英文:
Setting Navigation Value Based On What The Function Returns
问题
以下是翻译好的部分：
当我调用登录函数时，我收到以下错误：
错误：在使用对象作为参数调用导航时，您需要指定名称或键。请参阅 https://reactnavigation.org/docs/navigation-actions#navigate 以获取用法。
这是我的代码：
login.jsx
<TouchableOpacity
style={styles.button}
onPress={() => navigation.navigate(this.onClick())}>
<Icon
style={styles.buttonIcon}
name="arrow-right"
size={30}
color="white"
type="feather"
/>
</TouchableOpacity>
这是函数：
ParticleAuth.js
onClick = async () => {
this.init();
this.setLanguage();
this.setChainInfo();
this.login();
const result = await particleAuth.isLogin();
console.log('Result:', result);
if (result) {
return "LoggedIn"; // LoggedIn和Error是屏幕的名称
} else {
return "Error";
}
};
根据错误信息，我需要预设navigation.navigate的值。以下是一些可能的替代代码：
谢谢！
英文:
When I call a login function, this is the error I get:
Error: You need to specify name or key when calling navigate with an object as the argument. See https://reactnavigation.org/docs/navigation-actions#navigate for usage.
This is my code:
login.jsx
<TouchableOpacity
style={styles.button}
onPress={() => navigation.navigate(this.onClick())}>
<Icon
style={styles.buttonIcon}
name="arrow-right"
size={30}
color="white"
type="feather"
/>
This is the function:
ParticleAuth.js
onClick = async () => {
this.init();
this.setLanguage();
this.setChainInfo();
this.login();
const result = await particleAuth.isLogin();
console.log('Result:', result);
if (result) {
return "LoggedIn"; // LoggedIn & Error are names of screens
} else {
return "Error";
}
};
According to the error, I have to preset the value for navigation.navigate. What might be some alternate code to mine?
Thanks!
答案1
得分: 1
你可以将你的 navigation 移动到 onClick() 函数中，并在其中导航到屏幕，如下所示：
<TouchableOpacity
style={styles.button}
onPress={this.onClick}>
<Icon
style={styles.buttonIcon}
name="arrow-right"
size={30}
color="white"
type="feather"
/>
</TouchableOpacity>
onClick = async () => {
this.init();
this.setLanguage();
this.setChainInfo();
this.login();
const result = await particleAuth.isLogin();
console.log('Result:', result);
// 修改这些行
let screenName = '';
if (result) {
screenName = "LoggedIn"; // LoggedIn 和 Error 是屏幕的名称
} else {
screenName = "Error";
}
navigation.navigate(screenName);
};
英文:
You can move your navigation to onClick() function and navigate to the screen in there as below:
<TouchableOpacity
style={styles.button}
onPress={this.onClick}>
<Icon
style={styles.buttonIcon}
name="arrow-right"
size={30}
color="white"
type="feather"
/>
onClick = async () => {
this.init();
this.setLanguage();
this.setChainInfo();
this.login();
const result = await particleAuth.isLogin();
console.log('Result:', result);
// change this lines
let screenName = '';
if (result) {
screenName = "LoggedIn"; // LoggedIn & Error are names of screens
} else {
screenName = "Error";
}
navigation.navigate(screenName);
};
答案2
得分: 0
我找到了解决方法。这不是最高效的方法，但我使用了一个名为“加载中”的中间屏幕，当登录功能工作时。
英文:
I found the solution. It's not the most efficient, but I used an intermediary screen called Loading while the login function works
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