好吧，似乎每个人都把他的俏皮话写在这里，所以我的是:

const encoded = Object.entries(obj).map(([k, v]) => `${k}=${encodeURIComponent(v)}`).join("&");

如果有很多嵌套对象，前面的答案就不起作用了。

相反，你可以从jquery-param/jquery-param.js中选择函数参数。这对我来说非常有效!

    var param = function (a) {
    var s = [], rbracket = /\[\]$/,
        isArray = function (obj) {
            return Object.prototype.toString.call(obj) === '[object Array]';
        }, add = function (k, v) {
            v = typeof v === 'function' ? v() : v === null ? '' : v === undefined ? '' : v;
            s[s.length] = encodeURIComponent(k) + '=' + encodeURIComponent(v);
        }, buildParams = function (prefix, obj) {
            var i, len, key;

            if (prefix) {
                if (isArray(obj)) {
                    for (i = 0, len = obj.length; i < len; i++) {
                        if (rbracket.test(prefix)) {
                            add(prefix, obj[i]);
                        } else {
                            buildParams(prefix + '[' + (typeof obj[i] === 'object' ? i : '') + ']', obj[i]);
                        }
                    }
                } else if (obj && String(obj) === '[object Object]') {
                    for (key in obj) {
                        buildParams(prefix + '[' + key + ']', obj[key]);
                    }
                } else {
                    add(prefix, obj);
                }
            } else if (isArray(obj)) {
                for (i = 0, len = obj.length; i < len; i++) {
                    add(obj[i].name, obj[i].value);
                }
            } else {
                for (key in obj) {
                    buildParams(key, obj[key]);
                }
            }
            return s;
        };

    return buildParams('', a).join('&').replace(/%20/g, '+');
};

另一种方法(没有递归对象):

   getQueryString = function(obj)
   {
      result = "";

      for(param in obj)
         result += ( encodeURIComponent(param) + '=' + encodeURIComponent(obj[param]) + '&' );

      if(result) //it's not empty string when at least one key/value pair was added. In such case we need to remove the last '&' char
         result = result.substr(0, result.length - 1); //If length is zero or negative, substr returns an empty string [ref. http://msdn.microsoft.com/en-us/library/0esxc5wy(v=VS.85).aspx]

      return result;
   }

alert( getQueryString({foo: "hi there", bar: 123, quux: 2 }) );

PHP符号的TypeScript版本(没有URL转义版本)

/**
 * Converts an object into a Cookie-like string.
 * @param toSerialize object or array to be serialized
 * @param prefix used in deep objects to describe the final query parameter
 * @returns ampersand separated key=value pairs
 *
 * Example:
 * ```js
 * serialize({hello:[{world: "nice"}]}); // outputs  "hello[0][world]=nice"
 * ```
 * ---
 * Adapted to TS from a StackOverflow answer https://stackoverflow.com/a/1714899/4537906
 */
const serialize = (toSerialize: unknown = {}, prefix?: string) => {
  const keyValuePairs = [];

  Object.keys(toSerialize).forEach((attribute) => {
    if (Object.prototype.hasOwnProperty.call(toSerialize, attribute)) {
      const key = prefix ? `${prefix}[${attribute}]` : attribute;
      const value = toSerialize[attribute];
      const toBePushed =
        value !== null && typeof value === "object"
          ? serialize(value, key)
          : `${key}=${value}`;
      keyValuePairs.push(toBePushed);
    }
  });

  return keyValuePairs.join("&");
};

Use:

Object.keys(obj).reduce(function(a,k){a.push(k+'='+encodeURIComponent(obj[k]));return a},[]).join('&')

我喜欢这句话，但我敢打赌，如果它在语义上符合公认的答案，它将是一个更受欢迎的答案:

function serialize( obj ) {
    let str = '?' + Object.keys(obj).reduce(function(a, k){
        a.push(k + '=' + encodeURIComponent(obj[k]));
        return a;
    }, []).join('&');
    return str;
}

使用Node.js v6.6.3

const querystring = require('querystring')

const obj = {
  foo: 'bar',
  baz: 'tor'
}

let result = querystring.stringify(obj)
// foo=bar&baz=tor

参考:查询字符串