React Synthetic Event区分左击事件和右击事件 [英] React Synthetic Event distinguish Left and Right click events

问题描述

我试图区分OnClick函数中的左键单击和右键单击。但是，

I am trying to distinguish between left and right clicks in an OnClick function. But,

var r = React.createClass({
   handleClick : function(e){
       //left click
       if(e.which==1){ 
          //Do something
       }
   },
   render : function(){
       return <p onClick={this.handleClick}>Something </p>
   }
});

显示未为综合事件。我如何在这里区分左键和右键？

Turns out e.which is undefined for Synthetic Events. How can I distinguish between left and right clicks here?

推荐答案

您也可以执行类似的操作。同时拥有onClick和onContextMenu处理程序

You can do something like this too. Have both onClick and onContextMenu handlers

return <p onClick={this.handleClick} onContextMenu={this.handleClick}>Something </p>

您可以检查 nativeEvent 作为其他答案建议或检查类型。 （此外，如果单击鼠标右键，则阻止默认设置。）

You can either check for nativeEvent as the other answer suggests or check for type. (Also, prevent default if it is a right click.)

使用 type

handleClick: function(e) {
  if (e.type === 'click') {
    console.log('Left click');
  } else if (e.type === 'contextmenu') {
    console.log('Right click');
  }
}

使用 nativeEvent

handleClick: function(e) {
  if (e.nativeEvent.which === 1) {
    console.log('Left click');
  } else if (e.nativeEvent.which === 3) {
    console.log('Right click');
  }
}

这里是一个演示 http://jsbin.com/seyeliv/edit?html，输出

Here is a demo http://jsbin.com/seyeliv/edit?html,output

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