I want to filter the data of diaryItems that do not match the key value using the innerstate key value in JavaScript and put it in the answer constant.
I used the code like this, but I didn't get the desired value. How do I fix the code?
this is my code
const innerstate = {
instar: 'egg',
feedType: 'feed',
feedAmount: 0,
inputfeedtime: 0,
wormSize: 'third',
inputamount: 0,
}
const diaryItems =
[
{diaryItemId: 174, name: "instar"},
{diaryItemId: 175, name: "recordFeedPeriod"},
{diaryItemId: 176, name: "feedPeriod"},
{diaryItemId: 177, name: "feedAmount"},
{diaryItemId: 178, name: "feedType"},
{diaryItemId: 195, name: "totalFeedAmount"},
{diaryItemId: 196, name: "inputfeedtime"},
{diaryItemId: 197, name: "wormSize"},
{diaryItemId: 198, name: "inputamount"}
]
(expected answer)
const answer = [
{diaryItemId: 174, name: "instar"},
{diaryItemId: 178, name: "feedType"},
{diaryItemId: 177, name: "feedAmount"},
{diaryItemId: 196, name: "inputfeedtime"},
{diaryItemId: 197, name: "wormSize"},
{diaryItemId: 198, name: "inputamount"}
]
I tried this code but it doesn't work
const answer = diaryItems.filter((v) => v.name !==
Object.keys(innerState))
You were close. You need to check if the keys contain the current entry's name, instead of checking for equality to all keys:
const innerstate = {
instar: 'egg',
feedType: 'feed',
feedAmount: 0,
inputfeedtime: 0,
wormSize: 'third',
inputamount: 0,
};
const diaryItems = [
{diaryItemId: 174, name: "instar"},
{diaryItemId: 175, name: "recordFeedPeriod"},
{diaryItemId: 176, name: "feedPeriod"},
{diaryItemId: 177, name: "feedAmount"},
{diaryItemId: 178, name: "feedType"},
{diaryItemId: 195, name: "totalFeedAmount"},
{diaryItemId: 196, name: "inputfeedtime"},
{diaryItemId: 197, name: "wormSize"},
{diaryItemId: 198, name: "inputamount"}
];
const keys = Object.keys(innerstate);
const result = diaryItems.filter(({name}) => keys.includes(name));
console.log(result);You just need to check if the name is in the keys array
const answer = diaryItems.filter((v) =>
Object.keys(innerstate).includes(v.name)
)