我有一个组件将使用map来渲染多个复选框,并且每个复选框都有一个回调函数“onPress”获取道具,“onPress”函数将setState检查,但现在当我点击一个复选框时,所有复选框都将选择,它导致他们都使用相同的状态,目标我想选择每个复选框我只是ckick,我知道我可以写每个复选框的许多状态不同的“onPress”功能,但它看起来很愚蠢,我会添加更多未来的复选框,什么是解决任务的最佳和最灵活的方式?
import React, { Component } from 'react'
import { View } from 'react-native'
import { CheckBox } from 'react-native-elements'
const styles = {
CheckBox: {
borderBottomWidth: 0.3,
borderBottomColor: 'gray'
},
checkBox : {
backgroundColor: "#ffffff",
borderWidth: 0
},
text: {
flex: 0.95,
backgroundColor: "#ffffff"
}
}
const languages = ["中文","英文","日文","韓文"]
class Language extends Component {
constructor(props) {
super(props);
this.state = { checked: false };
}
onPress = () => {
this.setState({ checked: !this.state.checked })
}
renderlanguages = () => {
return languages.map((langauge) => {
return(
<View key = { langauge } style = { styles.CheckBox }>
<CheckBox
title = { langauge }
iconRight
containerStyle = { styles.checkBox }
textStyle = { styles.text }
checkedColor = 'red'
checked = { this.state.checked }
onPress = { this.onPress }
/>
</View>
)
})
}
render(){
return(
<View>
{ this.renderlanguages() }
</View>
)
}
}
export default Language;
答案 0 :(得分:2)
你可以将langauge
(注意这可能是language
)变量的错误传递给函数,并告诉我们确定正在检查哪一个1} p>
onPress = (langauge) => {
this.setState({ [langauge]: { checked: !this.state[langauge].checked } })
}
renderlanguages = () => {
return languages.map((langauge) => {
return(
<View key = { langauge } style = { styles.CheckBox }>
<CheckBox
title = { langauge }
iconRight
//component = { () => {return <TouchableOpacity></TouchableOpacity>}}
containerStyle = { styles.checkBox }
textStyle = { styles.text }
checkedColor = 'red'
checked = { this.state[langauge].checked }
onPress = { () => this.onPress(langauge) }
/>
</View>
)
})
}