javascript - 获取 "Illegal callback invocation from native module. This callback type only permits a single invocation from native code"

Question

我在两个类中使用react-native-image picker和react-native-image-crop picker库。

一个是启动库图像(react-native-image picker)，另一个包打开裁剪窗口(react-native-image-crop-picker)。

此处，启动库位于一个屏幕中，并且在从另一个屏幕上的库导航时打开裁剪器。

我的问题是，点击选择的裁剪窗口后，它再次重置裁剪窗口，需要再次裁剪，然后出现非法调用错误。

引用代码sniipet

 // Opens the library image in Library.js screen
import ImagePicker from 'react-native-image-picker';
     ImagePicker.launchImageLibrary(options, (response)  => {
            if (response.didCancel) {
              console.warn('User cancelled photo picker');
            }
            else if (response.error) {
              console.warn('ImagePicker Error: ', response.error);
            }
            else {
             this.props.navigation.navigate('CropWindow', { screenName: 'CropImage',uri: response.uri});
            }

下面是CropWindow.js中的裁剪窗口

import ImagePicker from 'react-native-image-crop-picker';
    ImagePicker.openCropper({
                path: response,
                width: deviceWidth,
                height: deviceWidth*5/4
             }).then(image => {
                this.props.navigation.navigate('ShowAllCroppedImage', {uri: response.uri, croppedImage: this.croppedImage.bind(this)});
              })
              .catch((err) => {
                console.log("openCropper error = " + err)
              });

Answer 1

Android 权限问题

import { PermissionsAndroid } from 'react-native';

async requestCameraPermission() {
  try {
    const granted = await PermissionsAndroid.request(
      PermissionsAndroid.PERMISSIONS.CAMERA,
      {
        'title': 'Cool Photo App Camera Permission',
        'message': 'Cool Photo App needs access to your camera ' +
                   'so you can take awesome pictures.'
      }
    )
    if (granted === PermissionsAndroid.RESULTS.GRANTED) {
      console.log("You can use the camera")
    } else {
      console.log("Camera permission denied")
    }
  } catch (err) {
    console.warn(err)
  }
}