I have an an object with two fields filter1
and filter2
having values in form of array
let filter = {filter1:["mine","your"]: filter2:["C","D"]}
//values not fixed
the data is in form of array of objects
let data = [
{ id:1, filter1:["mine"], filter2:["E","C"]},
{ id:2, filter1:["mine"], filter2:["E","C","F"]},
{ id:3, filter1:["your"], filter2:["C"]},
{ id:3, filter1:["your"], filter2:["D","C"]},
{ id:5, filter1:["other"], filter2:["F"]},
...
]
I have to filter out those objects which have any one of the fields present in specific key
for eg if filter
is {filter1:["mine"]: filter2:["F","D"]}
it will first search for any element of filter1 in filter1 of data object and after that search for any element of filter2 which is present in the filter2 of data object and return the object if any one them is found
few example
result for {filter1:["mine"]: filter2:["F","D"]}
result = [
{ id:1, filter1:["mine"], filter2:["E","C"]}, //since filter1 "mine"
{ id:2, filter1:["mine"], filter2:["E","C","F"]}, //since filter1 "mine"
{ id:3, filter1:["your"], filter2:["D","C"]}, //since from "F" and "D" from filter2 "D" is present
{ id:5, filter1:["other"], filter2:["F"]}, //since "F" from filter2 is present
]
result for {filter1:["your"]: filter2:["F","G"]}
result = [
{ id:2, filter1:["mine"], filter2:["E","C","F"]}, //since "F" from filter2 is present
{ id:3, filter1:["your"], filter2:["D","C"]}, //since filter1 is "your"
{ id:5, filter1:["other"], filter2:["F"]}, //since "F" from filter2 is present
]
result for {filter1:[]: filter2:["D"]}
result = [
{ id:3, filter1:["your"], filter2:["D","C"]}, //since filter2 has "D"
]
CodePudding user response:
You can use a combination of .filter()
, .some()
and .includes()
:
const data = [
{ id:1, filter1:["mine"], filter2:["E","C"]},
{ id:2, filter1:["mine"], filter2:["E","C","F"]},
{ id:3, filter1:["your"], filter2:["C"]},
{ id:3, filter1:["your"], filter2:["D","C"]},
{ id:5, filter1:["other"], filter2:["F"]}
];
const search = ({ filter1, filter2 }) =>
data.filter(item =>
item.filter1.some(fitem => filter1.includes(fitem)) ||
item.filter2.some(fitem => filter2.includes(fitem))
);
const result = search({ filter1:["mine"], filter2:["F","D"] });
console.log(result);
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CodePudding user response:
You can generalize the solution proposed by RoMilton by calling some()
on the Object.entries()
of the passed filter object, and then iterating each key
and filter_array
with nested some()
calls.
If you also Array#concat()
the currently iterated data element property into an array you can include non-array properties in the filter object ie id
in this case.
const data = [
{ id: 1, filter1: ["mine"], filter2: ["E", "C"] },
{ id: 2, filter1: ["mine"], filter2: ["E", "C", "F"] },
{ id: 3, filter1: ["your"], filter2: ["C"] },
{ id: 3, filter1: ["your"], filter2: ["D", "C"] },
{ id: 5, filter1: ["other"], filter2: ["F"] }
];
const search = (array, filter_object) =>
array.filter(item =>
Object.entries(filter_object).some(([key, filter_array]) =>
[].concat(item[key]).some(fitem => filter_array.includes(fitem)))
);
const filter = { filter1: ["mine"], filter2: ["F", "D"] };
const result = search(data, filter);
console.log(...result.map(({ id }) => ({ id })));
const filter2 = { id: [5], filter1: ["mine"] }
const result2 = search(data, filter2);
console.log(...result2.map(({ id }) => ({ id })));
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