React Native ios picker is always open

Question

I have two pickers on my screen. Whenever I navigate to the screen in iOS app I find that the pickers are always open and all options are visible.

It works

Answer 1

Extending the ActionSheetIOS answer, I created a custom component that is a drop-in replacement for Picker (I'm using Button from https://react-native-elements.github.io/react-native-elements/docs/overview.html):

import React from 'react';
import { ActionSheetIOS, Platform } from 'react-native';
import { Button } from 'react-native-elements';

class PickerDropDown extends React.Component {
  onIOSButton = () => {
    let options = this.props.children.map((item, i) => {
      return item.props.label;
    });
    options.push("Cancel");
    ActionSheetIOS.showActionSheetWithOptions(
      {
        options: options,
        cancelButtonIndex: options.length - 1,
      },
      this.onIOSButtonPick
    );
  }

  onIOSButtonPick = (buttonIndex) => {
    if (buttonIndex < this.props.children.length && buttonIndex != this.props.selectedValue) {
      if (typeof this.props.selectedValue === 'undefined' || (typeof this.props.selectedValue !== 'undefined' && buttonIndex != this.findIndexForValue(this.props.selectedValue))) {
        this.props.onValueChange(this.props.children[buttonIndex].props.value, buttonIndex);
      }
    }
  }

  findLabelForValue = (searchValue) => {
    for (let i = 0; i < this.props.children.length; i++) {
      if (this.props.children[i].props.value == searchValue) {
        return this.props.children[i].props.label;
      }
    }
    return null;
  }

  findIndexForValue = (searchValue) => {
    for (let i = 0; i < this.props.children.length; i++) {
      if (this.props.children[i].props.value == searchValue) {
        return i;
      }
    }
    return -1;
  }

  render() {
    if (Platform.OS === "ios") {
      let title = "";
      if (this.props.children && this.props.children.length > 0) {
        if (typeof this.props.selectedValue !== 'undefined') {
          title = this.findLabelForValue(this.props.selectedValue);
        } else {
          title = this.props.children[0].props.label;
        }
      }
      return (
      <View>
        <Button
          title={title}
          onPress={this.onIOSButton}
          type="clear"
          icon={{
            name: "arrow-drop-down",
            size: 15,
            color: "black"
          }}
          iconRight={true}
        />
     </View>
      );
    } else {
      return (
        <Picker {...this.props} />
      );
    }
  }
}

Answer 2

I don't know why you'd choose the answer with ActionSheet as accepted answer. However I'll give a workaround for this problem:

Put this values in your state:

this.state= {
    pickerOpacity: 0,
    opacityOfOtherItems: 1 //THIS IS THE OPACITY OF ALL OTHER ITEMS, WHICH COLLIDES WITH YOUR PICKER.
    label: 'Firstvalue'
}

In your render method do following:

{this.checkIfIOS()}
      <Picker
         selectedValue={this.state.selected}
         style={{ height: 50, alignSelf: 'center', opacity: this.state.pickerOpacity, marginBottom:30, width: 250}}
         onValueChange={(itemValue, itemIndex) =>{
            this.setState({
                selected: itemValue,
                label: itemValue
            });
            toggle();
            }
          }>
          <Picker.Item label="Your Label" value="yourValue"/>
      </Picker>

So now we've to check, whether our client is android or ios. Therefore import Platform and put the checkIfIos()-Method in your code:

import {Platform} from 'react-native'

checkIfIOS(){
        if(Platform.OS === 'ios'){ // check if ios
            console.log("IOS!!!");
            //this button will (onpress) set our picker visible
            return (<Button buttonStyle={{backgroundColor:'#D1D1D1', opacity: this.state.opacityOfOtherItems}} onPress={this.toggle()} color="#101010" title={this.state.label} onPress={this.changeOpacity}/>); 
        }else if(Platform.OS === 'android'){ //check if android
            this.setState({
                pickerOpacity: 1 //set picker opacity:1 -> picker is visible.
            });
            console.log("ANDROID!!!");
        }
    }

toggle(){
    if(Platform.OS === 'ios'){

        if(this.state.pickerOpacity == 0){
            this.setState({
                pickerOpacity: 1,
                opacityOfOtherItems: 0 // THIS WILL HIDE YOUR BUTTON!
            });
         }else{
             this.setState({
                 pickerOpacity: 0,
                 opacityOfOtherItems: 1
             });
          }
     }
}

EDIT: Screenshot with iOS (Click here for Video)

Answer 3

Use ActionSheet instead of Picker on iOS. https://facebook.github.io/react-native/docs/actionsheetios.html

As answered by jevakallio this is the default behaviour on iOS. But this doesn't give a good UX so remove all picker components and replace with ActionSheet.

I did and it works great. The reason I prefered ActionSheet over other components suggested by jevakallio because it is developed by the RN team and has a good native feeling. The last option suggested react-native-modal-picker is also very good.

Answer 4

use

<Picker itemStyle={{height:50}} > </Picker>

it only affect ios picker. change picker itemstyle height to your standard.

Answer 5

React-native-modal-picker was discontinued. react-native-modal-selector

Answer 6

react-native-dropdown has reference bugs If you don't want to use Action Sheets but a 'real' picker, this one I found is way better...

https://github.com/lawnstarter/react-native-picker-select