反应原生 TouchableOpacity onPress 问题 [英] React native TouchableOpacity onPress Problems
本文介绍了反应原生 TouchableOpacity onPress 问题的处理方法,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
我有一个简单的图标按钮,如下所示:
I have a simple icon button as follows :
class SideIcon extends Component {
render() {
return (
<TouchableOpacity onPress={console.log('puff')} style={styles.burgerButton}>
<Icon name="bars" style={styles.burgerIcon}/>
</TouchableOpacity>
);
}
}
它是从以下组件调用的:
It's called from the following component :
export default test = React.createClass({
getInitialState: function() {
return {
isModalOpen: false
}
},
_openModal() {
this.setState({
isModalOpen: true
});
},
_closeModal() {
this.setState({
isModalOpen: false
});
},
render() {
return (
<View style={styles.containerHead} keyboardShouldPersistTaps={true}>
**<SideIcon openModal={this._openModal} closeModal={this._closeModal} />**
<Text style={styles.logoName}>DareMe</Text>
<SideBar isVisible={this.state.isModalOpen} />
</View>
);
}
});
但是 TouchableOpacity
上的 onPress
永远不起作用.我哪里错了?虽然它在组件加载期间显示控制台语句.
But the onPress
on the TouchableOpacity
never works. Where I'm going wrong ? Although It shows console statements during the component load.
推荐答案
你应该绑定一个调用你的代码的函数.
You should bind a function that invokes your code.
例如:
<TouchableOpacity onPress={() => console.log('puff')} style={styles.burgerButton}>
<Icon name="bars" style={styles.burgerIcon}/>
</TouchableOpacity>
更好的方法是给它一个组件函数的引用
A better way is to give it a reference to a component function
class SideIcon extends Component {
handleOnPress = () => {
console.log('puff')
}
render() {
return (
<TouchableOpacity onPress={this.handleOnPress} style={styles.burgerButton}>
<Icon name="bars" style={styles.burgerIcon}/>
</TouchableOpacity>
);
}
}
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