如何在 React Native 中隐藏 Picker 的第一个 Picker Item？

Question

This is my sample code, in Android I cant select first Picker Item at first time, i have created one dummy item so i need to hide first field.

<Picker 
    selectedValue={ this.state.selectedValue }
    onValueChange={(item) => this.setState({selectedValue: item})}>
    <Picker.Item label={"Slect Any"} value={null} />
    <Picker.Item label={"item1"} value={'item1'} key={key} />
    <Picker.Item label={"item2"} value={'item2'} key={key} />
    <Picker.Item label={"item3"} value={'item3'} key={key} />
</Picker>

Any one help this.

Answer 1

You'll need your component to have an extra boolean field canSelectAny in the state:

<Picker 
    selectedValue={ this.state.selectedValue }
    onValueChange={(item) => this.setState({selectedValue: item})}>
    {this.state.canSelectAny && <Picker.Item label={"Select Any"} value={null} />}
    <Picker.Item label={"item1"} value={'item1'} key={key} />
    <Picker.Item label={"item2"} value={'item2'} key={key} />
    <Picker.Item label={"item3"} value={'item3'} key={key} />
</Picker>

Answer 2

这应该是你问题的答案

{this.state.loading = false ? (<Picker.Item label="Art" value="Art" />) :null}

Answer 3

This is my sample code, in Android I cant select first Picker Item at first time, i have created one dummy item so i need to hide first field.

<Picker 
    selectedValue={ this.state.selectedValue }
    onValueChange={(item) => this.setState({selectedValue: item})}>
    <Picker.Item label={"Slect Any"} value={null} />
    <Picker.Item label={"item1"} value={'item1'} key={key} />
    <Picker.Item label={"item2"} value={'item2'} key={key} />
    <Picker.Item label={"item3"} value={'item3'} key={key} />
</Picker>

Any one help this.