setState 始终(如果我转到上一个屏幕或转到下一个屏幕，我想始终将状态状态设置为成功)

Question

constructor(props) {
super();
this.state = {
Status: '',
};
}
Successful = () => {
console.log('Status' + this.state.Status);
if (this.state.Status === 'SUCCESS') {
this.props.navigation.navigate('OScreen');
}
};
render() {
return (
<View style={styles.container}>
<Text style={styles.text}>PAY</Text>
<TouchableOpacity style={styles.button}                                 
onPress=this.setState({Status: 'SUCCESS'});>
<View style={styles.item}>
<Text style={styles.pay}>Proceed</Text>
</View>
</TouchableOpacity>
</View>
);
}

Answer 1

用函数包装印刷机

onPress={() => { alert('doing something') } }

constructor(props) {
super();
this.state = {
Status: '',
};
}

render() {
return (
<View style={styles.container}>
<Text style={styles.text}>PAY</Text>
<TouchableOpacity style={styles.button}                                 
onPress={()=>{
this.setState({Status: 'SUCCESS'});
this.props.navigation.navigate('OScreen');
}}>
<View style={styles.item}>
<Text style={styles.pay}>Proceed</Text>
</View>
</TouchableOpacity>
</View>
);
}