The Mozilla docs say to return undefined (instead of "none"):

http://jsfiddle.net/userdude/rZ5Px/

function replacer(key,value)
{
    if (key=="privateProperty1") return undefined;
    else if (key=="privateProperty2") return undefined;
    else return value;
}

var x = {
    x:0,
    y:0,
    divID:"xyz",
    privateProperty1: 'foo',
    privateProperty2: 'bar'
};

alert(JSON.stringify(x, replacer));

Here is a duplication method, in case you decide to go that route (as per your comment).

http://jsfiddle.net/userdude/644sJ/

function omitKeys(obj, keys)
{
    var dup = {};
    for (var key in obj) {
        if (keys.indexOf(key) == -1) {
            dup[key] = obj[key];
        }
    }
    return dup;
}

var x = {
    x:0,
    y:0,
    divID:"xyz",
    privateProperty1: 'foo',
    privateProperty2: 'bar'
};

alert(JSON.stringify(omitKeys(x, ['privateProperty1','privateProperty2'])));

EDIT - I changed the function key in the bottom function to keep it from being confusing.

delete operator is used to remove an object property.

delete operator does not returns the new object, only returns a boolean: true or false.

In the other hand, after interpreter executes var updatedjsonobj = delete myjsonobj['otherIndustry']; , updatedjsonobj variable will store a boolean value.

How to remove Json object specific key and its value ?

You just need to know the property name in order to delete it from the object's properties.

delete myjsonobj['otherIndustry'];

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

If you want to remove a key when you know the value you can use Object.keys function which returns an array of a given object's own enumerable properties.

let value="test";
let myjsonobj = {
      "employeeid": "160915848",
      "firstName": "tet",
      "lastName": "test",
      "email": "[email protected]",
      "country": "Brasil",
      "currentIndustry": "aaaaaaaaaaaaa",
      "otherIndustry": "aaaaaaaaaaaaa",
      "currentOrganization": "test",
      "salary": "1234567"
}
Object.keys(myjsonobj).forEach(function(key){
  if (myjsonobj[key] === value) {
    delete myjsonobj[key];
  }
});
console.log(myjsonobj);

I think what you actually want is to do the following on service:

body: JSON.stringify({ ...bootStrapInputList }),

That line should result in:

{[{"resource":"Providers"},{"resource":"ServiceLocation"}]}

I saw jsN00b comment and while you said it worked and of course, I believe you and I'm happy you solved it, but doing:

body: JSON.stringify( bootStrapInputList ),

would result in:

[{"resource":"Providers"},{"resource":"ServiceLocation"}]

Which is not the same (no object wrapping the array).

Good luck!

You can pass a "replacer" function that returns the exact value you want.

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

var result = JSON.stringify(data, function(k, v) {
    switch (k) {
    case "": case "name": case "resistance":
    	return v
    case "inputPort": case "outputPort":
    	return v.id
    default:
    	return undefined;
  }
}, 2)

document.querySelector("pre").textContent = result

<pre></pre>

The "" represents the top level object. For that, "name", and "resistance", it simply returns the original value.

For "inputPort" and "outputPort" it returns the id property.

Anything else gets undefined, which means it gets omitted from the result.

delete operator is used to remove an object property.

delete operator does not returns the new object, only returns a boolean: true or false.

In the other hand, after interpreter executes var updatedjsonobj = delete myjsonobj['otherIndustry']; , updatedjsonobj variable will store a boolean value.

How to remove Json object specific key and its value ?

You just need to know the property name in order to delete it from the object's properties.

delete myjsonobj['otherIndustry'];

let myjsonobj = {
  "employeeid": "160915848",
  "firstName": "tet",
  "lastName": "test",
  "email": "test@email.com",
  "country": "Brasil",
  "currentIndustry": "aaaaaaaaaaaaa",
  "otherIndustry": "aaaaaaaaaaaaa",
  "currentOrganization": "test",
  "salary": "1234567"
}
delete myjsonobj['otherIndustry'];
console.log(myjsonobj);

If you want to remove a key when you know the value you can use Object.keys function which returns an array of a given object's own enumerable properties.

let value="test";
let myjsonobj = {
      "employeeid": "160915848",
      "firstName": "tet",
      "lastName": "test",
      "email": "test@email.com",
      "country": "Brasil",
      "currentIndustry": "aaaaaaaaaaaaa",
      "otherIndustry": "aaaaaaaaaaaaa",
      "currentOrganization": "test",
      "salary": "1234567"
}
Object.keys(myjsonobj).forEach(function(key){
  if (myjsonobj[key] === value) {
    delete myjsonobj[key];
  }
});
console.log(myjsonobj);

JSON.stringify has an often overlooked parameter called the replacer. It can accept an array of key names to include in the json output:

 JSON.stringify(data, ["ID","VERSION"], " ");  // FILE is excluded 

Snippet

Run the snippet to see the output with and without using the replacer.

let data = {
  "ID": "2196",
  "VERSION": "1-2022",
  "FILE": "2196.docx"
};

console.log("Without replacer: ",
  JSON.stringify(data, null, " ")
);

console.log("Using replacer: ",
  JSON.stringify(data, ["ID","VERSION"], " ")
);

This simple regular expression solution works to unquote JSON property names in most cases:

const object = { name: 'John Smith' };
const json = JSON.stringify(object);  // {"name":"John Smith"}
console.log(json);
const unquoted = json.replace(/"([^"]+)":/g, '$1:');
console.log(unquoted);  // {name:"John Smith"}

Extreme case:

var json = '{ "name": "J\\":ohn Smith" }'
json.replace(/\\"/g,"\uFFFF");  // U+ FFFF
json = json.replace(/"([^"]+)":/g, '$1:').replace(/\uFFFF/g, '\\\"');
// '{ name: "J\":ohn Smith" }'

Special thanks to Rob W for fixing it.

Limitations

In normal cases the aforementioned regexp will work, but mathematically it is impossible to describe the JSON format with a regular expression such that it will work in every single cases (counting the same number of curly brackets is impossible with regexp.) Therefore, I have create a new function to remove quotes by formally parsing the JSON string via native function and reserialize it:

function stringify(obj_from_json) {
    if (typeof obj_from_json !== "object" || Array.isArray(obj_from_json)){
        // not an object, stringify using native function
        return JSON.stringify(obj_from_json);
    }
    // Implements recursive object serialization according to JSON spec
    // but without quotes around the keys.
    let props = Object
        .keys(obj_from_json)
        .map(key => `${key}:${stringify(obj_from_json[key])}`)
        .join(",");
    return `{${props}}`;
}

Example: https://jsfiddle.net/DerekL/mssybp3k/

Your replacer function is not correct. It needs to return the value whenever you want to include the key/value in the JSON. Your code makes it only return undefined whatever the key/value.

So use this one:

function replacer(key, value) {
    if (key !== "preset") { return value; }
}

Or alternatively, you could test for the type of the value:

function replacer(key, value) {
    if (!(value instanceof Preset)) { return value; }
}

You can use the replacer. The second parameter provided by JSON.stringify.Replacer could be a function or array.

In your case we can create a function which replaces all the special characters with a blank space.The below example replaces the whitespaces and underscores.

function replacer(key, value) {
  return value.replace(/[^\w\s]/gi, '');
}

var foo = {"a":"1","b":2};
var jsonString = JSON.stringify(foo, replacer);

If you simply want to replace the one special character, use:

JSON.stringify({ a: 1, b: 2 }, null, '\t');

For more information on replacer, check the MDN page JSON.stringify().