[英]I want to show and hide a element (ActivityIndicator)
What I'm trying to do is when I click the button it disappears, and when it's already loaded in another component it shows again the button.我想要做的是当我点击按钮时它消失了,当它已经加载到另一个组件中时它再次显示按钮。
JSX code: JSX 代码:
import { ActivityIndicator } from 'react-native';
{......}
{/button besides loading/}
<LinearGradient
colors={['#D97B29', '#F2B47E']}
style={styles.signIn}
>
<Text style={[styles.textSign, {
color:'#fff'
}]}>Entrar</Text>
{/loading besides button /}
<ActivityIndicator size="large" color="#D97B29" />
</LinearGradient>
{...}
How can I handle it whit the useState?我该如何处理 useState?
Assuming you want to hide/show the indicator when the button is pressed, you need假设您想在按下按钮时隐藏/显示指示器,您需要
const [ showLoader, setShowLoader ] = useState(false);
Your button on press should show something like您按下的按钮应该显示类似
<Button onPress={() => setShowLoader(!showLoader)} text="Some text" />
and your code should look like你的代码应该看起来像
{showLoader && (
<ActivityIndicator size="large" color="#D97B29" />
)}
{!showLoader && <LinearGradient
colors={['#D97B29', '#F2B47E']}
style={styles.signIn}
>
<Text style={[styles.textSign, {
color:'#fff'
}]}>Entrar</Text>
</LinearGradient> }
```
something like this像这样的
{state && <ActivityIndicator size="large" color="#D97B29" />}
if state
is true
it will show the component如果
state
为true
,它将显示该组件
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