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"

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;
     };

如果您想递归地转换嵌套对象，并且对象可能包含也可能不包含数组(数组可能包含对象或数组等)，那么解决方案就会变得稍微复杂一些。这就是我的尝试。

我还添加了一些选项，用于选择是否要为每个对象成员记录它位于主对象中的深度，以及选择是否要向来自已转换数组的成员添加标签。

理想情况下，您应该测试thing参数是否真的接收到对象或数组。

function thingToString(thing,maxDepth,recordLevel,markArrays){
    //thing: object or array to be recursively serialized
    //maxDepth (int or false):
    // (int) how deep to go with converting objects/arrays within objs/arrays
    // (false) no limit to recursive objects/arrays within objects/arrays
    //recordLevel (boolean):
    //  true - insert "(level 1)" before transcript of members at level one (etc)
    //  false - just 
    //markArrays (boolean):
    //  insert text to indicate any members that came from arrays
    var result = "";
    if (maxDepth !== false && typeof maxDepth != 'number') {maxDepth = 3;}
    var runningDepth = 0;//Keeps track how deep we're into recursion

    //First prepare the function, so that it can call itself recursively
    function serializeAnything(thing){
        //Set path-finder values
        runningDepth += 1;
        if(recordLevel){result += "(level " + runningDepth + ")";}

        //First convert any arrays to object so they can be processed
        if (thing instanceof Array){
            var realObj = {};var key;
            if (markArrays) {realObj['type'] = "converted array";}
            for (var i = 0;i < thing.length;i++){
                if (markArrays) {key = "a" + i;} else {key = i;}
                realObj[key] = thing[i];
            }
            thing = realObj;
            console.log('converted one array to ' + typeof realObj);
            console.log(thing);
        }

        //Then deal with it
        for (var member in thing){
            if (typeof thing[member] == 'object' && runningDepth < maxDepth){
                serializeAnything(thing[member]);
                //When a sub-object/array is serialized, it will add one to
                //running depth. But when we continue to this object/array's
                //next sibling, the level must go back up by one
                runningDepth -= 1;
            } else if (maxDepth !== false && runningDepth >= maxDepth) {
                console.log('Reached bottom');
            } else 
            if (
                typeof thing[member] == "string" || 
                typeof thing[member] == 'boolean' ||
                typeof thing[member] == 'number'
            ){
                result += "(" + member + ": " + thing[member] + ") ";
            }  else {
                result += "(" + member + ": [" + typeof thing[member] + " not supported]) ";
            }
        }
    }
    //Actually kick off the serialization
    serializeAnything(thing);

    return result;

}

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;
}

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("&")

只需使用以下方法:

encodeURIComponent(JSON.stringify(obj))

// elastic search example let story ={ "query": { "bool": { "must": [ { "term": { "revision.published": 0, } }, { "term": { "credits.properties.by.properties.name": "Michael Guild" } }, { "nested": { "path": "taxonomy.sections", "query": { "bool": { "must": [ { "term": { "taxonomy.sections._id": "/science" } }, { "term": { "taxonomy.sections._website": "staging" } } ] } } } } ] } } } const whateva = encodeURIComponent(JSON.stringify(story)) console.log(whateva)