React Native Camera - 多张照片

Question

I am currently using react-native-camera as a library to take pictures. I managed to show and hide a one and only camera component depending on a specific state. I am working on an app that has multiple buttons to take a picture, for example:

• Button A (show camera -> take picture -> store value on local state A)
• Button B (show camera -> take picture -> store value on local state B)
• Button C (show camera -> take picture -> store value on local state C)

I have been breaking my head on how to do this, but can't figure it out.

My code is the following:

 import React, { Component } from 'react'; import { StyleSheet, Text, TouchableOpacity, View, Button } from 'react-native'; import { RNCamera } from 'react-native-camera'; export default class BadInstagramCloneApp extends Component { constructor(props){ super(props); this.state = { isVisible: false, value1: null, value2: null } } render() { return ( <View style={styles.subcontainer}> {this.state.isVisible === true ? <View style={styles.container}> <RNCamera ref={ref => { this.camera = ref; }} style={styles.preview} type={RNCamera.Constants.Type.back} flashMode={RNCamera.Constants.FlashMode.on} permissionDialogTitle={'Permission to use camera'} permissionDialogMessage={'We need your permission to use your camera phone'} onGoogleVisionBarcodesDetected={({ barcodes }) => { console.log(barcodes); }} /> <View style={{ flex: 0, flexDirection: 'row', justifyContent: 'center' }}> <TouchableOpacity onPress={this.takePicture.bind(this)} style={styles.capture}> <Text style={{ fontSize: 14 }}> SNAP </Text> </TouchableOpacity> </View> </View> : <View> <Button title='PHOTO 1' onPress={this.changeState}/> <Button title='PHOTO 2' onPress={this.changeState2}/> <Button title='SHOW RESULTS' onPress={this.showResults}/> </View> } </View> ); } changeState = () =>{ this.setState({isVisible: true}) } changeState2 = () =>{ this.setState({isVisible: true}) } showResults = () => { console.log('VALUE1: ' + this.state.value1); console.log('VALUE2: ' + this.state.value2); } takePicture = async function() { if (this.camera) { const options = { quality: 0.5, base64: true }; const data = await this.camera.takePictureAsync(options); console.log(data.uri); this.setState({isVisible: false}); } }; } const styles = StyleSheet.create({ container: { flex: 1, flexDirection: 'column', backgroundColor: 'black' }, subcontainer: { flex: 1, flexDirection: 'column', }, preview: { flex: 1, justifyContent: 'flex-end', alignItems: 'center', }, capture: { flex: 0, backgroundColor: '#fff', borderRadius: 5, padding: 15, paddingHorizontal: 20, alignSelf: 'center', margin: 20, }, });

Answer 1

I would use the state to distinguish which "camera" you are using.

Your initial state:

this.state = {
  isVisible: false,
  pictureType: null,
  value1: null,
  value2: null
}

The function to call when a button is called, where each button has a different pictureType :

initTakingPicture = (pictureType) => {
  this.setState({
    isVisible: true,
    pictureType: pictureType
  })
}

Your example button:

<Button title='PHOTO 1' onPress={() => this.initTakingPicture("A")}/>

Then in your takePicture function you can check the state to distinguish which type of picture you are taking and save it into the according field:

  takePicture = async function() {
    if (this.camera) {
      const options = { quality: 0.5, base64: true };
      const data = await this.camera.takePictureAsync(options);
      console.log(data.uri);
      let fieldToSave = "value1" // Fallback
      if (this.state.pictureType === "A") {
        // Operation you need to do for pictureType A
        fieldToSave = "value1"
      } else if (this.state.pictureType === "B") {
        // Operation you need to do for pictureType B
        fieldToSave = "value2"
      } 

      this.setState({
        isVisible: false,  
        pictureType: null,
        [fieldToSave]: data.uri
      });
    }
  };