[当用户单击屏幕上的json数据并将符号值传递给另一个类时如何重定向到另一页

Question

我正在使用react native和expo。屏幕上有一些json数据（类似iOS），例如

A
Acompany
B
Bcompany

这里A是符号，Acompany是名称

[当用户单击它时，它应该重定向到另一个屏幕，例如（Stock.js），并同时通过symbol吗？当用户单击并发送该数据时，如何将其重定向到另一个屏幕（symbol）？

我的代码：

import React, { useState, useEffect } from "react";
import {
  StyleSheet,
  View,
  TouchableWithoutFeedback,
  Keyboard,
  FlatList,
  TextInput,
  Button,
  Text,
} from "react-native";
import { useStocksContext } from "../contexts/StocksContext";
import { scaleSize } from "../constants/Layout";
import { Ionicons } from "@expo/vector-icons";

import { ListItem } from "react-native";

export default function SearchScreen({ navigation }) {
  const { ServerURL, addToWatchlist } = useStocksContext();

  const [state, setState] = useState({
    /*  initial state here */
    myListData: [],
  });
  const [search, setSearch] = useState("");

  useEffect(() => {
    renderWithData();
    // FixMe: fetch symbol names from the servner and save in local SearchScreen state
  }, []);

  const updateSearch = (text) => {
    setSearch(text);
  };

  renderWithData = () => {
    return fetch("http://131.181.190.87:3001/all")
      .then((res) => res.json())
      .then((json) => {
        setState({
          isLoaded: true,
          myListData: json,
        });
        setTimeout(() => {
          console.log(state.myListData);
        }, 10000);
      });
  };

  let filteredItems = state.myListData.filter((item) => {
    return (
      item.symbol.toUpperCase().indexOf(search.toUpperCase()) !== -1 ||
      item.name.indexOf(search) !== -1
    );
  });

  let movies = filteredItems.map((val) => {
    return (
      <View key={val.symbol} style={styles.text}>
        <Text style={styles.text}>{val.symbol}</Text>
        <Text style={styles.text}>{val.name}</Text>
      </View>
    );
  });

  return (
    <TouchableWithoutFeedback onPress={Keyboard.dismiss}>
      <View style={styles.container}>
        <TextInput
          style={styles.textinput}
          placeholder="Search here"
          placeholderTextColor="white"
          value={search}
          onChangeText={(text) => updateSearch(text)}
        />

        <Text>csdn</Text>
        <View style={styles.text}>{movies}</View>
      </View>
    </TouchableWithoutFeedback>
  );
}

const styles = StyleSheet.create({
  textinput: {
    color: "white",
    height: "20",
    fontSize: 18,
  },
  text: {
    color: "white",
    backgroundColor: "black",
  },
  flatstuff: {
    color: "white",
  },

  // use scaleSize(x) to adjust sizes for small/large screens
});

Answer 1

您必须使用导航库来支持应用程序的导航您可以参考react-native-navigation here

一旦您具有如下所示的基本堆栈设置

    <Stack.Screen name="Home" component={SearchScreen} />
    <Stack.Screen name="Details" component={DetailsScreen} />

您可以像下面一样导航

navigation.navigate('Details',{text:'123'})

您可以从详细信息屏幕访问如下所示的参数

const { text } = route.params;