How to delete property from spread operator?
You could use Rest syntax in Object Destructuring to get all the properties except drugName to a rest variable like this:
const transformedResponse = [{
drugName: 'HYDROCODONE-HOMATROPINE MBR',
drugStrength: '5MG-1.5MG',
drugForm: 'TABLET',
brand: false
},
{
drugName: 'HYDROCODONE ABC',
drugStrength: '10MG',
drugForm: 'SYRUP',
brand: true
}]
const output = transformedResponse.map(({ drugName, ...rest }) => rest)
console.log(output)Also, when you spread an array inside {}, you get an object with indices of the array as key and the values of array as value. This is why you get an object with 0 as key in loggerResponse:
const array = [{ id: 1 }, { id: 2 }]
console.log({ ...array })Another option is to write a generic function, removeKey -
const removeKey = (k, { [k]:_, ...o }) =>
o
const values =
[ { a: 1, x: 1 }
, { a: 1, y: 1 }
, { a: 1, z: 1 }
]
console .log (values .map (v => removeKey ("a", v)))
// [ { x: 1 }, { y: 1 }, { z: 1 } ]The function can be easily adapted to remove multiple keys, if necessary -
const removeKey = (k = "", { [k]:_, ...o } = {}) =>
o
const removeKeys = (keys = [], o = {}) =>
keys .reduce ((r, k) => removeKey (k, r), o)
const values =
[ { a: 1, x: 1 }
, { a: 1, y: 1 }
, { a: 1, z: 1 }
]
console .log (values .map (v => removeKeys (['a', 'z'], v)))
// [ { x: 1 }, { y: 1 }, {} ]1 line solution using ES9 Object Rest Operator
const loggerResponse = {
"0": {
isBrand: false,
drugName: "test drug",
drugStrength: "5 mg 1 5 mg",
drugForm: "Tablet",
},
};
const { drugName, ...newResponse } = loggerResponse["0"];
console.log(newResponse);