【发布时间】:2018-09-20 14:07:39
【问题描述】:
我在 HomeScreen.js 中有一个组件 LatestAdded.js,当按下它时,我想导航到另一个名为 ProductDetailScreen.js 的屏幕,但是当 HomeScreen.js 渲染它直接导航到 ProductDetailScreen.js 和如果我按下后退按钮并导航到 HomeScreen.js 并按下 LatestAdded.js 组件,它会显示 错误 true 不是函数(评估 this.props.viewdetail())
LastedAdded.js
<TouchableWithoutFeedback onPress={() => {
this.props.viewdetail();
}}>
<View style={{
flex: 1,
flexDirection: 'row',
borderRadius: 4,
borderWidth: 0.5,
borderColor: '#d6d7da', width: 180, padding: 5, marginRight: 5, height: 120
}}>
<View style={{}}>
<Text style={{ fontSize: 15, marginTop: 20 }}> {this.props.title}</Text>
<Text style={{ fontSize: 15, fontWeight: '700', marginTop: 10, marginBottom:10 }}> {this.props.author}</Text>
<StarRating
disabled={false}
maxStars={5}
starSize= {10}
rating={3}
fullStarColor= '#EDC430'
selectedStar={(rating) => this.onStarRatingPress(rating)}
/>
</View>
<Image style={{ width: 80, height: 100, margin: 2 }} source={{ uri: this.props.image }} />
</View>
</TouchableWithoutFeedback>
在 HomeScreen.js 中渲染方法
renderItem = ({ item }) => {
return (
<LatestAdded title={item.pro_name} author={item.pro_price} image={item.img_path}
viewdetail = { this.props.navigation.navigate('ProductDetail',{pro:item.pro_id})} />
)}
ProductDetailScreen.js
class ProductDetailScreen extends Component {
render() {
const { navigation } = this.props;
const itemId = navigation.getParam('pro', 'none');
return (
<View style={styles.container}>
<Text>{itemId}</Text>
</View>
);
}
}
【问题讨论】:
-
() => this.props.navigation.navigate('ProductDetail',{pro:item.pro_id})
标签: javascript reactjs react-native react-native-navigation