在本机反应中以编程方式单击按钮

Question

In my react-native app I have a screen where it contains a form. I wanna submit the form on loading the screen because the values in form is gonna be constant. So i wanna submit the form on loading. Is it possible? Can someone help me..

<PrimaryButton
          // onPress={onSubmit(siteUrl)}
          onPress={() => onSubmit(siteUrl)}
          text={translate("general.enter")}
          style={styles.buttonEnter}
          mode={isSiteUrlSubmitted ? "loading" : undefined}
        />

i tried this onPress={onSubmit(siteUrl)} but it shows Invariant Violation: Too many re-renders. React limits the number of renders to prevent an infinite loop.

Answer 1

you can call onSubmit() in React.useEffect

as such

React.useEffect(() => {
   onSubmit() 
}, []);

When you add blank parentheses It will only be performed once in loading