Use:

const objectToQueryParams = (o = {}) =>
  Object.entries(o)
    .map((p) => `${encodeURIComponent(p[0])}=${encodeURIComponent(p[1])}`)
    .join("&");

参考以下要点了解更多信息: https://gist.github.com/bhaireshm

 let data = {
    id:1,
    name:'Newuser'
    };
const getqueryParam = data => {
  let datasize = Object.keys(data).length;
  let initial = '?';

  Object.keys(data).map(function (key, index) {
    initial = initial.concat(`${key}=${data[key]}`);
    index != datasize - 1 && (initial = initial.concat('&'));
  });
  console.log(initial, 'MyqueryString');
  return initial;
};

console.log(getqueryParam(data))//You can get the query string here

If you have baseUrl means to get full query use 

baseUrl concat (getqueryParam(数据)。

以下是已接受答案的CoffeeScript版本。

serialize = (obj, prefix) ->
  str = []
  for p, v of obj
    k = if prefix then prefix + "[" + p + "]" else p
    if typeof v == "object"
      str.push(serialize(v, k))
    else
      str.push(encodeURIComponent(k) + "=" + encodeURIComponent(v))

  str.join("&")

Use:

const objectToQueryParams = (o = {}) =>
  Object.entries(o)
    .map((p) => `${encodeURIComponent(p[0])}=${encodeURIComponent(p[1])}`)
    .join("&");

参考以下要点了解更多信息: https://gist.github.com/bhaireshm

下面是ES6中的一行代码:

Object.keys(obj).map(k => `${encodeURIComponent(k)}=${encodeURIComponent(obj[k])}`).join('&');

我的实现编码一个对象作为一个查询字符串，使用减少:

export const encodeAsQueryString = (params) => (
  Object.keys(params).reduce((acc, key)=>(
    params.hasOwnProperty(key) ? (
      [...acc, encodeURIComponent(key) + '=' + encodeURIComponent(params[key])]
    ) : acc
  ), []).join('&')
);