I'm struggling to find the index of the array that has id === 3.
const id = 3;
const bigList = [
[
{ id: "1", foo: "a" },
{ id: "2", foo: "b" }
],
[
{ id: "3", foo: "c" },
{ id: "4", foo: "d" }
],
[
{ id: "5", foo: "e" },
{ id: "6", foo: "f" }
]
];
I tried this to filter out that array.
const filteredList = bigList.filter(list => list.filter(item => item.id === id))
But it was wrong.
I really appreciate if you show me how it works. Thanks!
You're really close.
One way to do it is this,
bigList.filter(list => list.filter(item => item.id === "3").length !== 0);
If you look at the documentation of filter, you'll see that it returns a new array. So, the result of your inner filter would yield an array that needs to be evaluated to a true or false condition.
One way to do that is to check the length of the array.
You can use findIndex with a some(). Also notice that id variable is number but property of the object is a string
const id = 3;
const bigList = [
[
{ id: "1", foo: "a" },
{ id: "2", foo: "b" }
],
[
{ id: "3", foo: "c" },
{ id: "4", foo: "d" }
],
[
{ id: "5", foo: "e" },
{ id: "6", foo: "f" }
]
];
const res = bigList.findIndex(x => x.some(y => y.id === id.toString()));
console.log(res)You could use some to check if the list has the element with id of 3, and remember to cast the id to string before compare
const id = 3;
const bigList = [
[
{ id: "1", foo: "a" },
{ id: "2", foo: "b" },
],
[
{ id: "3", foo: "c" },
{ id: "4", foo: "d" },
],
[
{ id: "5", foo: "e" },
{ id: "6", foo: "f" },
],
];
const filteredList = bigList.filter((list) =>
list.some((item) => Number(item.id) === id)
);
console.log(filteredList);or you could use Abstract Equality Comparison (==) to compare, then item.id will be coerced to number
const id = 3;
const bigList = [
[
{ id: "1", foo: "a" },
{ id: "2", foo: "b" },
],
[
{ id: "3", foo: "c" },
{ id: "4", foo: "d" },
],
[
{ id: "5", foo: "e" },
{ id: "6", foo: "f" },
],
];
const filteredList = bigList.filter((list) =>
list.some((item) => item.id == id)
);
console.log(filteredList);Use the filter but only once so that inside it you can control the return value, like this:
console.log(bigList.filter((el) => {
var returnVal = false;
for (var j = 0; j < el.length; j++) {
if (el[j].id == 3) {
returnVal = true;
break;
}
else { returnVal = false; }
}
return returnVal;
}));
You can flatten your array using array#flat and then filter it based on searchId.
const searchId = 3,
bigList = [ [ { id: "1", foo: "a" }, { id: "2", foo: "b" } ], [ { id: "3", foo: "c" }, { id: "4", foo: "d" } ], [ { id: "5", foo: "e" }, { id: "6", foo: "f" } ] ],
result = bigList
.flat()
.filter(({id }) => id === String(searchId));
console.log(result);You can use findIndex method to find the index of the array containing id = 3.
bigList.findIndex((item) => (item.findIndex(el => el.id === id.toString())) > -1)
indexor all the matched elements in the list?