设置状态时在本机比较颜色对象
it1352
帮助3人
问题说明
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:)
正确答案
#1
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,
});
}
....
};
这篇好文章是转载于：学新通技术网
