[javascript] turn typescript object into json string

I'm trying to initialize an object in typescript which requires a JSON string for the "options" parameter. To be precise it is the object here. The options parameter is required to be a JSON string and not an object for initializing the dijit.

Is there a way to create a JSON string from a typescript object without it being a manual process?

Please DO NOT link any questions which don't specifically say "TypeScript" as this question specifically relates to TypeScript. While a derivative of JavaScript the way that you write code is different and therefore this is the only post asking this question currently relating to TypeScript.

Be careful when using these JSON.(parse/stringify) methods. I did the same with complex objects and it turned out that an embedded array with some more objects had the same values for all other entities in the object tree I was serializing.

const temp = [];
const t = {
    name: "name",
    etc: [
        {
            a: 0,
        },
    ],
};
for (let i = 0; i < 3; i++) {
    const bla = Object.assign({}, t);
    bla.name = bla.name + i;
    bla.etc[0].a = i;
    temp.push(bla);
}

console.log(JSON.stringify(temp));

If you're using fs-extra, you can skip the JSON.stringify part with the writeJson function:

const fsExtra = require('fs-extra');

fsExtra.writeJson('./package.json', {name: 'fs-extra'})
.then(() => {
  console.log('success!')
})
.catch(err => {
  console.error(err)
})

You can use the standard JSON object, available in Javascript:

var a: any = {};
a.x = 10;
a.y='hello';
var jsonString = JSON.stringify(a);

TS gets compiled to JS which then executed. Therefore you have access to all of the objects in the JS runtime. One of those objects is the JSON object. This contains the following methods:

• JSON.parse() method parses a JSON string, constructing the JavaScript value or object described by the string.
• JSON.stringify() method converts a JavaScript object or value to a JSON string.

Example:

const jsonString = '{"employee":{ "name":"John", "age":30, "city":"New York" }}';_x000D_
_x000D_
_x000D_
const JSobj = JSON.parse(jsonString);_x000D_
_x000D_
console.log(JSobj);_x000D_
console.log(typeof JSobj);_x000D_
_x000D_
const JSON_string = JSON.stringify(JSobj);_x000D_
_x000D_
console.log(JSON_string);_x000D_
console.log(typeof JSON_string);