
    else if (this.state.isHovered) btnClass += ' btn-over';
    return <button className={btnClass}>{this.props.label}</button>;
  }

Or else, you can use classnames NPM package to make dynamic and conditional className props simpler to work with (especially more so than conditional string manipulation).

classNames('foo', 'bar'); // => 'foo bar'
classNames('foo', { bar: true }); // => 'foo bar'
classNames({ 'foo-bar': true }); // => 'foo-bar'
classNames({ 'foo-bar': false }); // => ''
classNames({ foo: true }, { bar: true }); // => 'foo bar'
classNames({ foo: true, bar: true }); // => 'foo bar'

Answer 12

In case someone uses Typescript (which does not except null values for style) and wants to use react styling, I would suggest this hack:

<p 
style={choiceStyle ? styles.choiceIsMade : styles.none}>
{question}
</p>

const styles = {
choiceIsMade: {...},
 none: {}
}

Answer 13

Change Inline CSS Conditionally Based on Component State.

This is also the Correct way to handle conditional styling in React.

condition ? expressionIfTrue : expressionIfFalse;

example =>

{this.state.input.length > 15 ? inputStyle={border: '3px solid red'}: inputStyle }

This code means that if the character is more than 15 entered in the input field, then our input field's border will be red and the length of the border will be 3px.