Disabling buttons on react native

Question

I'm making an android app using react native and I've used TouchableOpacity component to create buttons.
I use a text input component to accept text from the user and the button should only be enabled once the text input matches a certain string.
I can think of a way to do this by initially rendering the button without the TouchableOpactiy wrapper and re-rendering with the wrapper once the input string matches.
But I'm guessing there is a much better way to do this. Can anyone help?

Answer 1

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>

Answer 2

Just do this

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

Answer 3

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/

Answer 4

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>

Answer 5

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.

Answer 6

To disable Text you have to set the opacity:0 in Text style like this:

<TouchableOpacity style={{opacity:0}}>
  <Text>I'm disabled</Text>
</TouchableOpacity>

Answer 7

You can build an CustButton with TouchableWithoutFeedback, and set the effect and logic you want with onPressIn, onPressout or other props.

Answer 8

I think the most efficient way is to wrap the touchableOpacity with a view and add the prop pointerEvents with a style condition.

<View style={this.state.disabled && commonStyles.buttonDisabled}        
      pointerEvents={this.state.disabled ? "none" : "auto"}>
  <TouchableOpacity
    style={styles.connectButton}>
  <Text style={styles.connectButtonText}">CONNECT </Text>
  </TouchableOpacity>
</View>

CSS:

buttonDisabled: {
    opacity: 0.7
  }

Answer 9

So it is very easy to disable any button in react native

<TouchableOpacity disabled={true}>
    <Text> 
          This is disabled button
   </Text>
</TouchableOpacity>

disabled is a prop in react native and when you set its value to "true" it will disable your button

Happy Cooding

Answer 10

I was able to fix this by putting a conditional in the style property.

const startQuizDisabled = () => props.deck.cards.length === 0;

<TouchableOpacity
  style={startQuizDisabled() ? styles.androidStartQuizDisable : styles.androidStartQuiz}
  onPress={startQuiz}
  disabled={startQuizDisabled()}
>
  <Text 
    style={styles.androidStartQuizBtn} 
  >Start Quiz</Text>
</TouchableOpacity>

const styles = StyleSheet.create({
androidStartQuiz: {
    marginTop:25,
    backgroundColor: "green",
    padding: 10,
    borderRadius: 5,
    borderWidth: 1
},
androidStartQuizDisable: {
    marginTop:25,
    backgroundColor: "green",
    padding: 10,
    borderRadius: 5,
    borderWidth: 1,
    opacity: 0.4
},
androidStartQuizBtn: {
    color: "white",
    fontSize: 24
}
})

Answer 11

Here's my work around for this I hope it helps :

<TouchableOpacity
    onPress={() => {
        this.onSubmit()
    }}
    disabled={this.state.validity}
    style={this.state.validity ?
          SignUpStyleSheet.inputStyle :
          [SignUpStyleSheet.inputAndButton, {opacity: 0.5}]}>
    <Text style={SignUpStyleSheet.buttonsText}>Sign-Up</Text>
</TouchableOpacity>

in SignUpStyleSheet.inputStyle holds the style for the button when it disabled or not, then in style={this.state.validity ? SignUpStyleSheet.inputStyle : [SignUpStyleSheet.inputAndButton, {opacity: 0.5}]} I add the opacity property if the button is disabled.

Answer 12

You can enable and disable button or by using condition or directly by default it will be disable : true

 // in calling function of button 
    handledisableenable()
        {
         // set the state for disabling or enabling the button
           if(ifSomeConditionReturnsTrue)
            {
                this.setState({ Isbuttonenable : true })
            }
          else
          {
             this.setState({ Isbuttonenable : false})
          }
        }

<TouchableOpacity onPress ={this.handledisableenable} disabled= 
     {this.state.Isbuttonenable}>

    <Text> Button </Text>
</TouchableOpacity>

Answer 13

Use disabled true property

<TouchableOpacity disabled={true}> </TouchableOpacity>

Answer 14

Here is the simplest solution ever:

You can add onPressOut event to the TouchableOpcaity and do whatever you want to do. It will not let user to press again until onPressOut is done