Disabling buttons on react native

Question

TouchableOpacity extents TouchableWithoutFeedback, so you can just use the disabled property :

<TouchableOpacity disabled={true}>
  <Text>I'm disabled</Text>
</TouchableOpacity>

React Native TouchableWithoutFeedback #disabled documentation

The new Pressable API has a disabled option too :

<Pressable disable={true}>
  {({ pressed }) => (
    <Text>I'm disabled</Text>
  )}
</Pressable>

Just do this

<TouchableOpacity activeOpacity={disabled ? 1 : 0.7} onPress={!disabled && onPress}>
  <View>
    <Text>{text}</Text>
  </View>
</TouchableOpacity>

This seems like the kind of thing that could be solved using a Higher Order Component. I could be wrong though because I'm struggling to understand it 100% myself, but maybe it'll be helpful to you (here's a couple links)...

• http://www.bennadel.com/blog/2888-experimenting-with-higher-order-components-in-reactjs.htm
• http://jamesknelson.com/structuring-react-applications-higher-order-components/

this native-base there is solution:

<Button
    block
    disabled={!learnedWordsByUser.length}
    style={{ marginTop: 10 }}
    onPress={learnedWordsByUser.length && () => {
      onFlipCardsGenerateNewWords(learnedWordsByUser)
      onFlipCardsBtnPress()
    }}
>
    <Text>Let's Review</Text>
</Button>

TouchableOpacity receives activeOpacity. You can do something like this

<TouchableOpacity activeOpacity={enabled ? 0.5 : 1}>
</TouchableOpacity>

So if it's enabled, it will look normal, otherwise, it will look just like touchablewithoutfeedback.