我在反应本机和编程方面仍然是新手,我正在尝试将项目从我的平面列表传递到模式中。我要传递的是图标、状态和描述。我该怎么做?
这是我的平面列表
buildPanel(index, item) {
let panel = [];
let keys = DBkeys['Requests'].MyRequest;
let status = item[keys['status']];
panel.push(<View style={{ position: 'absolute', right: 0, bottom: 0, padding: normalize(5), alignItems: 'center' }} key={'status'}>
<TouchableOpacity onPress={this.handleShowModal}>
<Icon name={img.itemStatus[status].name} type={img.itemStatus[status].type} color={img.itemStatus[status].color} size={normalize(38)} />
</TouchableOpacity>
</View>);
return panel;
}
<View style={[styles.panelContainer, status === 'success' ? {} : { backgroundColor: color.white }]}>
<FlatList
showsVerticalScrollIndicator={false}
progressViewOffset={-10}
refreshing={this.state.refreshing}
onRefresh={this.onRefresh.bind(this)}
onMomentumScrollEnd={(event) => event.nativeEvent.contentOffset.y === 0 ? this.onRefresh() : null}
data={content}
renderItem={({ item }) => item}
keyExtractor={(item, key) => key.toString()} …Run Code Online (Sandbox Code Playgroud)我仍在学习 React Native,并且尝试从 API 检索数据并将其作为自定义单选按钮返回,但如果我确实调用 API,则会收到此错误:
null 不是对象(正在评估 'this.props.activities.map')
{this.props.activities.map((val, index) => {
let { key, type, placeholder } = val;
if (type === "selection") {
var buttons = [];
placeholder.forEach((e, index) => {
var selectedButton = img.findButton(e, true);
var normalButton = img.findButton(e);
buttons.push(
<RadioButton
key={index}
value={e}
element={<Image source={selectedButton} />}
selectedElement={<Image source={normalButton} />}
onPress={() => this.changeSelection(key, e)}
selected={this.state[key]["value"]}
/>
);
});
var rows = [],
columns = [];
var i = 0;
buttons.forEach((e, index) => {
rows.push(e);
i++;
if (i …Run Code Online (Sandbox Code Playgroud)