Javascript 将 typescript 对象转换为 json 字符串
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turn typescript object into json string
提问by user1567453
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.
我正在尝试在打字稿中初始化一个对象，它需要一个 JSON 字符串作为“选项”参数。准确的说是这里的对象。options 参数必须是 JSON 字符串，而不是用于初始化 dijit 的对象。
Is there a way to create a JSON string from a typescript object without it being a manual process?
有没有办法从打字稿对象创建 JSON 字符串而不需要手动处理？
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.
请不要链接任何没有特别提到“TypeScript”的问题，因为这个问题特别与 TypeScript 相关。虽然是 JavaScript 的衍生物，但你编写代码的方式是不同的，因此这是目前唯一一个与 TypeScript 相关的问题。
回答by Luka Jacobowitz
Just use JSON.stringify(object). It's built into Javascript and can therefore also be used within Typescript.
只需使用JSON.stringify(object). 它内置于 Javascript 中，因此也可以在 Typescript 中使用。
回答by Giovanni P.
You can use the standard JSON object, available in Javascript:
您可以使用 Javascript 中可用的标准 JSON 对象：
var a: any = {};
a.x = 10;
a.y='hello';
var jsonString = JSON.stringify(a);
回答by Willem van der Veen
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 JSONobject. This contains the following methods:
TS 被编译成 JS，然后执行。因此，您可以访问 JS 运行时中的所有对象。这些对象之一是JSON对象。这包含以下方法：
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.
JSON.parse()方法解析 JSON 字符串，构造字符串描述的 JavaScript 值或对象。JSON.stringify()方法将 JavaScript 对象或值转换为 JSON 字符串。
Example:
例子：
const jsonString = '{"employee":{ "name":"John", "age":30, "city":"New York" }}';
const JSobj = JSON.parse(jsonString);
console.log(JSobj);
console.log(typeof JSobj);
const JSON_string = JSON.stringify(JSobj);
console.log(JSON_string);
console.log(typeof JSON_string);回答by Oliver Gro?
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.
使用这些 JSON.(parse/stringify) 方法时要小心。我对复杂的对象做了同样的事情，结果发现包含更多对象的嵌入数组对于我正在序列化的对象树中的所有其他实体具有相同的值。
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));
