reactjs - 为 onClick 事件列出的函数在 ReactJS 上加载时触发

Question

我有一个组件正在接收记录列表作为 Prop 。我正在尝试向 onClick 事件添加一个函数，但是一旦组件加载，要在 click 事件上执行的代码集就会触发。在本例中，我指的是 processM 函数。我是 ReactJS 的新手，希望得到任何帮助。

var TempJobTaskActions = React.createClass({
	processM: function(url){

		console.log('in processMove: ', url);

		$.ajax({
			type: 'GET',
			url: url,
			success: function(data){
			}
		});

	},
	render: function(){

		var action = null;
		var downUrl = '../MoveDown?id='+this.props.tid;
		var upUrl = '../MoveUp?id='+this.props.tid;

		if(this.props.position == "solo"){
			action = (<span></span>);
		}

		if(this.props.position == "first"){
			action = (<span><a className='cdc-link cdc-icon glyphicon glyphicon-arrow-down' 
				href='#' onClick={this.processM(downUrl)} title='Move Down'></a></span>);
		}

		if(this.props.position == "last"){
			action = (<span><a className='cdc-link cdc-icon glyphicon glyphicon-arrow-up' 
				href='#' onClick={this.processM(upUrl)} title='Move Up'></a></span>);
		}

		if(this.props.position == "middle"){
			action = (
				<div>
					<span><a className='cdc-link cdc-icon glyphicon glyphicon-arrow-up' 
						href='#' onClick={this.processM(upUrl)} title='Move Up'></a></span>
					<span><a className='cdc-link cdc-icon glyphicon glyphicon-arrow-down' 
						href='#' onClick={this.processM(downUrl)} title='Move Down'></a></span>
				</div>
			);
		}	

		return (
			<div>{action}</div>
		);
	}
});

Answer 1

this.processM(upUrl) 将调用该函数。要发送变量，您应该像这样绑定(bind)

this.processM.bind(this, upUrl)

Here is a good example with bind