React Synthetic Event distinguish Left and Right click events

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

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

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

Using type

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

Using nativeEvent

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

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

The property you're looking for is e.button or e.buttons.

The button number that was pressed when the mouse event was fired: Left button=0, middle button=1 (if present), right button=2.
_{– MDN:Web/Events/click}

However, with or without react, I'm only getting click events with the left mouse button (trackpad). You could use onMouseDown which works for me.

Here's a demo using e.buttons. You may want to preventDefault in onContextMenu also.

Use:

if (e.button === 0) { // or e.nativeEvent.which === 1    // do something on left click}

Here is a DEMO