这个函数跳过null/undefined值

export function urlEncodeQueryParams(data) {
    const params = Object.keys(data).map(key => data[key] ? `${encodeURIComponent(key)}=${encodeURIComponent(data[key])}` : '');
    return params.filter(value => !!value).join('&');
}

URLSearchParams看起来不错，但是对于嵌套对象来说行不通。

试着使用

encodeURIComponent(JSON.stringify(object))

好吧，这是一个旧的帖子，但我正面临这个问题，我已经找到了我的个人解决方案。也许可以帮助别人…

     function objToQueryString(obj){
        var k = Object.keys(obj);
        var s = "";
        for(var i=0;i<k.length;i++) {
            s += k[i] + "=" + encodeURIComponent(obj[k[i]]);
            if (i != k.length -1) s += "&";
        }
        return s;
     };

Ramda:

    R.pipe(R.toPairs, R.map(R.join('=')), R.join('&'))({a: 'b', b: 'a'})

ES6解决方案用于JavaScript对象的查询字符串编码

const params = {
  a: 1,
  b: 'query stringify',
  c: null,
  d: undefined,
  f: '',
  g: { foo: 1, bar: 2 },
  h: ['Winterfell', 'Westeros', 'Braavos'],
  i: { first: { second: { third: 3 }}}
}

static toQueryString(params = {}, prefix) {
  const query = Object.keys(params).map((k) => {
    let key = k;
    const value = params[key];

    if (!value && (value === null || value === undefined || isNaN(value))) {
      value = '';
    }

    switch (params.constructor) {
      case Array:
        key = `${prefix}[]`;
        break;
      case Object:
        key = (prefix ? `${prefix}[${key}]` : key);
        break;
    }

    if (typeof value === 'object') {
      return this.toQueryString(value, key); // for nested objects
    }

    return `${key}=${encodeURIComponent(value)}`;
  });

  return query.join('&');
}

toQueryString (params)

"a=1&b=query%20stringify&c=&d=&f=&g[foo]=1&g[bar]=2&h[]=Winterfell&h[]=Westeros&h[]=Braavos&i[first][second][third]=3"

下面是Object.entries的简洁递归版本。它处理任意嵌套的数组，但不处理嵌套的对象。它还会删除空元素:

const format = (k,v) => v !== null ? `${k}=${encodeURIComponent(v)}` : ''

const to_qs = (obj) => {
    return [].concat(...Object.entries(obj)
                       .map(([k,v]) => Array.isArray(v) 
                          ? v.map(arr => to_qs({[k]:arr})) 
                          : format(k,v)))
           .filter(x => x)
           .join('&');
}

例如:

let json = { 
    a: [1, 2, 3],
    b: [],              // omit b
    c: 1,
    d: "test&encoding", // uriencode
    e: [[4,5],[6,7]],   // flatten this
    f: null,            // omit nulls
    g: 0
};

let qs = to_qs(json)

=> "a=1&a=2&a=3&c=1&d=test%26encoding&e=4&e=5&e=6&e=7&g=0"