如何在呈现React组件时添加逻辑if语句?

时间:2015-02-01 01:22:45

标签: reactjs

如果我的代码看起来像这样......

var Lounge = React.createClass({displayName: "Lounge",
  render: function() {
    return (
            React.createElement("a", {href:"/lounge/detail/" + this.props.id +  "/"},
            React.createElement("div", {className: "lounge"},
            React.createElement("h2", {className: "loungeAuthor"},
              this.props.author.name
            ),
            React.createElement("p", {className: "loungeArticle"},
              this.props.article
            ),
            React.createElement("img", {className: "loungeImage", src: this.props.image})
          )
        )
    );
  }
});

我需要做一个逻辑,如果检查只渲染" img"如果图像数据存在,则为component。有人知道使用React解决这个问题的最佳方法吗?

5 个答案:

答案 0 :(得分:37)

如果你想内联,你可以这样做:

{this.props.image && <img className="loungeImage" src={this.props.image}/>}

this.props.image && React.createElement("img", {className: "loungeImage", src: this.props.image})

如果要检查的值是假的,但会导致某些由React呈现,就像空字符串一样,您可能希望通过使用{将其转换为检查中的布尔等效值{1}}:

!!

答案 1 :(得分:10)

保持此逻辑内联以及使用JSX可能有助于提高可读性

import React from 'react';
import PropTypes from 'prop-types';
import s from './Lounge.css'; // See CSS Modules

// Stateless functional component (since there is no state)
function Lounge({ id, article, author, imageUrl }) {
  return (
    <a href={`/lounge/detail/${id}/`}>
      <span className={s.lounge}>
        <span className={s.author}>{author.name}</span>
        <span className={s.article}>{article}</span>
        {imageUrl && <img className={s.image} src={imageUrl} />} // <==
      </span>
    </a>
  );
}

// Props validation
Lounge.propTypes = {
  id: PropTypes.number.isRequired,
  article: PropTypes.string.isRequired,
  imageUrl: PropTypes.string,
  author: PropTypes.shape({
    name: PropTypes.string.isRequired
  }).isRequired
};

export default Lounge;

答案 2 :(得分:2)

好吧所以这对很多人来说可能很清楚,由于某些原因我没有看到它,然后我又快速浏览文档而我没有使用JSX(真的不喜欢它)

因此,在渲染反应组件时添加逻辑if的JavaScript方式就像这样完成。 (请记住undefined的工作正常,所以假设if语句没有被命中,它仍然可以正常渲染)

var Lounge = React.createClass({displayName: "Lounge",
  render: function() {
    if (this.props.image != "") {
       var imageElement = React.createElement("img", {className: "loungeImage", src: this.props.image});
    }
    return (
            React.createElement("a", {href:"/lounge/detail/" + this.props.id +  "/"},
            React.createElement("div", {className: "lounge"},
            React.createElement("h2", {className: "loungeAuthor"},
              this.props.author.name
            ),
            React.createElement("p", {className: "loungeArticle"},
              this.props.article
            ),
            imageElement
          )
        )
    );
  }
});

答案 3 :(得分:2)

您可以使用React if component

var Node = require('react-if-comp');

var Lounge = React.createClass({
    displayName: 'Lounge',
    render: function() {
        return (
            <a href={'/lounge/detail/' + this.props.id + '/'}>
                <div className='lounge'>
                    <h2 className='loungeAuthor'>{author.name}</h2>
                    <p className='loungeArticle'>{article}</p>
                    <Node if={this.props.image}
                        then={<img className='loungeImage'
                        src={this.props.image} />} />
                </div>
            </a>
        );
    }
});

答案 4 :(得分:1)

使用JSX + IIFE(立即调用函数)可以非常方便和明确:

{(() => {
  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})}/>
  }
})()}