脚本加载反应

Question

我想从CDN加载脚本,然后在React中执行该脚本公开的函数:

componentWillMount() {
    console.log('componentWillMount is called');
    const script = document.createElement('script');
    script.src = 'https://foo.azurewebsites.net/foo.js';
    document.body.appendChild(script);
}


componentDidMount() {
    console.log('componentDidMount is called');
    window.foo.render({
        formId: '77fd8848-791a-4f13-9c82-d24f9290edd7',
    }, '#container');
}


render() {
    console.log('render is called');
    return (
        <div id="container"></div>
    );
}

脚本有时需要花时间加载(通常是第一次),什么时候componentDidMount()被称为"foo"不可用,我得到这样的错误:

TypeError:无法读取未定义的属性'render'

如何componentDidMount()在脚本成功加载后确保调用？

Answer 1

根据React组件规范和生命周期,我不认为在componentWillMount()或componentDidMount()中加载脚本是个好主意.

以下代码可能对您有所帮助.

function new_script(src) {
  return new Promise(function(resolve, reject){
    var script = document.createElement('script');
    script.src = src;
    script.addEventListener('load', function () {
      resolve();
    });
    script.addEventListener('error', function (e) {
      reject(e);
    });
    document.body.appendChild(script);
  })
};
// Promise Interface can ensure load the script only once.
var my_script = new_script('http://example.com/aaa.js');

class App extends React.Component {

  constructor(props) {
    super(props);
    this.state = {
      status: 'start'
    };
  }

  do_load = () => {
    var self = this;
    my_script.then(function() {
      self.setState({'status': 'done'});
    }).catch(function() {
      self.setState({'status': 'error'});
    })
  }

  render() {
    var self = this;
    if (self.state.status === 'start') {
      self.state.status = 'loading';
      setTimeout(function () {
        self.do_load()
      }, 0);
    }

    return (
      <div>{self.state.status}   {self.state.status === 'done' && 'here you can use the script loaded'}</div>
    );
  }
}