[英]How to dynamically Load API Data on a button Click in React Native
I am a starter in React Native. 我是React Native的入门者。 I have an API that dynamically displays the number of levels on a page. 我有一个API,可以在页面上动态显示级别数。 Every time the user clicks on the button, I want to fetch an API and navigate to another page. 每次用户单击按钮时,我要获取一个API并导航到另一个页面。 I have defined three on click events for this, but I want to clean up my code and write with the help of one OnClick event. 我为此定义了三个on click事件,但是我想清理代码并借助一个OnClick事件进行编写。
My levels are as follows 我的等级如下
//Level1
async Level1() {
const datavalue = await getResultValue('url1');
this.props.navigation.navigate('Final','getSomething');
}
//Level2
async Level12() {
const datavalue = await getResultValue('url2');
this.props.navigation.navigate('Final','getSomething');
}
//Level3
async Level3() {
const datavalue = await getResultValue('url3');
this.props.navigation.navigate('Final','getSomething');
}
Each an every level, there is a corresponding API. 每个每个级别,都有一个对应的API。 Is there a way I can achieve it with the help of One OnClick Event? 有没有一种方法可以借助One OnClick Event来实现?
Any help would be appreciated. 任何帮助,将不胜感激。 Thank you in advance. 先感谢您。
async navigateToFinalScreen(params){
data = await fetch(params.url);
this.props.navigation.navigate(params.screenToNavigate, 'getSomething');
}
/* calling This method */ / *调用此方法* /
this.navigateToNextScreen(
{ url:"url",
screenToNavigate:"Final" }
)
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