问题
How do I optionally include an element in JSX? Here is an example using a banner that should be in the component if it has been passed in. What I want to avoid is having to duplicate HTML tags in the if statement.
render: function () {
var banner;
if (this.state.banner) {
banner = <div id="banner">{this.state.banner}</div>;
} else {
banner = ?????
}
return (
<div id="page">
{banner}
<div id="other-content">
blah blah blah...
</div>
</div>
);
}
回答1:
Just leave banner as being undefined and it does not get included.
回答2:
What about this. Let's define a simple helping If
component.
var If = React.createClass({
render: function() {
if (this.props.test) {
return this.props.children;
}
else {
return false;
}
}
});
And use it this way:
render: function () {
return (
<div id="page">
<If test={this.state.banner}>
<div id="banner">{this.state.banner}</div>
</If>
<div id="other-content">
blah blah blah...
</div>
</div>
);
}
UPDATE: As my answer is getting popular, I feel obligated to warn you about the biggest danger related to this solution. As pointed out in another answer, the code inside the <If />
component is executed always regardless of whether the condition is true or false. Therefore the following example will fail in case the banner
is null
(note the property access on the second line):
<If test={this.state.banner}>
<div id="banner">{this.state.banner.url}</div>
</If>
You have to be careful when you use it. I suggest reading other answers for alternative (safer) approaches.
UPDATE 2: Looking back, this approach is not only dangerous but also desperately cumbersome. It's a typical example of when a developer (me) tries to transfer patterns and approaches he knows from one area to another but it doesn't really work (in this case other template languages).
If you need a conditional element, do it like this:
render: function () {
return (
<div id="page">
{this.state.banner &&
<div id="banner">{this.state.banner}</div>}
<div id="other-content">
blah blah blah...
</div>
</div>
);
}
If you also need the else branch, just use a ternary operator:
{this.state.banner ?
<div id="banner">{this.state.banner}</div> :
<div>There is no banner!</div>
}
It's way shorter, more elegant and safe. I use it all the time. The only disadvantage is that you cannot do else if
branching that easily but that is usually not that common.
Anyway, this is possible thanks to how logical operators in JavaScript work. The logical operators even allow little tricks like this:
<h3>{this.state.banner.title || 'Default banner title'}</h3>
回答3:
Personally, I really think the ternary expressions show in (JSX In Depth) are the most natural way that conforms with the ReactJs standards.
See the following example. It's a little messy at first sight but works quite well.
<div id="page">
{this.state.banner ? (
<div id="banner">
<div class="another-div">
{this.state.banner}
</div>
</div>
) :
null}
<div id="other-content">
blah blah blah...
</div>
</div>
回答4:
You may also write it like
{ this.state.banner && <div>{...}</div> }
If your state.banner
is null
or undefined
, the right side of the condition is skipped.
回答5:
The If
style component is dangerous because the code block is always executed regardless of the condition. For example, this would cause a null exception if banner
is null
:
//dangerous
render: function () {
return (
<div id="page">
<If test={this.state.banner}>
<img src={this.state.banner.src} />
</If>
<div id="other-content">
blah blah blah...
</div>
</div>
);
}
Another option is to use an inline function (especially useful with else statements):
render: function () {
return (
<div id="page">
{function(){
if (this.state.banner) {
return <div id="banner">{this.state.banner}</div>
}
}.call(this)}
<div id="other-content">
blah blah blah...
</div>
</div>
);
}
Another option from react issues:
render: function () {
return (
<div id="page">
{ this.state.banner &&
<div id="banner">{this.state.banner}</div>
}
<div id="other-content">
blah blah blah...
</div>
</div>
);
}
回答6:
&& + code-style + small components
This simple test syntax + code-style convention + small focused components is for me the most readable option out there. You just need to take special care of falsy values like false
, 0
or ""
.
render: function() {
var person= ...;
var counter= ...;
return (
<div className="component">
{person && (
<Person person={person}/>
)}
{(typeof counter !== 'undefined') && (
<Counter value={counter}/>
)}
</div>
);
}
do notation
ES7 stage-0 do notation syntax is also very nice and I'll definitively use it when my IDE supports it correctly:
const Users = ({users}) => (
<div>
{users.map(user =>
<User key={user.id} user={user}/>
)}
</div>
)
const UserList = ({users}) => do {
if (!users) <div>Loading</div>
else if (!users.length) <div>Empty</div>
else <Users users={users}/>
}
More details here: ReactJs - Creating an "If" component... a good idea?
回答7:
Simple, create a function.
renderBanner: function() {
if (!this.state.banner) return;
return (
<div id="banner">{this.state.banner}</div>
);
},
render: function () {
return (
<div id="page">
{this.renderBanner()}
<div id="other-content">
blah blah blah...
</div>
</div>
);
}
This is a pattern I personally follow all the time. Makes code really clean and easy to understand. What's more it allows you to refactor Banner
into its own component if it gets too large (or re-used in other places).
回答8:
The experimental ES7 do
syntax makes this easy. If you're using Babel, enable the es7.doExpressions
feature then:
render() {
return (
<div id="banner">
{do {
if (this.state.banner) {
this.state.banner;
} else {
"Something else";
}
}}
</div>
);
}
See http://wiki.ecmascript.org/doku.php?id=strawman:do_expressions
回答9:
As already mentioned in the answers, JSX presents you with two options
Ternary operator
{ this.state.price ? <div>{this.state.price}</div> : null }
Logical conjunction
{ this.state.price && <div>{this.state.price}</div> }
However, those don't work for price == 0
.
JSX will render the false branch in the first case and in case of logical conjunction, nothing will be rendered. If the property may be 0, just use if statements outside of your JSX.
回答10:
This component works when you have more than one element inside "if" branch:
var Display = React.createClass({
render: function() {
if (!this.props.when) {
return false;
}
return React.DOM.div(null, this.props.children);
}
});
Usage:
render: function() {
return (
<div>
<Display when={this.state.loading}>
Loading something...
<div>Elem1</div>
<div>Elem2</div>
</Display>
<Display when={!this.state.loading}>
Loaded
<div>Elem3</div>
<div>Elem4</div>
</Display>
</div>
);
},
P.s. someone think that this components are not good for code reading. But in my mind html with javascript is worse
回答11:
Most examples are with one line of "html" that is rendered conditionally. This seems readable for me when I have multiple lines that needs to be rendered conditionally.
render: function() {
// This will be renered only if showContent prop is true
var content =
<div>
<p>something here</p>
<p>more here</p>
<p>and more here</p>
</div>;
return (
<div>
<h1>Some title</h1>
{this.props.showContent ? content : null}
</div>
);
}
First example is good because instead of null
we can conditionally render some other content like {this.props.showContent ? content : otherContent}
But if you just need to show/hide content this is even better since Booleans, Null, and Undefined Are Ignored
render: function() {
return (
<div>
<h1>Some title</h1>
// This will be renered only if showContent prop is true
{this.props.showContent &&
<div>
<p>something here</p>
<p>more here</p>
<p>and more here</p>
</div>
}
</div>
);
}
回答12:
There is another solution, if component for React:
var Node = require('react-if-comp');
...
render: function() {
return (
<div id="page">
<Node if={this.state.banner}
then={<div id="banner">{this.state.banner}</div>} />
<div id="other-content">
blah blah blah...
</div>
</div>
);
}
回答13:
I use a more explicit shortcut: A Immediately-Invoked Function Expression (IIFE):
{(() => {
if (isEmpty(routine.queries)) {
return <Grid devices={devices} routine={routine} configure={() => this.setState({configured: true})}/>
} else if (this.state.configured) {
return <DeviceList devices={devices} routine={routine} configure={() => this.setState({configured: false})}/>
} else {
return <Grid devices={devices} routine={routine} configure={() => this.setState({configured: true})}/>
}
})()}
回答14:
I made https://www.npmjs.com/package/jsx-control-statements to make it a bit easier, basically it allows you to define <If>
conditionals as tags and then compiles them into ternary ifs so that the code inside the <If>
only gets executed if the condition is true.
回答15:
There is also a really clean one line version... { this.props.product.title || "No Title" }
Ie:
render: function() {
return (
<div className="title">
{ this.props.product.title || "No Title" }
</div>
);
}
回答16:
I made https://github.com/ajwhite/render-if recently to safely render elements only if the predicate passes.
{renderIf(1 + 1 === 2)(
<span>Hello!</span>
)}
or
const ifUniverseIsWorking = renderIf(1 + 1 === 2);
//...
{ifUniverseIsWorking(
<span>Hello!</span>
)}
回答17:
You can conditionally include elements using the ternary operator like so:
render: function(){
return <div id="page">
//conditional statement
{this.state.banner ? <div id="banner">{this.state.banner}</div> : null}
<div id="other-content">
blah blah blah...
</div>
</div>
}
回答18:
You can use a function and return the component and keep thin the render function
class App extends React.Component {
constructor (props) {
super(props);
this._renderAppBar = this._renderAppBar.bind(this);
}
render () {
return <div>
{_renderAppBar()}
<div>Content</div>
</div>
}
_renderAppBar () {
if (this.state.renderAppBar) {
return <AppBar />
}
}
}
回答19:
Here is my approach using ES6.
import React, { Component } from 'react';
// you should use ReactDOM.render instad of React.renderComponent
import ReactDOM from 'react-dom';
class ToggleBox extends Component {
constructor(props) {
super(props);
this.state = {
// toggle box is closed initially
opened: false,
};
// http://egorsmirnov.me/2015/08/16/react-and-es6-part3.html
this.toggleBox = this.toggleBox.bind(this);
}
toggleBox() {
// check if box is currently opened
const { opened } = this.state;
this.setState({
// toggle value of `opened`
opened: !opened,
});
}
render() {
const { title, children } = this.props;
const { opened } = this.state;
return (
<div className="box">
<div className="boxTitle" onClick={this.toggleBox}>
{title}
</div>
{opened && children && (
<div class="boxContent">
{children}
</div>
)}
</div>
);
}
}
ReactDOM.render((
<ToggleBox title="Click me">
<div>Some content</div>
</ToggleBox>
), document.getElementById('app'));
Demo: http://jsfiddle.net/kb3gN/16688/
I'm using code like:
{opened && <SomeElement />}
That will render SomeElement
only if opened
is true. It works because of the way how JavaScript resolve logical conditions:
true && true && 2; // will output 2
true && false && 2; // will output false
true && 'some string'; // will output 'some string'
opened && <SomeElement />; // will output SomeElement if `opened` is true, will output false otherwise
As React
will ignore false
, I find it very good way to conditionally render some elements.
回答20:
Maybe it helps someone who comes across the question: All the Conditional Renderings in React It's an article about all the different options for conditional rendering in React.
Key takeaways of when to use which conditional rendering:
** if-else
- is the most basic conditional rendering
- beginner friendly
- use if to opt-out early from a render method by returning null
** ternary operator
- use it over an if-else statement
- it is more concise than if-else
** logical && operator
- use it when one side of the ternary operation would return null
** switch case
- verbose
- can only be inlined with self invoking function
- avoid it, use enums instead
** enums
- perfect to map different states
- perfect to map more than one condition
** multi-level/nested conditional renderings
- avoid them for the sake of readability
- split up components into more lightweight components with their own simple conditional rendering
- use HOCs
** HOCs
- use them to shield away conditional rendering
- components can focus on their main purpose
** external templating components
- avoid them and be comfortable with JSX and JavaScript
回答21:
With ES6 you can do it with a simple one-liner
const If = ({children, show}) => show ? children : null
"show" is a boolean and you use this class by
<If show={true}> Will show </If>
<If show={false}> WON'T show </div> </If>
回答22:
I don't think this has been mentioned. This is like your own answer but I think it's even simpler. You can always return strings from the expressions and you can nest jsx inside expressions, so this allows for an easy to read inline expression.
render: function () {
return (
<div id="page">
{this.state.banner ? <div id="banner">{this.state.banner}</div> : ''}
<div id="other-content">
blah blah blah...
</div>
</div>
);
}
<script src="http://dragon.ak.fbcdn.net/hphotos-ak-xpf1/t39.3284-6/10574688_1565081647062540_1607884640_n.js"></script>
<script src="http://dragon.ak.fbcdn.net/hphotos-ak-xpa1/t39.3284-6/10541015_309770302547476_509859315_n.js"></script>
<script type="text/jsx;harmony=true">void function() { "use strict";
var Hello = React.createClass({
render: function() {
return (
<div id="page">
{this.props.banner ? <div id="banner">{this.props.banner}</div> : ''}
<div id="other-content">
blah blah blah...
</div>
</div>
);
}
});
var element = <div><Hello /><Hello banner="banner"/></div>;
React.render(element, document.body);
}()</script>
回答23:
I like the explicitness of Immediately-Invoked Function Expressions (IIFE
) and if-else
over render callbacks
and ternary operators
.
render() {
return (
<div id="page">
{(() => (
const { banner } = this.state;
if (banner) {
return (
<div id="banner">{banner}</div>
);
}
// Default
return (
<div>???</div>
);
))()}
<div id="other-content">
blah blah blah...
</div>
</div>
);
}
You just need to get acquainted of the IIFE
syntax, {expression}
is the usual React syntax, inside it just consider that you're writing a function that is invoking itself.
function() {
}()
that need to be wrapped inside parens
(function() {
}())
回答24:
There is also a technique using render props to conditional render a component. It's benefit is that the render wouldn't evaluate until the condition is met, resulting in no worries for null and undefined values.
const Conditional = ({ condition, render }) => {
if (condition) {
return render();
}
return null;
};
class App extends React.Component {
constructor() {
super();
this.state = { items: null }
}
componentWillMount() {
setTimeout(() => { this.setState({ items: [1,2] }) }, 2000);
}
render() {
return (
<Conditional
condition={!!this.state.items}
render={() => (
<div>
{this.state.items.map(value => <p>{value}</p>)}
</div>
)}
/>
)
}
}
回答25:
When having to only render something if passed condition is fullfilled, you can use syntax:
{ condition && what_to_render }
The code in this manner would look like this :
render() {
const { banner } = this.state;
return (
<div id="page">
{ banner && <div id="banner">{banner}</div> }
<div id="other-content">
blah blah blah...
</div>
</div>
);
}
There are, of course, other valid ways to do this, it's all up to preferences and the occassion. You can learn more ways on how to do conditional rendering in React in this article if you're interested!
回答26:
Just to add another option - if you like/tolerate coffee-script you can use coffee-react to write your JSX in which case if/else statements are usable as they are expressions in coffee-script and not statements:
render: ->
<div className="container">
{
if something
<h2>Coffeescript is magic!</h2>
else
<h2>Coffeescript sucks!</h2>
}
</div>
回答27:
Just to extend @Jack Allan answer with references to docs.
React basic (Quick Start) documentation suggests null
in such case.
However, Booleans, Null, and Undefined Are Ignored as well, mentioned in Advanced guide.
来源:https://stackoverflow.com/questions/22538638/how-to-have-conditional-elements-and-keep-dry-with-facebook-reacts-jsx