我正在为我们的产品使用'Native Base'组件并且很好用, 但是我一度陷入困境,而是把物品放在Nativebase Picker中。我的代码就像这样
渲染方法代码 -
render(){
return (
<View style={{marginTop: 20, flexDirection:'row', flexWrap:'wrap', justifyContent:'space-around', alignItems:'center'}}>
<View style={{flex:1, justifyContent:'center', alignItems:'flex-end' }}>
<Button
style={{ backgroundColor: '#6FAF98', }}
onPress={this._showDateTimePicker}
>
<Text>Select Date</Text>
</Button>
</View>
<View style={{flex:1, justifyContent:'center', alignItems:'stretch'}}>
<Picker
style={{borderWidth: 1, borderColor: '#2ac', alignSelf:'stretch'}}
supportedOrientations={['portrait','landscape']}
iosHeader="Select one"
mode="dropdown"
selectedValue={this.state.leaveType}
onValueChange={(value)=>this.setState({leaveType:value,})
//this.onValueChange.bind(this)
}>
<Item label="Full Day" value="leave1" />
{
this.showStartDateFirstHalf() // Here I want to show this picker item on the basis of a condition
}
<Item label="2nd half" value="leave3" />
</Picker>
</View>
<DateTimePicker
isVisible={this.state.isStartDatePickerPickerVisible}
onConfirm={this._handleDatePicked}
onCancel={this._hideDateTimePicker}
mode='date'
/>
</View>
);
}
showStartDateFirstHalf()
{
if(!this.state.isMultipleDays)
{
return(
<Item label="1st Half" value="leave2" />
);
}
}
所以,如果this.state.isMultipleDays
为false,此代码工作正常,但当this.state.isMultipleDays
为真时,则表示当它在else
部分时,我收到此错误 - < / p>
答案 0 :(得分:1)
我认为这是一个更容易的答案。不要创建单独的showStartDateFirstHalf()函数,而是尝试:
render() {
const pickerItems = [
{
label: 'Full Day',
value: 'leave1',
},
{
label: '1st Half',
value: 'leave2',
},
{
label: '2nd Half',
value: 'leave3',
},
];
const filteredItems = pickerItems.filter(item => {
if (item.value === 'leave2' && this.state.isMultipleDays) {
return false;
}
return true;
});
// The 'return' statement of your render function
return (
...
<Picker ...>
{(() =>
filteredItems.map(item =>
<Item label={item.label} value={item.value} />
)()}
</Picker>
...
);
}
这样,您已经拥有了在渲染周期的return
语句之前确定的项目列表。另外,如果条件不满足,则使用filter
代替map
不会仅将null
作为第二项,而是会完全删除该项目。