[英]Redux Reducer: Update object in array by key immutably
Question: Update the object in devData array with key = 1 this way: -> update data and dateTime 问题:通过以下方式用键= 1更新devData数组中的对象:->更新数据和dateTime
//Redux state:
{
info: {
success: true,
devData: [
{
key: 1,
data: { <- update this
power: 48 ,
unit: "kWh"
},
dateTime: "2019-08-24T18:21:21.000Z" <- update this
},
{
key: 2,
data: {
power: 48,
unit: "kWh"
},
dateTime: "2019-08-24T18:21:01.000Z"
}
]
}
}
My update method in reducer: 我在reducer中的更新方法:
switch (action.type) {
case 'update':
return {
...state,
info: {
...state.info,
devData: state.info.devData.map(currentValue => {
if (currentValue.key === 1) {
currentValue.data = action.payload.data;
currentValue.dateTime = action.payload.dateTime;
}
return currentValue;
})
}
};
}
My problems: 我的问题:
My update method is complicating and ugly, difficult to understand 我的更新方法复杂且丑陋,难以理解
I am not sure, if I am really doing this update in an immutable way 我不确定,如果我真的以不变的方式进行此更新
I think my update method is computationally expensive , not efficient 我认为我的更新方法计算量很大 ,效率不高
I need a method for updating to address the problems above. 我需要一种更新方法来解决上述问题。
First install the immer
. 首先安装
immer
。
and just update a part of your state
which you want. 并只更新您想要的
state
的一部分。
import produce from "immer"
(...)
switch (action.type) {
case 'update':
return produce(state, draft => {
draft.info.devData[0].data = action.payload.data;
draft.info.devData[0].dateTime = action.payload.dateTime;
//This code is working ONLY with devData[0]
//So, You SHOULD change this above code to work as dynamically.
});
}
immer
is a super simple way to update the state
with immutability. immer
是使用不变性更新state
的超级简单方法。
I strongly recommend using it. 我强烈建议使用它。
case “update”:
const index = this.state.devData.findIndex(v => v.key === 1);
return {
...state,
info: {
...state.info,
devData:
[
return [
...state.info.devData.slice(0, index),
{
data: action.payload.data,
dateTime: action.payload.dateTime
},
...state.slice(index + 1)
]
]
}
};
the idea is to first find an index of an element neede to update. 这个想法是首先找到需要更新的元素的索引。 after that use spread operators to create new areay made of:
slice
of everything before element to be updated + updated element + everything after updated element. 之后,使用传播运算符创建新的区域,该区域由以下组成:待更新元素之前的所有内容的
slice
+更新元素+已更新元素之后的所有内容。
ps. ps。 writing from my mobile so didnt test it
从我的手机写,所以没有测试
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