为什么我的React复选框onChange处理程序在渲染时触发

为什么我的React复选框onChange处理程序在渲染时触发

本文介绍了为什么我的React复选框onChange处理程序在渲染时触发,然后在单击该框时不触发?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

限时删除!!

已经阅读了React文档并将问题归结为一个简单的案例,但仍然不太了解我在做什么错.

Have read through the React docs and boiled the problem down to a simple case, still can't quite understand what I'm doing wrong.

JSFiddle: https://jsfiddle.net/justin_levinson/pyn7fLq5/或编写如下:

JSFiddle: https://jsfiddle.net/justin_levinson/pyn7fLq5/ or written below:

var TestForm = React.createClass({
    render : function() {
        return (
            <div>
                <p>Test Form</p>
                <form>
                    <TestBox />
                </form>
            </div>
        )
    }
});

var TestBox = React.createClass({
    render : function() {
        return (<input type="checkbox" name={"testbox"} defaultChecked={true} onChange={this.handleCheck()}/>)
    },
    handleCheck : function(event) {
        console.log("check");
        console.log(event);
    }
});

页面加载后,我在日志中得到一个检查",随后是该事件的未定义",然后在随后的点击中它不会再次触发.尝试过使用onClick和onChange进行此操作,还创建了一个受控的(checked = {true})而不是上面的不受控制的(defaultChecked = {true}).

When the page loads, I get a 'check' in the log followed by 'undefined' for the event, then it doesn't fire again on subsequent clicks. Have tried this with both onClick and onChange as well as creating a controlled (checked={true}) instead of the uncontrolled (defaultChecked={true}) above.

谢谢!

推荐答案

因为您已经在render上调用了该方法.

Because you're already calling the method on render.

onChange = {this.handleCheck()} 更改为 onChange = {this.handleCheck}

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09-06 12:27