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我正在开发一个包含React Native的移动应用程序。我想在按下按钮后打开WebView。这是我的代码,但它不工作。按下onPress方法不起作用。用reactNative按下按钮后打开WebView
import React, { Component } from 'react';
import { View, StyleSheet, Button, WebView } from 'react-native';
import { Constants } from 'expo';
export default class webView extends Component {
onNavigationStateChange = navState => {
if (navState.url.indexOf('https://www.google.com') === 0) {
const regex = /#access_token=(.+)/;
const accessToken = navState.url.match(regex)[1];
console.log(accessToken);
}
};
renderContent() {
return (
<WebView
source={{
uri: '',
}}
onNavigationStateChange={this.onNavigationStateChange}
startInLoadingState
scalesPageToFit
javaScriptEnabled
style={{ flex: 1 }}
/>
);
}
render() {
return (
<View style={styles.container}>
<Button
style={styles.paragraph}
title="Login"
onPress={() => this.renderContent()}
/>
</View>
);
}
}
const styles = StyleSheet.create({
container: {
flex: 1,
alignItems: 'center',
justifyContent: 'center',
paddingTop: Constants.statusBarHeight,
backgroundColor: '#ecf0f1',
},
});
我试过onPress={this.renderContent()}
这也是,但它给了一个例外。我能做什么 ?