React Native - Navigation.navigate() 仅适用于 onPress 函数

Question

在我的应用中，

onPress={() => navigation.navigate('<ScreenName>')}

工作完美，而如果我尝试在点击按钮时调用的函数中更改屏幕，例如

const signup = (mobileNumber, {navigation}) => {
Object.keys(mobileNumber).forEach((key) => {
    console.log(mobileNumber[key]);
});

if (mobileNumber.mobileNumber == null) {
    Alert.alert("Oops!", "Please insert your mobile number.");
    return;
}

let dataToSend = {mobile_number: mobileNumber.mobileNumber};
let formBody = [];
for (let key in dataToSend) {
    let encodedKey = encodeURIComponent(key);
    let encodedValue = encodeURIComponent(dataToSend[key]);
    formBody.push(encodedKey + '=' + encodedValue);
}
formBody = formBody.join('&');

fetch('https://mywebsite.com/api/v1/createAccount.php', {
    method: 'POST',
    body: formBody,
    headers: {
    //Header Defination
    'Content-Type':
    'application/x-www-form-urlencoded;charset=UTF-8',
    },
})
.then((response) => response.json())
.then((responseJson) => {
console.log(responseJson);
// If server response message same as Data Matched
if (responseJson.status === 'success') {
    //Alert.alert("Excellent!", "Please go on");
    navigation.navigate('<ScreenName>');
} else {
    console.log('Please check your mobileNumber');
}
})
.catch((error) => {
//Hide Loader
console.error(error);
});
}


function JoinScreen({ navigation }) {
const [mobileNumber, setMobileNumber] = useState(null);
const [isLoading, setLoading] = useState(true);

const [data, newData] = useState(null);

useEffect(() => {
    fetch("https://mywebsite.com/api/v1/createAccount.php")
        .then((response) => response.text())
        .then((response) => newData(response));
}, []);

return (
    <>
        <StatusBar hidden />
        <SafeAreaView style={{ flex: 1, alignItems: 'center', backgroundColor: '#262423' }}>

            <TextInput
                style={styles.input}
                placeholder="Your mobile number"
                placeholderTextColor="#fff"
                autoCapitalize="none"
                autoCorrect={false}
                onChangeText={(val) => setMobileNumber(val)}
                keyboardType="phone-pad"
            />

            <TouchableOpacity onPress={() => signup({ mobileNumber })}>
                <Text style={{ textAlign: "center", paddingTop: 20, color: "#babf26", fontSize: 20 }} >Create account</Text>
            </TouchableOpacity>
        </SafeAreaView>
    </>
);

}

应用程序 returns“TypeError: undefined 不是一个对象（正在计算'_ref6.navigation'）”

如果有帮助，如果在“注册”函数中我没有将“导航”括在花括号中，我会得到“undefined is not an object (evaluating 'navigation.navigate') "

Answer 1

您没有将 navigation 对象传递给 signup 函数。语句

const signup = (mobileNumber, {navigation})

无效。花括号用于解构传递给函数的对象的属性。 JoinScreen({ navigation }) 起作用的原因（很可能）是因为 JoinScreen 被定义为导航器内的屏幕。因此，导航框架确实将一个对象传递给在导航器中定义为屏幕的所有屏幕，并且该对象的一部分是 navigation 对象，您可以使用大括号对其进行解构。

对于signup，情况并非如此，因为这只是一个函数。但是，您可以按如下方式从 JoinScreen 传递它。

function JoinScreen({ navigation }) {
 ...
  <TouchableOpacity onPress={() => signup(mobileNumber, navigation)}>
                <Text style={{ textAlign: "center", paddingTop: 20, color: "#babf26", fontSize: 20 }} >Create account</Text>
            </TouchableOpacity>
}

然后，你的注册功能。

const signup = (mobileNumber, navigation) => {

...

}

Answer 2

您的注册函数应该在您的屏幕组件中，或者您应该将导航对象作为参数传递。

React Native - Navigation.navigate() 仅适用于 onPress 函数

React Native - Navigation.navigate() only works within an onPress function

javascript

mobile-application

react-native