在 React Native 中更新变量和设置状态时遇到问题

Question

我有这个功能：

async function search() {
     try {
    const response = await places.nearbysearch({
      location: text, // LatLon delimited by ,
      radius: distance,  // Radius cannot be used if rankBy set to DISTANCE
      type: ['cafe','bakery','meal_delivery','meal_takeaway','restaurant'], // Undefined type will return all types
      // rankby: "distance" // See google docs for different possible values
    });

    const { status, results, next_page_token, html_attributions } = response;

    var index = Math.floor(Math.random() * response.results.length);
    console.log(response.results[index].name);

    restName = response.results[index].name;
    console.log(restName);

  } catch (error) {
    console.log(error);
  }


 }

在export default function App() {}

之内

我在视图中创建了一个文本组件，如下所示：

return (
    <View style={styles.container}>

    <View style={{minWidth: '100%', minHeight: '100%', backgroundColor: '#f0fbff', alignFit:'stretch', marginTop: 0, alignItems:'center', justifyContent:'space-evenly'}}>
    <Card style={{padding: 5, margin: 1}}>
      <Text style={{ fontWeight: 'bold', fontSize: 25,  }}>{restName}{"\n"}</Text>

即使 restName 更新（我制作了一个按下时运行 search() 函数的按钮），文本也没有更新，我将如何用它实现设置状态或刷新变量。当我 console.log restName 时，它显示正确的内容，所以我知道变量已更新，只是文本没有更新。

简而言之，实际变量发生变化但显示没有任何不同

UPDATE 我尝试使用 setstate，但它仍然没有更新这里是功能

    var restName = "";
 async function search() {
     try {
    const response = await places.nearbysearch({
      location: text, // LatLon delimited by ,
      radius: distance,  // Radius cannot be used if rankBy set to DISTANCE
      type: ['cafe','bakery','meal_delivery','meal_takeaway','restaurant'], // Undefined type will return all types
      // rankby: "distance" // See google docs for different possible values
    });

    const { status, results, next_page_token, html_attributions } = response;

    var index = Math.floor(Math.random() * response.results.length);
    console.log(response.results[index].name);

    this.setState({ restName : response.results[index].name })
    console.log(restName);

  } catch (error) {
    console.log(error);
  }


 }

Answer 1

要更新状态，您不能只重新分配

restName = response.results[index].name

但必须打电话setState

this.setState({ restName : response.results[index].name })

在 React Native 中更新变量和设置状态时遇到问题

Trouble with updating variable and set state in react native

setstate

react-native