设置状态时如何在本机中比较颜色对象 [英] How to compare color object in react native while setting state

问题描述

有人可以帮我比较一下本机中的彩色对象吗?通常我会很快这样做:

Can some one please help me in comparing color object in react native. Generally in swift I would have done so:

if aColorObj == UIColorClass.white{
  aColorObj = UIColorClass.gray
}

如何在下面的本机反应中执行类似操作:

How to do similar in react native below:

onPressLearnMore() {

  this.setState = {
    /*
      how to write this:
       if mbackgroundColor == 'white'{
      mbackgroundColor = 'red'
      }else if mbackgroundColor == 'gray'{
      mbackgroundColor = 'white'
      } 
    */
  };
}

谢谢:)

推荐答案

这里最简单的解决方案也是，将初始背景色保存为一种状态，并具有可用于与颜色进行比较的颜色对象.

The simplest solution here is too, save the initial background color in a state, have colors object that you can use to compare color with.

const Colors = {
  Grey: '#DCDCDC',
  White: '#FFFFFF',
  Blue: '#0000FF',
  Black: '#000000',
};


state = {
  backgroundColor: Colors.White,
};

<View
    style={[
      styles.container,
      { backgroundColor: this.state.backgroundColor }, // set background color here from state
    ]}>

然后您可以使用一个功能来检查背景色.

Then you can use a function to check background color.

checkBackgroundColor = () => {

  if (this.state.backgroundColor === Colors.Blue) {
    console.log("It's blue");

    this.setState({
      backgroundColor: Colors.White,
    });
  }
  ....
};

零食示例

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