React Native Navigation 传回参数两屏弹出

问题描述

我有一个主屏幕，然后它被推到下一个进行选择，并将参数传递给下一个被推的屏幕，当你在那个屏幕中选择某些东西时，我需要 pop(2) 并将选定的详细信息发送到主屏幕.

I have a Screen for Main then it's push to Next for selection and passes parameters to next pushed screen and when you select something in that screen I need to pop(2) and send selected details to the main screen.

但是 pop 不接受参数.如何实现这一点.

But pop didn't accept parameters. How to accomplish this.

我不使用 Redux 或 MobX.

I don't use Redux or MobX.

推荐答案

你需要这样做

屏幕 A:

this.props.navigation.navigate('ScreenB', {
              onPressScreenAFun: (params) => {
                this.screenAFun(params)
              },
            })

screenAFun = (params) => {
console.log(params)
}

屏幕 B:

this.props.navigation.navigate('ScreenC', {
                  onPressScreenBFun: (params) => {
                    this.screenBFun(params)
                  },
                })

    screenBFun = (params) => {
       const { onPressScreenAFun } = this.props.navigation.navigate.state.params

      onPressScreenAFun(params)
      this.props.navigation.goBack()
    }

屏幕 C:

    moveBack = (params) => {
       const { onPressScreenBFun } = this.props.navigation.navigate.state.params

      onPressScreenBFun(params)
      this.props.navigation.goBack()
    }

这是如何使用两个 pop() 将参数从屏幕 C 传递到屏幕 A

this is how you can pass params from Screen C to Screen A with two pop()

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