在这里查看了本地的反应文档和各种答案之后,我可以看到:
this.refs[WEBVIEW_REF].stopLoading();
用于阻止webView继续加载。
我试过这个,但总是得到:
undefined is not an object (evaluating 'this.refs[WEBVIEW_REF]')
我的代码如下,我试图阻止应用程序加载帮助页面并在浏览器而不是应用程序中启动。
import React, { Component } from 'react';
import {
AppRegistry,
StyleSheet,
View,
WebView,
Linking,
} from 'react-native';
const HEADER = '#3b5998';
const BGWASH = 'white';
const WEBVIEW_REF = 'webview';
const DEFAULT_URL = 'https://somesite.com/';
export default class reactNativeApp extends Component {
state = {
url: DEFAULT_URL,
scalesPageToFit: true,
};
_onShouldStartLoadWithRequest(e) {
if (e.url.indexOf('assistenza') >= 0) {
Linking.openURL(e.url);
return false
}
return true
}
_onNavigationStateChange(e) {
if (e.url.indexOf('assistenza') >= 0) {
this.refs[WEBVIEW_REF].stopLoading();
Linking.openURL(e.url);
return false
}
return true
}
render() {
return (
<View>
<WebView
ref={WEBVIEW_REF}
automaticallyAdjustContentInsets={false}
style={styles.webView}
source={{uri: this.state.url}}
javaScriptEnabled={true}
domStorageEnabled={true}
decelerationRate="normal"
startInLoadingState={true}
scalesPageToFit={this.state.scalesPageToFit}
onShouldStartLoadWithRequest={this._onShouldStartLoadWithRequest}
onNavigationStateChange={this._onNavigationStateChange}
/>
</View>
);
}
}
AppRegistry.registerComponent('reactNativeApp', () => reactNativeApp);
对此为何的任何想法?我见过的每个例子都使用它:/
答案 0 :(得分:2)
您需要使用以下任一方式绑定函数:
constructor (props) {
super(props)
this._onNavigationStateChange = this._onNavigationStateChange.bind(this)
}
或render
:
onNavigationStateChange={this._onNavigationStateChange.bind(this)}