如何在onsubmit函数react native中导航到新页面

Question

我有一个表单，它应该存储数据，并在导航到新屏幕时将其作为参数传递。我有这样的结构。formik形式：

return (
<View style={styles.container}>
<Formik
initialValues={{
first_name: '',
last_name: '',
}}
// Form submission action
onSubmit={(values) => {
const d = addData(values);
this.props.navigation.navigate('Cart', {
screen: 'Process Payment',
params: {data: d},
});
}}>
{(props) => (
<KeyboardAvoidingView
behavior={Platform.OS === 'ios' ? 'padding' : null}
style={{flex: 1, width: '100%'}}>
<ScrollView style={styles.inner}>
<TextInput
style={styles.input}
placeholder="first name"
onChangeText={props.handleChange('first_name')}
value={props.values.first_name}
/>
<TextInput
style={styles.input}
placeholder="last name"
onChangeText={props.handleChange('last_name')}
value={props.values.last_name}
/>
<View style={[{width: '90%', margin: 10, alignSelf: 'center'}]}>
<Button
title="place order"
color="maroon"
onPress={props.handleSubmit}
style={{padding: 3, width: '80%'}}
/>
</View>
</ScrollView>
</KeyboardAvoidingView>
)}
</Formik>
</View>
);
}

提交功能：

onSubmit={(values) => {
const d = addData(values);
this.props.navigation.navigate('Cart', {
screen: 'Process Payment',
params: {data: d},
});
}}>

我的错误：Warning: An unhandled error was caught from submitForm() [TypeError: undefined is not an object (evaluating '_this.props.navigation')]我不知道如何在onsubmit功能中导航到新屏幕

Answer 1

onSubmit={(values) => {
const d = addData(values);
props.navigation.navigate('Cart', {
screen: 'Process Payment',
params: {data: d},
});
}}>

我拿出"；这个"；刚刚打电话给props.navigation.navigation