“未定义不是（评价‘child.props.value’）的对象”在反应天然

Question

我所遇到当我用选择器组件，如下所示该错误，

<Picker 
    style={{ flex: 1 }}> 
    selectedValue={this.props.shift} 
    onValueChange={value => this.props.employeeFormAction({ 
              prop:'shift', value })} 
> 
     <Picker.Item label='Monday' value='Monday' /> 
     <Picker.Item label='Tuesday' value='Tuesday' /> 
     <Picker.Item label='Wednesday' value='Wednesday' /> 
     <Picker.Item label='Thursday' value='Thursday' /> 
     <Picker.Item label='Friday' value='Friday' /> 
     <Picker.Item label='Saturday' value='Saturday' /> 
     <Picker.Item label='Sunday' value='Sunday' /> 
</Picker> 

余误差选择器部件已经厌倦了来自同一社区

react-native - Picker - undefined is not an object (evaluating 'this.props.children[position].props)

这一解决方案，但它并没有为我工作。任何机构能否为这个问题提出解决方案。

Answer 1

尽量不要硬编码值。这种方式更清洁：

// inside your component (supposing is not a functional component) 
_renderShifts =() => { 
    const shifts = ['Monday', 'Tuesday', 'Wednesday','Thursday', 
       'Friday','Saturday', 'Sunday']; 

    return shifts.map((shift) => { 
    <Picker.Item key={"shift_"+i} label={shift} value={shift} /> 
    }); 
} 


// now your picker should render this way 
_renderPicker =() => { 
    <Picker 
    style={{ flex: 1 }}> 
    selectedValue={this.props.shift} 
    onValueChange={value => this.props.employeeFormAction({prop:'shift', value })} 
    > 
    {this._renderShifts()} 
    </Picker> 
}