我有一个React组件，在组件的渲染方法中，我有这样的东西:
render() {
return (
<div>
<div>
// removed for brevity
</div>
{ switch(...) {} }
<div>
// removed for brevity
</div>
</div>
);
}
Now the point is that I have two div elements, one at the top and one at the bottom, that are fixed. In the middle I want to have a switch statement, and according to a value in my state I want to render a different component. So basically, I want the two div elements to be fixed always, and just in the middle to render a different component each time. I'm using this to implement a multi-step payment procedure). Though, as is the code currently it doesn't work, as it gives me an error saying that switch is unexpected. Any ideas how to achieve what I want?
这个助手应该可以做到这一点。
使用示例:
{componentSwitch(3, (switcher => switcher
.case(1, () =>
<p>It is one</p>
)
.case(2, () =>
<p>It is two</p>
)
.default(() =>
<p>It is something different</p>
)
))}
助手:
interface SwitchCases<T> {
case: (value: T, result: () => React.ReactNode) => SwitchCases<T>;
default: (result: () => React.ReactNode) => SwitchCases<T>;
}
export function componentSwitch<T>(value: T, cases: (cases: SwitchCases<T>) => void) {
var possibleCases: { value: T, result: () => React.ReactNode }[] = [];
var defaultResult: (() => React.ReactNode) | null = null;
var getSwitchCases: () => SwitchCases<T> = () => ({
case: (value: T, result: () => React.ReactNode) => {
possibleCases.push({ value: value, result });
return getSwitchCases();
},
default: (result: () => React.ReactNode) => {
defaultResult = result;
return getSwitchCases();
},
})
// getSwitchCases is recursive and will add all possible cases to the possibleCases array and sets defaultResult.
cases(getSwitchCases());
// Check if one of the cases is met
for(const possibleCase of possibleCases) {
if (possibleCase.value === value) {
return possibleCase.result();
}
}
// Check if the default case is defined
if (defaultResult) {
// Typescript wrongly assumes that defaultResult is always null.
var fixedDefaultResult = defaultResult as (() => React.ReactNode);
return fixedDefaultResult();
}
// None of the cases were met and default was not defined.
return undefined;
}
你不能在渲染中有开关。放置访问一个元素的对象文字的伪切换方法并不理想，因为它会导致所有视图都要处理，并且可能导致在该状态下不存在的道具的依赖错误。
这里有一个很好的干净的方法来做到这一点，不需要每个视图提前渲染:
render () {
const viewState = this.getViewState();
return (
<div>
{viewState === ViewState.NO_RESULTS && this.renderNoResults()}
{viewState === ViewState.LIST_RESULTS && this.renderResults()}
{viewState === ViewState.SUCCESS_DONE && this.renderCompleted()}
</div>
)
如果视图状态的条件不只基于一个简单的属性——比如每行有多个条件，那么枚举和getViewState函数封装条件是分离条件逻辑和清理呈现的好方法。
