在Javascript中将Map转换为JSON对象

时间:2016-05-25 12:54:05

标签: javascript arrays json object dictionary

所以我得到了以下javascript,其中包含一个键/值对,用于将嵌套路径映射到目录。

function createPaths(aliases, propName, path) {
    aliases.set(propName, path);
}

map = new Map();

createPaths(map, 'paths.aliases.server.entry', 'src/test');
createPaths(map, 'paths.aliases.dist.entry', 'dist/test');

现在我想要做的是从地图中的键创建一个JSON对象。

必须是,

paths: {
  aliases: {
    server: {
      entry: 'src/test'
    },
    dist: {
      entry: 'dist/test'
    }
  }
}

不确定是否有一种方法可以做到这一点。任何帮助表示赞赏。

5 个答案:

答案 0 :(得分:4)

您可以循环遍历地图和键,并指定值

function createPaths(aliases, propName, path) {
    aliases.set(propName, path);
}

var map = new Map(),
    object = {};

createPaths(map, 'paths.aliases.server.entry', 'src/test');
createPaths(map, 'paths.aliases.dist.entry', 'dist/test');

map.forEach((value, key) => {
    var keys = key.split('.'),
        last = keys.pop();
    keys.reduce((r, a) => r[a] = r[a] || {}, object)[last] = value;
});

console.log(object);

答案 1 :(得分:3)

我希望这个功能足够自我解释。这就是我以前做的工作。

/*
 * Turn the map<String, Object> to an Object so it can be converted to JSON
 */
function mapToObj(inputMap) {
    let obj = {};

    inputMap.forEach(function(value, key){
        obj[key] = value
    });

    return obj;
}


JSON.stringify(returnedObject)

答案 2 :(得分:2)

另一种方法。我很好奇哪个有更好的表现,但jsPerf失败了:(。

var obj = {};

function createPaths(map, path, value)
{
	if(typeof path === "string") path = path.split(".");
	
	if(path.length == 1)
	{
		map[path[0]] = value;
		return;
	}
	else
	{
		if(!(path[0] in map)) map[path[0]] = {};
		return createPaths(map[path[0]], path.slice(1), value);
	}
}

createPaths(obj, 'paths.aliases.server.entry', 'src/test');
createPaths(obj, 'paths.aliases.dist.entry', 'dist/test');

console.log(obj);

没有递归:

var obj = {};

function createPaths(map, path, value)
{
    var map = map;
    var path = path.split(".");
    for(var i = 0, numPath = path.length - 1; i < numPath; ++i)
    {
        if(!(path[i] in map)) map[path[i]] = {};
        map = map[path[i]];
    }
    map[path[i]] = value;
}

createPaths(obj, 'paths.aliases.server.entry', 'src/test');
createPaths(obj, 'paths.aliases.dist.entry', 'dist/test');
createPaths(obj, 'paths.aliases.dist.dingo', 'dist/test');
createPaths(obj, 'paths.bingo.dist.entry', 'dist/test');

console.log(obj);

var obj = {};

function createPaths(map, path, value)
{
    var map = map;
    var path = path.split(".");
    
    while(path.length > 1)
    {
        map = map[path[0]] = map[path.shift()] || {};
    }
    
    map[path.shift()] = value;
  
}

createPaths(obj, 'paths.aliases.server.entry', 'src/test');
createPaths(obj, 'paths.aliases.dist.entry', 'dist/test');
createPaths(obj, 'paths.aliases.dist.dingo', 'dist/test');
createPaths(obj, 'paths.bingo.dist.entry', 'dist/test');

console.log(obj);

答案 3 :(得分:2)

仅使用ES6方式

  1. Object.fromEntries

const log = console.log;

const map = new Map();
// undefined
map.set(`a`, 1);
// Map(1) {"a" => 1}
map.set(`b`, 2);
// Map(1) {"a" => 1, "b" => 2}
map.set(`c`, 3);
// Map(2) {"a" => 1, "b" => 2, "c" => 3}

// Object.fromEntries ✅
const obj = Object.fromEntries(map);

log(`\nobj`, obj);
// obj { a: 1, b: 2, c: 3 }

  1. ...spreaddestructuring assignment

const log = console.log;

const map = new Map();
// undefined
map.set(`a`, 1);
// Map(1) {"a" => 1}
map.set(`b`, 2);
// Map(1) {"a" => 1, "b" => 2}
map.set(`c`, 3);
// Map(2) {"a" => 1, "b" => 2, "c" => 3}

const autoConvertMapToObject = (map) => {
  const obj = {};
  for (const item of [...map]) {
    const [
      key,
      value
    ] = item;
    obj[key] = value;
  }
  return obj;
}

const obj = autoConvertMapToObject(map)

log(`\nobj`, obj);
// obj { a: 1, b: 2, c: 3 }

裁判

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Object/fromEntries

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_syntax

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Destructuring_assignment

https://2ality.com/2015/08/es6-map-json.html

答案 4 :(得分:1)

鉴于{v3},fromEntries()自节点v12开始可用:

const entries = new Map([
  ['foo', 'bar'],
  ['baz', 42]
]);

const obj = Object.fromEntries(entries);

console.log(obj);
// expected output: Object { foo: "bar", baz: 42 }