Another good solution: (requires underscore)

x.toJSON = function () {
    return _.omit(this, [ "privateProperty1", "privateProperty2" ]);
};

The benefit of this solution is that anyone calling JSON.stringify on x will have correct results - you don't have to alter the JSON.stringify calls individually.

Non-underscore version:

x.toJSON = function () {
    var result = {};
    for (var x in this) {
        if (x !== "privateProperty1" && x !== "privateProperty2") {
            result[x] = this[x];
        }
    }
    return result;
};

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

1. delete method: The most common way

const myObject = {
    "employeeid": "160915848",
    "firstName": "tet",
    "lastName": "test",
    "email": "[email protected]",
    "country": "Brasil",
    "currentIndustry": "aaaaaaaaaaaaa",
    "otherIndustry": "aaaaaaaaaaaaa",
    "currentOrganization": "test",
    "salary": "1234567"
};

delete myObject['currentIndustry'];
// OR delete myObject.currentIndustry;
  
console.log(myObject);

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

let myObject = {
    "employeeid": "160915848",
    "firstName": "tet",
    "lastName": "test",
    "email": "[email protected]",
    "country": "Brasil",
    "currentIndustry": "aaaaaaaaaaaaa",
    "otherIndustry": "aaaaaaaaaaaaa",
    "currentOrganization": "test",
    "salary": "1234567"
  };

myObject.currentIndustry = undefined;
myObject = JSON.parse(JSON.stringify(myObject));

console.log(myObject);

3. With es6 spread Operator:

const myObject = {
    "employeeid": "160915848",
    "firstName": "tet",
    "lastName": "test",
    "email": "[email protected]",
    "country": "Brasil",
    "currentIndustry": "aaaaaaaaaaaaa",
    "otherIndustry": "aaaaaaaaaaaaa",
    "currentOrganization": "test",
    "salary": "1234567"
};


const {currentIndustry, ...filteredObject} = myObject;
console.log(filteredObject);

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 :)

You can use a replacer function for this.

var obj = {
  "id": 1,
  "name": "Light Switch",
  "lightStatus": true,
  "inputPort": {
    "id": 2,
    "value": 0
  },
  "outputPort": {
    "id": 2,
    "value": false
  },
  "resistance": 100
};

var stringified = JSON.stringify(obj, function(key, val) {
  if (key === 'id' || key === 'lightStatus') {
    return void(0);
  }
  if (key === 'inputPort' || key === 'outputPort') {
    return val.id;
  }
  return val;
});

console.log(stringified);

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

1. delete method: The most common way

const myObject = {
    "employeeid": "160915848",
    "firstName": "tet",
    "lastName": "test",
    "email": "test@email.com",
    "country": "Brasil",
    "currentIndustry": "aaaaaaaaaaaaa",
    "otherIndustry": "aaaaaaaaaaaaa",
    "currentOrganization": "test",
    "salary": "1234567"
};

delete myObject['currentIndustry'];
// OR delete myObject.currentIndustry;
  
console.log(myObject);

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

let myObject = {
    "employeeid": "160915848",
    "firstName": "tet",
    "lastName": "test",
    "email": "test@email.com",
    "country": "Brasil",
    "currentIndustry": "aaaaaaaaaaaaa",
    "otherIndustry": "aaaaaaaaaaaaa",
    "currentOrganization": "test",
    "salary": "1234567"
  };

myObject.currentIndustry = undefined;
myObject = JSON.parse(JSON.stringify(myObject));

console.log(myObject);

3. With es6 spread Operator:

const myObject = {
    "employeeid": "160915848",
    "firstName": "tet",
    "lastName": "test",
    "email": "test@email.com",
    "country": "Brasil",
    "currentIndustry": "aaaaaaaaaaaaa",
    "otherIndustry": "aaaaaaaaaaaaa",
    "currentOrganization": "test",
    "salary": "1234567"
};


const {currentIndustry, ...filteredObject} = myObject;
console.log(filteredObject);

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 :)

You should not stringify the object before removing the key.

function test() {
  var json = {
    "ID": "2196",
    "VERSION": "1-2022",
    "FILE": "2196.docx"
  };
  // stringify to show in console as string
  json = JSON.stringify(json)
  console.log("json " + json);

  // convert back to object so you can remove the key
  json = JSON.parse(json);
  delete json['FILE'];
  
  // stringify to show new object in console as string
  json = JSON.stringify(json);
  console.log("json " + json);
  return;
}

test();

Delete the key on a copy of the objecrt before stringifying the object to get your result. You can create some sort of function that deletes a key from an object if it exists.

var obj = {
    id: '1',
    presetKey: 'abc',
    preset: 'preset',
}

function filterPreset(obj, key) {
  if (typeof obj === 'object' && obj[key]) {
    var copy = Object.assign({}, obj);
    delete copy[key];
    return copy;
  }
  return obj;
}

console.log('result', JSON.stringify(filterPreset(obj, 'preset')));
console.log('no mutation', obj);

It looks like this is a simple Object toString method that you are looking for.

In Node.js this is solved by using the util object and calling util.inspect(yourObject). This will give you all that you want. follow this link for more options including depth of the application of method. http://nodejs.org/api/util.html#util_util_inspect_object_options

So, what you are looking for is basically an object inspector not a JSON converter. JSON format specifies that all properties must be enclosed in double quotes. Hence there will not be JSON converters to do what you want as that is simply not a JSON format.Specs here: https://developer.mozilla.org/en-US/docs/Using_native_JSON

Object to string or inspection is what you need depending on the language of your server.

Yes, it is possible to remove the quotes from the keys. Doing so will render it invalid as JSON, but still valid when used directly in JavaScript code, only if the keys you use are also valid variable names. When you paste a JSON object in JavaScript code, it may be useful to have the quotes removed, as they can in some cases take up a lot of data. If that's relevant to your project, then this is the answer for you.

The function below works for any JavaScript variable and works fine with objects/arrays containing objects/arrays. I've added comments to explain what's going on. Please note that this code does not check for possible reserved keywords that may not be allowed as JavaScript keys, such as "for".

function toUnquotedJSON(param){ // Implemented by Frostbolt Games 2022
    if(Array.isArray(param)){ // In case of an array, recursively call our function on each element.
        let results = [];
        for(let elem of param){
            results.push(toUnquotedJSON(elem));
        }
        return "[" + results.join(",") + "]";
    }
    else if(typeof param === "object"){ // In case of an object, loop over its keys and only add quotes around keys that aren't valid JavaScript variable names. Recursively call our function on each value.
        let props = Object
            .keys(param)
            .map(function(key){
                // A valid JavaScript variable name starts with a dollar sign (?), underscore (_) or letter (a-zA-Z), followed by zero or more dollar signs, underscores or alphanumeric (a-zA-Z\d) characters.
                if(key.match(/^[a-zA-Z_$][a-zA-Z\d_$]*$/) === null) // If the key isn't a valid JavaScript variable name, we need to add quotes.
                    return `"${key}":${toUnquotedJSON(param[key])}`;
                else
                    return `${key}:${toUnquotedJSON(param[key])}`;
            })
            .join(",");
        return `{${props}}`;
    }
    else{ // For every other value, simply use the native JSON.stringify() function.
        return JSON.stringify(param);
    }
}