[javascript] how to remove json object key and value.?

There are several ways to do this, lets see them one by one:

  1. delete method: The most common way

_x000D_
_x000D_
const myObject = {_x000D_
    "employeeid": "160915848",_x000D_
    "firstName": "tet",_x000D_
    "lastName": "test",_x000D_
    "email": "[email protected]",_x000D_
    "country": "Brasil",_x000D_
    "currentIndustry": "aaaaaaaaaaaaa",_x000D_
    "otherIndustry": "aaaaaaaaaaaaa",_x000D_
    "currentOrganization": "test",_x000D_
    "salary": "1234567"_x000D_
};_x000D_
_x000D_
delete myObject['currentIndustry'];_x000D_
// OR delete myObject.currentIndustry;_x000D_
  _x000D_
console.log(myObject);
_x000D_
_x000D_
_x000D_

  1. By making key value undefined: Alternate & a faster way:

_x000D_
_x000D_
let myObject = {_x000D_
    "employeeid": "160915848",_x000D_
    "firstName": "tet",_x000D_
    "lastName": "test",_x000D_
    "email": "[email protected]",_x000D_
    "country": "Brasil",_x000D_
    "currentIndustry": "aaaaaaaaaaaaa",_x000D_
    "otherIndustry": "aaaaaaaaaaaaa",_x000D_
    "currentOrganization": "test",_x000D_
    "salary": "1234567"_x000D_
  };_x000D_
_x000D_
myObject.currentIndustry = undefined;_x000D_
myObject = JSON.parse(JSON.stringify(myObject));_x000D_
_x000D_
console.log(myObject);
_x000D_
_x000D_
_x000D_

  1. With es6 spread Operator:

_x000D_
_x000D_
const myObject = {_x000D_
    "employeeid": "160915848",_x000D_
    "firstName": "tet",_x000D_
    "lastName": "test",_x000D_
    "email": "[email protected]",_x000D_
    "country": "Brasil",_x000D_
    "currentIndustry": "aaaaaaaaaaaaa",_x000D_
    "otherIndustry": "aaaaaaaaaaaaa",_x000D_
    "currentOrganization": "test",_x000D_
    "salary": "1234567"_x000D_
};_x000D_
_x000D_
_x000D_
const {currentIndustry, ...filteredObject} = myObject;_x000D_
console.log(filteredObject);
_x000D_
_x000D_
_x000D_

Or if you can use omit() of underscore js library:

const filteredObject = _.omit(currentIndustry, 'myObject');
console.log(filteredObject);

When to use what??

If you don't wanna create a new filtered object, simply go for either option 1 or 2. Make sure you define your object with let while going with the second option as we are overriding the values. Or else you can use any of them.

hope this helps :)

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