Question

我想点击其中一个数据列表，并在id与我们在React Native中点击的内容匹配时更改绿色。所以，如果我在id 1中单击callCircleButton而只是在id 1中单击更改带有名称的图标

关注我的代码：

constructor(props) {
    super(props);
    this.state = {
        statusButton: true,
        dataList: [
        {id: 1, name: 'te1'},
        {id: 2, name: 'test2'},
        ]
    }

callCircleButton(){
    if(this.state.statusButton == true){
      this.setState({statusButton: false})
    }else{
      this.setState({statusButton: true})
    }
  }

render() {
    return(
        {this.state.dataList.map((data, i) => {
            return (
                <View style={{flex:1, flexDirection: 'row'}}>
                    <Text>data.name</Text>
                    <TouchableOpacity onPress={() => this.callCircleButton()}>
                        { this.state.statusButton ?
                            <Icon active size={25} name='panorama-fish-eye' style={{marginRight:10, color: 'green'}}/>
                            :
                            <Icon active size={25} name='lens' style={{marginRight:10, color: 'white'}}/>
                        }
                    </TouchableOpacity>
                </View>
            )
        })}
    )
}

Answer 1

单击时将活动按钮的ID设置为状态，并检查循环中的状态值。例如：

127.0.0.1 5432 <nil>
::1 2345 <nil>
localhost 1234 <nil>

Answer 2

更改您的json数据结构。

insert into customer_price_list  
(customer_price_id, customer_account_id, product_id, currency_id, customer_price, source) 
select 0, customer_account_id, :product_id, :currency_id, :price, 'TRADE_DEFAULT' from customer_account
ON DUPLICATE KEY UPDATE customer_price=customer_price;

如何仅根据在React-Native中单击的ID单击按钮？

2 个答案: