想象一篇文章,当按下按钮时可以被点赞。这个按钮会修改远程数据库,所以将花费一些时间将喜欢的内容关联到具体的帖子上。
现在,如果用户使用以下代码快速按下按钮:
state = {
isLiked: false,
}
handlePress = () => {
this.setState(
{
isLiked: !this.state.isLiked,
},
this.handleLike
);
};
handleLike = async () => {
const { postId } = this.props;
try {
console.log(isLiked ? "Liking" : "Disliking")
await db.processLike(postId);
} catch (err) {
// If an error has occurred, reverse the 'isLiked' state
this.setState({
isLiked: !this.state.isLiked,
});
// TODO - Alert the error to the user in a toast
console.log(err);
}
console.log("DONE");
};
由于所有操作都是异步的,可能会出现以下情况:
喜欢
不喜欢
DONE <---------- 不喜欢已完成
DONE <---------- 喜欢已完成
我考虑创建一个“isLiking”状态来避免在所有异步作业完成之前运行代码。类似这样:
state = {
isLiking: false,
isLiked: false,
}
handlePress = () => {
if (this.state.isLiking) return; <------------------------------------
this.setState(
{
isLiking: true, <------------------------------------
isLiked: !this.state.isLiked,
},
this.handleLike
);
};
handleLike = async () => {
const { postId } = this.props;
try {
console.log(isLiked ? "Liking" : "Disliking");
await db.processLike(postId);
} catch (err) {
// If an error has occurred, reverse the 'isLiked' state
this.setState({
isLiked: !this.state.isLiked,
});
// TODO - Alert the error to the user in a toast
console.log(err);
}
this.setState({ isLiking: false }); <------------------------------------
console.log("DONE");
};
如果用户快速按下按钮,那么在代码上述过程完成之前,他将无法看到GUI更改(喜欢的按钮颜色(如果已喜欢,则为红色;如果未喜欢,则为白色))。
我还考虑编写一个防抖函数(用于handlePress),如下所示:
export const debounce = (func, wait, immediate) => {
/*
Returns a function, that, as long as it continues to be invoked, will not
be triggered. The function will be called after it stops being called for
N milliseconds. If `immediate` is passed, trigger the function on the
leading edge, instead of the trailing.
*/
let timeout;
return function () {
let context = this,
args = arguments;
let later = function () {
timeout = null;
if (!immediate) func.apply(context, args);
};
let callNow = immediate && !timeout;
clearTimeout(timeout);
timeout = setTimeout(later, wait);
if (callNow) func.apply(context, args);
};
};
...
debuncedHandlePress = debounce(this.handlePress, 500); // Now, when the button is pressed, it will call this function, instead of the original handlePress
但是这样做,我唯一能做的就是减少出现混乱结果的机会。也就是说,我仍然面临着与第一个代码相同的问题。
有没有什么办法可以让我按顺序获取结果,并避免在写入数据库时等待过长?
谢谢。